How can I pad a value with leading zeros?

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301
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What is the recommended way to zerofill a value in JavaScript? I imagine I could build a custom function to pad zeros on to a typecasted value, but I'm wondering if there is a more direct way to do this?

Note: By "zerofilled" I mean it in the database sense of the word (where a 6-digit zerofilled representation of the number 5 would be "000005").

edited Apr 27 '17 at 13:16

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4 revs, 3 users 100%
Wilco

This really isn't enough information to answer. The most common instance of "zero padding" is probably probably prepending zeroes onto dates: 5/1/2008 > 05/01/2008. Is that what you mean?
– Triptych
Aug 12 '09 at 16:38

5

For node apps, use npm install sprintf-js, and require it in the file you need: sprintf('%0d6', 5);
– Droogans
Nov 12 '14 at 19:27

function padWithZeroes(n, width) { while(n.length<width) n = '0' + n; return n;} ...assuming n not negative
– Paolo
Feb 18 '16 at 23:52

@Paolo that function doesn't work if n is numeric. You'd need to convert n to a String before the while in order to access n.length
– Nick F
Jun 3 '16 at 10:28

@NickF of course... and assuming 'n' is a string.
– Paolo
Jun 3 '16 at 10:46

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up vote
301
down vote

favorite

Note: By "zerofilled" I mean it in the database sense of the word (where a 6-digit zerofilled representation of the number 5 would be "000005").

edited Apr 27 '17 at 13:16

community wiki

4 revs, 3 users 100%
Wilco

This really isn't enough information to answer. The most common instance of "zero padding" is probably probably prepending zeroes onto dates: 5/1/2008 > 05/01/2008. Is that what you mean?
– Triptych
Aug 12 '09 at 16:38

5

For node apps, use npm install sprintf-js, and require it in the file you need: sprintf('%0d6', 5);
– Droogans
Nov 12 '14 at 19:27

function padWithZeroes(n, width) { while(n.length<width) n = '0' + n; return n;} ...assuming n not negative
– Paolo
Feb 18 '16 at 23:52

@Paolo that function doesn't work if n is numeric. You'd need to convert n to a String before the while in order to access n.length
– Nick F
Jun 3 '16 at 10:28

@NickF of course... and assuming 'n' is a string.
– Paolo
Jun 3 '16 at 10:46

|
show 2 more comments

up vote
301
down vote

favorite

Note: By "zerofilled" I mean it in the database sense of the word (where a 6-digit zerofilled representation of the number 5 would be "000005").

edited Apr 27 '17 at 13:16

community wiki

4 revs, 3 users 100%
Wilco

Note: By "zerofilled" I mean it in the database sense of the word (where a 6-digit zerofilled representation of the number 5 would be "000005").

javascript zerofill

edited Apr 27 '17 at 13:16

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4 revs, 3 users 100%
Wilco

edited Apr 27 '17 at 13:16

community wiki

4 revs, 3 users 100%
Wilco

edited Apr 27 '17 at 13:16

community wiki

4 revs, 3 users 100%
Wilco

community wiki

4 revs, 3 users 100%
Wilco

community wiki

4 revs, 3 users 100%
Wilco

This really isn't enough information to answer. The most common instance of "zero padding" is probably probably prepending zeroes onto dates: 5/1/2008 > 05/01/2008. Is that what you mean?
– Triptych
Aug 12 '09 at 16:38

5

For node apps, use npm install sprintf-js, and require it in the file you need: sprintf('%0d6', 5);
– Droogans
Nov 12 '14 at 19:27

function padWithZeroes(n, width) { while(n.length<width) n = '0' + n; return n;} ...assuming n not negative
– Paolo
Feb 18 '16 at 23:52

@Paolo that function doesn't work if n is numeric. You'd need to convert n to a String before the while in order to access n.length
– Nick F
Jun 3 '16 at 10:28

@NickF of course... and assuming 'n' is a string.
– Paolo
Jun 3 '16 at 10:46

|
show 2 more comments

This really isn't enough information to answer. The most common instance of "zero padding" is probably probably prepending zeroes onto dates: 5/1/2008 > 05/01/2008. Is that what you mean?
– Triptych
Aug 12 '09 at 16:38

5

For node apps, use npm install sprintf-js, and require it in the file you need: sprintf('%0d6', 5);
– Droogans
Nov 12 '14 at 19:27

function padWithZeroes(n, width) { while(n.length<width) n = '0' + n; return n;} ...assuming n not negative
– Paolo
Feb 18 '16 at 23:52

@Paolo that function doesn't work if n is numeric. You'd need to convert n to a String before the while in order to access n.length
– Nick F
Jun 3 '16 at 10:28

@NickF of course... and assuming 'n' is a string.
– Paolo
Jun 3 '16 at 10:46

This really isn't enough information to answer. The most common instance of "zero padding" is probably probably prepending zeroes onto dates: 5/1/2008 > 05/01/2008. Is that what you mean?
– Triptych
Aug 12 '09 at 16:38

For node apps, use npm install sprintf-js, and require it in the file you need: sprintf('%0d6', 5);
– Droogans
Nov 12 '14 at 19:27

function padWithZeroes(n, width) { while(n.length<width) n = '0' + n; return n;} ...assuming n not negative
– Paolo
Feb 18 '16 at 23:52

@Paolo that function doesn't work if n is numeric. You'd need to convert n to a String before the while in order to access n.length
– Nick F
Jun 3 '16 at 10:28

@NickF of course... and assuming 'n' is a string.
– Paolo
Jun 3 '16 at 10:46

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68 Answers
68

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up vote
172
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accepted

Note to readers!

As commenters have pointed out, this solution is "clever", and as
clever solutions often are, it's memory intensive and relatively
slow. If performance is a concern for you, don't use this solution!

Potentially outdated: ECMAScript 2017 includes String.prototype.padStart and Number.prototype.toLocaleString is there since ECMAScript 3.1. Example:
var n=-0.1;

n.toLocaleString('en', {minimumIntegerDigits:4,minimumFractionDigits:2,useGrouping:false})
...will output "-0000.10".
// or 

const padded = (.1+"").padStart(6,"0");

`-${padded}`
...will output "-0000.1".

A simple function is all you need

function zeroFill( number, width )

{

  width -= number.toString().length;

  if ( width > 0 )

  {

    return new Array( width + (/./.test( number ) ? 2 : 1) ).join( '0' ) + number;

  }

  return number + ""; // always return a string

}

you could bake this into a library if you want to conserve namespace or whatever. Like with jQuery's extend.

edited Dec 6 '17 at 15:21

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9 revs, 5 users 66%
Peter Bailey

1

Now you just need to take care of numbers like 50.1234 and you've got a readable version of my solution below! I did, however, assume that we were just left padding, not overall padding.
– coderjoe
Aug 12 '09 at 20:16

5

Not a fan of creating a new array each time you want to pad a value. Readers should be aware of the the potential memory and GC impact if they're doing large #'s of these (e.g. on a node.js server that's doing 1000's of these operations per second. @profitehlolz's answer would seem to be the better solution.)
– broofa
Sep 24 '12 at 12:21

6

This does not work for negative values: zeroFill(-10, 7) -> "0000-10"
– digitalbath
Nov 7 '12 at 15:20

2

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:35

8

As of ECMAScript 2017, there's a build-in function padStart.
– Tomas Nikodym
Sep 11 '16 at 21:23

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up vote
277
down vote

Simple way. You could add string multiplication for the pad and turn it into a function.

var pad = "000000";

var n = '5';

var result = (pad+n).slice(-pad.length);

As a function,

function paddy(num, padlen, padchar) {

    var pad_char = typeof padchar !== 'undefined' ? padchar : '0';

    var pad = new Array(1 + padlen).join(pad_char);

    return (pad + num).slice(-pad.length);

}

var fu = paddy(14, 5); // 00014

var bar = paddy(2, 4, '#'); // ###2

edited Mar 23 at 14:49

community wiki

4 revs, 2 users 83%
profitehlolz

84

This is a very nice solution, best I've seen. For clarity, here's a simpler version I'm using to pad the minutes in time formatting: ('0'+minutes).slice(-2)
– Luke Ehresman
Sep 14 '12 at 0:22

14

This is 5 times slower than the implementation with a while loop: gist.github.com/4382935
– andrewrk
Dec 26 '12 at 20:37

36

This has a FATAL FLAW with larger than expected numbers -- which is an extremely common occurrence. For example, paddy (888, 2) yields 88 and not 888 as required. This answer also does not handle negative numbers.
– Brock Adams
Jun 15 '13 at 22:20

7

@BrockAdams, zerofill is typically used for fixed width number/formatting cases -- therefore I think it'd actually be less likely (or even a non-issue) if given a 3 digit number when trying to do 2 digit zerofill.
– Seaux
Dec 8 '13 at 23:15

7

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

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up vote
228
down vote

I can't believe all the complex answers on here... Just use this:

var zerofilled = ('0000'+n).slice(-4);

edited Sep 11 at 7:40

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2 revs, 2 users 67%
Seaux

17

Good except for negative numbers and numbers longer than 4 digits.
– wberry
Sep 19 '14 at 22:40

7

@wberry this was not intended to be production ready code, but more a simple explanation of the basics to solving the question at hand. Aka it's very easy to determine if a number is positive or negative and add the appropriate sign if needed, so i didn't want to cluster my easy example with that. Also, it's just as easy to make a function taking a digit length as an arg and using that instead of my hardcoded values.
– Seaux
Sep 21 '14 at 4:13

4

This seems like a good solution for simple cases. Easier to understand.
– Artur Carvalho
Jan 28 '15 at 12:23

2

@OmShankar by "complex answers", I wasn't referring to explained or detailed answers (which are obviously encouraged on this site even though my answer isn't), but rather answers that contained unnecessary code complexity.
– Seaux
Aug 10 '15 at 23:10

3

I like this answer. In my own specific case, I know that the numbers are always positive and never more than 3 digits, and this is often the case when you're padding with leading 0's and your input is well known. Nice!
– Patrick Chu
Dec 16 '16 at 22:12

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up vote
104
down vote

I actually had to come up with something like this recently.
I figured there had to be a way to do it without using loops.

This is what I came up with.

function zeroPad(num, numZeros) {

    var n = Math.abs(num);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( num < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Then just use it providing a number to zero pad:

> zeroPad(50,4);

"0050"

If the number is larger than the padding, the number will expand beyond the padding:

> zeroPad(51234, 3);

"51234"

Decimals are fine too!

> zeroPad(51.1234, 4);

"0051.1234"

If you don't mind polluting the global namespace you can add it to Number directly:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

And if you'd rather have decimals take up space in the padding:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - n.toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Cheers!

XDR came up with a logarithmic variation that seems to perform better.

WARNING: This function fails if num equals zero (e.g. zeropad(0, 2))

function zeroPad (num, numZeros) {

    var an = Math.abs (num);

    var digitCount = 1 + Math.floor (Math.log (an) / Math.LN10);

    if (digitCount >= numZeros) {

        return num;

    }

    var zeroString = Math.pow (10, numZeros - digitCount).toString ().substr (1);

    return num < 0 ? '-' + zeroString + an : zeroString + an;

}

Speaking of performance, tomsmeding compared the top 3 answers (4 with the log variation). Guess which one majorly outperformed the other two? :)

edited May 23 '17 at 12:34

community wiki

6 revs, 3 users 78%
coderjoe

9

This is good, I like how it's readable and robust. The only thing I would change is the name of the numZeros parameter, since it's misleading. It's not the number of zeros you want to add, it's the minimum length the number should be. A better name would be numLength or even length.
– Senseful
Jan 2 '12 at 4:23

12

+1. Unlike the higher voted answers, this one handles negative numbers and does not return gross errors on overflow!!?!
– Brock Adams
Jun 15 '13 at 22:39

10

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

4

I improved the performance of this answer by over 100% by using logarithms. Please see the logarithmic test case at http://jsperf.com/left-zero-pad/10
– XDR
Aug 18 '14 at 4:30

2

The original solution is better, logarithmic variation doesn't work if num can be zero...
– Ondra C.
Mar 17 '15 at 10:03

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up vote
55
down vote

Here's what I used to pad a number up to 7 characters.

("0000000" + number).slice(-7)

This approach will probably suffice for most people.

Edit: If you want to make it more generic you can do this:

("0".repeat(padding) + number).slice(-padding)

edited Jun 27 '17 at 9:46

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6 revs
chetbox

This fails, badly, on large numbers and negative numbers. number = 123456789, with the code, above, for example.
– Brock Adams
Jun 15 '13 at 22:49

This solution is the best possible ever for certains scenarios! I've got no clue why we didn't come up with this before. The scenario is: you've got a number which is always positive [as Brock mentioned] and you know that it won't take more characters than your limit. In our case we have a score which we store in Amazon SDB. As you know SDB can't compare numbers, so all scores have to be zero padded. This is extremely easy and effective!
– Krystian
Oct 10 '13 at 11:27

2

Apologies, I should have mentioned that this won't work for negative numbers. I think this deals with the more common positive number case, though. It does what I need, at least!
– chetbox
Oct 16 '13 at 11:33

Needed this again and realised you don't need new String. You can just use string literals.
– chetbox
Mar 4 '16 at 15:56

2

(new Array(padding + 1).join("0") can be replaced with "0".repeat(padding)
– Pavel
Jun 22 '17 at 16:17

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up vote
23
down vote

Unfortunately, there are a lot of needless complicated suggestions for this problem, typically involving writing your own function to do math or string manipulation or calling a third-party utility. However, there is a standard way of doing this in the base JavaScript library with just one line of code. It might be worth wrapping this one line of code in a function to avoid having to specify parameters that you never want to change like the local name or style.

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumIntegerDigits: 3,

    useGrouping: false

});

This will produce the value of "005" for text. You can also use the toLocaleString function of Number to pad zeros to the right side of the decimal point.

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumFractionDigits: 2,

    useGrouping: false

});

This will produce the value of "5.00" for text. Change useGrouping to true to use comma separators for thousands.

Note that using toLocaleString() with locales and options arguments is standardized separately in ECMA-402, not in ECMAScript. As of today, some browsers only implement basic support, i.e. toLocaleString() may ignore any arguments.

Complete Example

edited May 28 '15 at 21:01

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2 revs, 2 users 95%
Daniel Barbalace

Wow, this is a super clean solution and works in node.js.
– jishi
Feb 6 '16 at 22:21

I chuckled when I read the intro -- "needless complicated suggestions". Software development, as in all engineering disciplines, involves weighing trade-offs. I tried this "standard way" myself -- and timed it. It definitely works, but I would never choose to use it for any kind of repetitive application due to how slow it is compared to many of the other "home rolled" solutions.
– bearvarine
Sep 16 '17 at 20:47

True, a lot of built-in JavaScript functions perform poorly when looping over lots of data, but I would argue you should not use JavaScript for that, even server-side like Node.js. If you have lots of data to process server-side, you should use a better platform like .NET or Java. If you are processing client-side data for display to the end user, you should only process the data for what you are currently rendering to the end user's screen. For example, render only the visible rows of a table and don't process data for other roads.
– Daniel Barbalace
Jul 23 at 21:03

add a comment |

up vote
20
down vote

If the fill number is known in advance not to exceed a certain value, there's another way to do this with no loops:

var fillZeroes = "00000000000000000000";  // max number of zero fill ever asked for in global



function zeroFill(number, width) {

    // make sure it's a string

    var input = number + "";  

    var prefix = "";

    if (input.charAt(0) === '-') {

        prefix = "-";

        input = input.slice(1);

        --width;

    }

    var fillAmt = Math.max(width - input.length, 0);

    return prefix + fillZeroes.slice(0, fillAmt) + input;

}

Test cases here: http://jsfiddle.net/jfriend00/N87mZ/

edited Jun 16 '13 at 1:08

community wiki

jfriend00

@BrockAdams - fixed for both cases you mention. If the number is already as wide as the fill width, then no fill is added.
– jfriend00
Jun 16 '13 at 1:08

Thanks; +1. The negative numbers are slightly off from the usual convention, though. In your test case, -88 should yield "-00088", for example.
– Brock Adams
Jun 16 '13 at 1:57

1

@BrockAdams - I wasn't sure whether calling zeroFill(-88, 5) should produce -00088 or -0088? I guess it depends upon whether you want the width argument to be the entire width of the number or just the number of digits (not including the negative sign). An implementer can easily switch the behavior to not include the negative sign by just removing the --width line of code.
– jfriend00
Jun 16 '13 at 3:07

add a comment |

up vote
18
down vote

In a proposed (stage 3) ES2017 method .padStart() you can simply now do (when implemented/supported):

string.padStart(maxLength, "0"); //max length is the max string length, not max # of fills

edited May 16 '16 at 14:25

community wiki

2 revs
Sterling Archer

I've been looking for this one for some time now. Thank you for giving some visibility. I hope in 2030 we have it working.
– dinigo
Jun 20 '17 at 10:55

1

Its works now @dinigo ^^ it is live
– Paulo Roberto
Nov 22 '17 at 12:47

yep, I've used it a couple of times
– dinigo
Nov 22 '17 at 13:02

add a comment |

up vote
17
down vote

The quick and dirty way:

y = (new Array(count + 1 - x.toString().length)).join('0') + x;

For x = 5 and count = 6 you'll have y = "000005"

edited May 24 '12 at 22:47

community wiki

Johann Philipp Strathausen

1

I got y="0005" with the above, y = (new Array(count + 1 - x.toString().length)).join('0') + x; is what gave me y="000005"
– forforf
May 23 '12 at 15:30

@forforf thanks, corrected it now!
– Johann Philipp Strathausen
May 24 '12 at 22:48

Thanks. I always like the shortest code solution.
– Orr Siloni
Nov 12 '12 at 14:08

3

@OrrSiloni: the shortest code solution is actually ('0000'+n).slice(-4)
– Seaux
Dec 8 '13 at 23:17

add a comment |

up vote
12
down vote

Here's a quick function I came up with to do the job. If anyone has a simpler approach, feel free to share!

function zerofill(number, length) {

    // Setup

    var result = number.toString();

    var pad = length - result.length;



    while(pad > 0) {

        result = '0' + result;

        pad--;

    }



    return result;

}

answered Aug 12 '09 at 16:54

community wiki

Wilco

I don't think there will be a "simpler" solution - it will always be a function of some sort. We could have a algorithm discussion, but at the end of the day, you'll still end up with a function that zerofills a number.
– Peter Bailey
Aug 12 '09 at 17:54

My general advice is to use "for" loops as they are generally more stable and faster in ECMAScript languages (ActionScript 1-3, and JavaScript)
– cwallenpoole
Aug 12 '09 at 18:03

Or, skip iteration altogether ;)
– Peter Bailey
Aug 12 '09 at 20:05

Simpler function? Probably not. One that doesn't loop? That is possible... though the algorithm isn't entirely trivial.
– coderjoe
Aug 12 '09 at 20:09

1

Wow, pretty much all the above comments are incorrect. @profitehlolz's solution is simpler and suitable for inlining (doesn't need to be a function). Meanwhile, for .vs. while performance is not different enough to be interesting: jsperf.com/fors-vs-while/34
– broofa
Sep 24 '12 at 12:07

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up vote
11
down vote

Late to the party here, but I often use this construct for doing ad-hoc padding of some value n, known to be a positive, decimal:

(offset + n + '').substr(1);

Where offset is 10^^digits.

E.g. Padding to 5 digits, where n = 123:

(1e5 + 123 + '').substr(1); // => 00123

The hexidecimal version of this is slightly more verbose:

(0x100000 + 0x123).toString(16).substr(1); // => 00123

Note 1: I like @profitehlolz's solution as well, which is the string version of this, using slice()'s nifty negative-index feature.

answered Sep 24 '12 at 12:25

community wiki

broofa

Very elegant and neat!
– Marc
Feb 13 '13 at 9:42

add a comment |

up vote
11
down vote

I really don't know why, but no one did it in the most obvious way. Here it's my implementation.

Function:

/** Pad a number with 0 on the left */

function zeroPad(number, digits) {

    var num = number+"";

    while(num.length < digits){

        num='0'+num;

    }

    return num;

}

Prototype:

Number.prototype.zeroPad=function(digits){

    var num=this+"";

    while(num.length < digits){

        num='0'+num;

    }

    return(num);

};

Very straightforward, I can't see any way how this can be any simpler. For some reason I've seem many times here on SO, people just try to avoid 'for' and 'while' loops at any cost. Using regex will probably cost way more cycles for such a trivial 8 digit padding.

answered Feb 5 '13 at 14:45

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Vitim.us

add a comment |

up vote
9
down vote

I use this snipet to get a 5 digits representation

(value+100000).toString().slice(-5) // "00123" with value=123

answered Sep 3 '13 at 11:08

community wiki

jasto salto

this is a smart way to add leading zero . I have done something similar to add leading zero for fixed length binary integer
– Chris.Huang
May 11 '15 at 2:14

add a comment |

up vote
8
down vote

The power of Math!

x = integer to pad

y = number of zeroes to pad

function zeroPad(x, y)

{

   y = Math.max(y-1,0);

   var n = (x / Math.pow(10,y)).toFixed(y);

   return n.replace('.','');  

}

answered Mar 21 '13 at 1:40

community wiki

tir

Wow, that is a very clever way to solve this problem. I already had my own solution for this that is exactly functionally equivalent, and even relatively short, but this is concise and probably quite fast.
– pseudosavant
Jan 9 '14 at 21:10

add a comment |

up vote
7
down vote

First parameter is any real number, second parameter is a positive integer specifying the minimum number of digits to the left of the decimal point and third parameter is an optional positive integer specifying the number if digits to the right of the decimal point.

function zPad(n, l, r){

    return(a=String(n).match(/(^-?)(d*).?(d*)/))?a[1]+(Array(l).join(0)+a[2]).slice(-Math.max(l,a[2].length))+('undefined'!==typeof r?(0<r?'.':'')+(a[3]+Array(r+1).join(0)).slice(0,r):a[3]?'.'+a[3]:''):0

}

           zPad(6, 2) === '06'

          zPad(-6, 2) === '-06'

       zPad(600.2, 2) === '600.2'

        zPad(-600, 2) === '-600'

         zPad(6.2, 3) === '006.2'

        zPad(-6.2, 3) === '-006.2'

      zPad(6.2, 3, 0) === '006'

        zPad(6, 2, 3) === '06.000'

    zPad(600.2, 2, 3) === '600.200'

zPad(-600.1499, 2, 3) === '-600.149'

edited Feb 4 '14 at 22:52

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4 revs
Jack Allan

add a comment |

up vote
7
down vote

Don't reinvent the wheel, use underscore string:

jsFiddle

var numToPad = '5';



alert(_.str.pad(numToPad, 6, '0')); // yields: '000005'

answered Sep 5 '14 at 12:34

community wiki

Benny Bottema

1

Pure JavaScript! Short and simple, and a jsFiddle example! EXCELLENT! :D
– tfont
Oct 10 '14 at 23:41

add a comment |

up vote
7
down vote

This is the ES6 solution.

function pad(num, len) {

  return '0'.repeat(len - num.toString().length) + num;

}

alert(pad(1234,6));

answered May 22 '15 at 3:38

community wiki

Lewis

Clearly the most elegant solution I would just modify it to avoid 2 string conversions: const numStr = String(num) and return '0'.repeat(len - numStr.length) + numStr;
– ngryman
Sep 26 '16 at 21:52

add a comment |

up vote
7
down vote

I didn't see anyone point out the fact that when you use String.prototype.substr() with a negative number it counts from the right.

A one liner solution to the OP's question, a 6-digit zerofilled representation of the number 5, is:

console.log(("00000000" + 5).substr(-6));

Generalizing we'll get:

function pad(num, len) { return ("00000000" + num).substr(-len) };



console.log(pad(5, 6));

console.log(pad(45, 6));

console.log(pad(345, 6));

console.log(pad(2345, 6));

console.log(pad(12345, 6));

edited Sep 21 '17 at 22:16

community wiki

2 revs
Stephen Quan

nice one liner solution
– rave
Sep 21 '17 at 21:06

add a comment |

up vote
6
down vote

After a, long, long time of testing 15 different functions/methods found in this questions answers, I now know which is the best (the most versatile and quickest).

I took 15 functions/methods from the answers to this question and made a script to measure the time taken to execute 100 pads. Each pad would pad the number 9 with 2000 zeros. This may seem excessive, and it is, but it gives you a good idea about the scaling of the functions.

The code I used can be found here:
https://gist.github.com/NextToNothing/6325915

Feel free to modify and test the code yourself.

In order to get the most versatile method, you have to use a loop. This is because with very large numbers others are likely to fail, whereas, this will succeed.

So, which loop to use? Well, that would be a while loop. A for loop is still fast, but a while loop is just slightly quicker(a couple of ms) - and cleaner.

Answers like those by Wilco, Aleksandar Toplek or Vitim.us will do the job perfectly.

Personally, I tried a different approach. I tried to use a recursive function to pad the string/number. It worked out better than methods joining an array but, still, didn't work as quick as a for loop.

My function is:

function pad(str, max, padder) {

  padder = typeof padder === "undefined" ? "0" : padder;

  return str.toString().length < max ? pad(padder.toString() + str, max, padder) : str;

}

You can use my function with, or without, setting the padding variable. So like this:

pad(1, 3); // Returns '001'

// - Or -

pad(1, 3, "x"); // Returns 'xx1'

Personally, after my tests, I would use a method with a while loop, like Aleksandar Toplek or Vitim.us. However, I would modify it slightly so that you are able to set the padding string.

So, I would use this code:

function padLeft(str, len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    str = str + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

padLeft(1, 3); // Returns '001'

// - Or -

padLeft(1, 3, "x"); // Returns 'xx1'

You could also use it as a prototype function, by using this code:

Number.prototype.padLeft = function(len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    var str = this + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

var num = 1;



num.padLeft(3); // Returns '001'

// - Or -

num.padLeft(3, "x"); // Returns 'xx1'

answered Aug 24 '13 at 4:12

community wiki

Jack B

I put this in a jsfiddle to make it quick for others to test, adding your version of the code: jsfiddle.net/kevinmicke/vnvghw7y/2 Your version is always very competitive, and sometimes the fastest. Thanks for the fairly exhaustive testing.
– kevinmicke
Oct 5 '15 at 19:17

add a comment |

up vote
6
down vote

ECMAScript 2017:
use padStart or padEnd

'abc'.padStart(10);         // "       abc"

'abc'.padStart(10, "foo");  // "foofoofabc"

'abc'.padStart(6,"123465"); // "123abc"

More info:

https://github.com/tc39/proposal-string-pad-start-end

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Nov 28 '16 at 8:49

community wiki

juFo

Voted up because it works in my scenario and would be nice to have in the specs. (Although it doesn't really answer the question, as it is not about "0"s :-) )
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:28

add a comment |

up vote
4
down vote

The latest way to do this is much simpler:

var number = 2

number.toLocaleString(undefined, {minimumIntegerDigits:2})

output: "02"

answered Feb 29 '16 at 4:00

community wiki

MrE

(8).toLocaleString(undefined, {minimumIntegerDigits: 5}) returns "00.008" for me, based on my locale.
– le_m
Mar 18 '17 at 21:58

add a comment |

up vote
4
down vote

Just an another solution, but I think it's more legible.

function zeroFill(text, size)

{

  while (text.length < size){

  	text = "0" + text;

  }

  

  return text;

}

edited Jul 22 '16 at 14:47

community wiki

3 revs
Arthur

Same as which other?
– Fabio Turati
Jul 22 '16 at 14:14

add a comment |

up vote
4
down vote

In all modern browsers you can use

numberStr.padStart(numberLength, "0");

function zeroFill(num, numLength) {

  var numberStr = num.toString();



  return numberStr.padStart(numLength, "0");

}



var numbers = [0, 1, 12, 123, 1234, 12345];



numbers.forEach(

  function(num) {

    var numString = num.toString();

    

    var paddedNum = zeroFill(numString, 5);



    console.log(paddedNum);

  }

);

Here is the MDN reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Aug 21 '17 at 22:07

community wiki

Lifehack

add a comment |

up vote
3
down vote

This one is less native, but may be the fastest...

zeroPad = function (num, count) {

    var pad = (num + '').length - count;

    while(--pad > -1) {

        num = '0' + num;

    }

    return num;

};

answered Sep 6 '09 at 23:05

community wiki

Jonathan Neal

i think you want to do pad = count - (num + '').length. negative numbers aren't handled well, but apart from that, it's not bad. +1
– nickf
Sep 6 '09 at 23:29

add a comment |

up vote
3
down vote

My solution

Number.prototype.PadLeft = function (length, digit) {

    var str = '' + this;

    while (str.length < length) {

        str = (digit || '0') + str;

    }

    return str;

};

Usage

var a = 567.25;

a.PadLeft(10); // 0000567.25



var b = 567.25;

b.PadLeft(20, '2'); // 22222222222222567.25

answered Dec 27 '12 at 17:15

community wiki

Aleksandar Toplek

add a comment |

up vote
3
down vote

Not that this question needs more answers, but I thought I would add the simple lodash version of this.

_.padLeft(number, 6, '0')

answered Jul 13 '15 at 15:21

community wiki

Art

1

Better: var zfill = _.partialRight(_.padStart, '0');
– Droogans
Mar 14 '16 at 18:03

2

Some people will probably protest "I don't want to use lodash" but it's hardly a valid argument anymore since you can install only this function: npm install --save lodash.padleft, then import padLeft from 'lodash.padleft' and omit the _.
– Andy
Oct 14 '16 at 20:44

add a comment |

up vote
3
down vote

Why not use recursion?

function padZero(s, n) {

    s = s.toString(); // in case someone passes a number

    return s.length >= n ? s : padZero('0' + s, n);

}

edited Oct 18 '16 at 12:12

community wiki

2 revs
lex82

padZero(223, 3) fails with '0223'
– Serginho
Oct 18 '16 at 10:55

Well, I assumed that this function is called with a string as the first parameter. However, I fixed it.
– lex82
Oct 18 '16 at 12:12

I love recursions :-)
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:33

add a comment |

up vote
2
down vote

Some monkeypatching also works

String.prototype.padLeft = function (n, c) {

  if (isNaN(n))

    return null;

  c = c || "0";

  return (new Array(n).join(c).substring(0, this.length-n)) + this; 

};

var paddedValue = "123".padLeft(6); // returns "000123"

var otherPadded = "TEXT".padLeft(8, " "); // returns "    TEXT"

answered Sep 7 '09 at 0:07

community wiki

Rodrigo

1

-1 monkeypatching the toplevel namespacing in javascript is bad practice.
– Jeremy Wall
Sep 7 '09 at 0:37

2

True for Object and Array objects, but String is not bad
– Rodrigo
Sep 11 '09 at 16:56

add a comment |

up vote
2
down vote

function pad(toPad, padChar, length){

    return (String(toPad).length < length)

        ? new Array(length - String(toPad).length + 1).join(padChar) + String(toPad)

        : toPad;

}

pad(5, 0, 6) = 000005

pad('10', 0, 2) = 10 // don't pad if not necessary

pad('S', 'O', 2) = SO

...etc.

Cheers

edited May 23 '12 at 3:50

community wiki

Madbreaks

This doesn't work. pad(100, '0', 4) => 000100
– drewish
May 22 '12 at 20:23

I think it should be more like: var s = String(toPad); return (s.length < length) ? new Array(length - s.length + 1).join('0') + s : s; if you fix it I'll remove my down vote.
– drewish
May 22 '12 at 20:24

@drewish Thanks for catching that, I agree with your edit (except for the hardcoded 0 join char :) -- Answer updated.
– Madbreaks
May 22 '12 at 21:25

Ah yeah I'd hard coded it in my usage.
– drewish
May 23 '12 at 2:04

I think it'd be better to store the string into a temporary variable. I did some benchmarking of that here: jsperf.com/padding-with-memoization it's also got benchmarks for the use of the new in "new Array" the results for new are all over the place but memiozation seems like the way to go.
– drewish
May 23 '12 at 16:28

add a comment |

up vote
2
down vote

The simplest, most straight-forward solution you will find.

function zerofill(number,length) {

    var output = number.toString();

    while(output.length < length) {

      output = '0' + output;

    }

    return output;

}

answered Jan 7 '15 at 9:22

community wiki

d4nyll

add a comment |

protected by Ionică Bizău Dec 2 '14 at 18:19

Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

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68 Answers
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oldest

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68 Answers
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active

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votes

up vote
172
down vote

accepted

Note to readers!

As commenters have pointed out, this solution is "clever", and as
clever solutions often are, it's memory intensive and relatively
slow. If performance is a concern for you, don't use this solution!

Potentially outdated: ECMAScript 2017 includes String.prototype.padStart and Number.prototype.toLocaleString is there since ECMAScript 3.1. Example:
var n=-0.1;

n.toLocaleString('en', {minimumIntegerDigits:4,minimumFractionDigits:2,useGrouping:false})
...will output "-0000.10".
// or 

const padded = (.1+"").padStart(6,"0");

`-${padded}`
...will output "-0000.1".

A simple function is all you need

function zeroFill( number, width )

{

  width -= number.toString().length;

  if ( width > 0 )

  {

    return new Array( width + (/./.test( number ) ? 2 : 1) ).join( '0' ) + number;

  }

  return number + ""; // always return a string

}

you could bake this into a library if you want to conserve namespace or whatever. Like with jQuery's extend.

edited Dec 6 '17 at 15:21

community wiki

9 revs, 5 users 66%
Peter Bailey

1

Now you just need to take care of numbers like 50.1234 and you've got a readable version of my solution below! I did, however, assume that we were just left padding, not overall padding.
– coderjoe
Aug 12 '09 at 20:16

5

Not a fan of creating a new array each time you want to pad a value. Readers should be aware of the the potential memory and GC impact if they're doing large #'s of these (e.g. on a node.js server that's doing 1000's of these operations per second. @profitehlolz's answer would seem to be the better solution.)
– broofa
Sep 24 '12 at 12:21

6

This does not work for negative values: zeroFill(-10, 7) -> "0000-10"
– digitalbath
Nov 7 '12 at 15:20

2

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:35

8

As of ECMAScript 2017, there's a build-in function padStart.
– Tomas Nikodym
Sep 11 '16 at 21:23

|
show 9 more comments

up vote
172
down vote

accepted

Note to readers!

As commenters have pointed out, this solution is "clever", and as
clever solutions often are, it's memory intensive and relatively
slow. If performance is a concern for you, don't use this solution!

Potentially outdated: ECMAScript 2017 includes String.prototype.padStart and Number.prototype.toLocaleString is there since ECMAScript 3.1. Example:
var n=-0.1;

n.toLocaleString('en', {minimumIntegerDigits:4,minimumFractionDigits:2,useGrouping:false})
...will output "-0000.10".
// or 

const padded = (.1+"").padStart(6,"0");

`-${padded}`
...will output "-0000.1".

A simple function is all you need

function zeroFill( number, width )

{

  width -= number.toString().length;

  if ( width > 0 )

  {

    return new Array( width + (/./.test( number ) ? 2 : 1) ).join( '0' ) + number;

  }

  return number + ""; // always return a string

}

you could bake this into a library if you want to conserve namespace or whatever. Like with jQuery's extend.

edited Dec 6 '17 at 15:21

community wiki

9 revs, 5 users 66%
Peter Bailey

1

Now you just need to take care of numbers like 50.1234 and you've got a readable version of my solution below! I did, however, assume that we were just left padding, not overall padding.
– coderjoe
Aug 12 '09 at 20:16

5

Not a fan of creating a new array each time you want to pad a value. Readers should be aware of the the potential memory and GC impact if they're doing large #'s of these (e.g. on a node.js server that's doing 1000's of these operations per second. @profitehlolz's answer would seem to be the better solution.)
– broofa
Sep 24 '12 at 12:21

6

This does not work for negative values: zeroFill(-10, 7) -> "0000-10"
– digitalbath
Nov 7 '12 at 15:20

2

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:35

8

As of ECMAScript 2017, there's a build-in function padStart.
– Tomas Nikodym
Sep 11 '16 at 21:23

|
show 9 more comments

up vote
172
down vote

accepted

Note to readers!

As commenters have pointed out, this solution is "clever", and as
clever solutions often are, it's memory intensive and relatively
slow. If performance is a concern for you, don't use this solution!

Potentially outdated: ECMAScript 2017 includes String.prototype.padStart and Number.prototype.toLocaleString is there since ECMAScript 3.1. Example:
var n=-0.1;

n.toLocaleString('en', {minimumIntegerDigits:4,minimumFractionDigits:2,useGrouping:false})
...will output "-0000.10".
// or 

const padded = (.1+"").padStart(6,"0");

`-${padded}`
...will output "-0000.1".

A simple function is all you need

function zeroFill( number, width )

{

  width -= number.toString().length;

  if ( width > 0 )

  {

    return new Array( width + (/./.test( number ) ? 2 : 1) ).join( '0' ) + number;

  }

  return number + ""; // always return a string

}

you could bake this into a library if you want to conserve namespace or whatever. Like with jQuery's extend.

edited Dec 6 '17 at 15:21

community wiki

9 revs, 5 users 66%
Peter Bailey

Note to readers!

As commenters have pointed out, this solution is "clever", and as
clever solutions often are, it's memory intensive and relatively
slow. If performance is a concern for you, don't use this solution!

Potentially outdated: ECMAScript 2017 includes String.prototype.padStart and Number.prototype.toLocaleString is there since ECMAScript 3.1. Example:
var n=-0.1;

n.toLocaleString('en', {minimumIntegerDigits:4,minimumFractionDigits:2,useGrouping:false})
...will output "-0000.10".
// or 

const padded = (.1+"").padStart(6,"0");

`-${padded}`
...will output "-0000.1".

A simple function is all you need

function zeroFill( number, width )

{

  width -= number.toString().length;

  if ( width > 0 )

  {

    return new Array( width + (/./.test( number ) ? 2 : 1) ).join( '0' ) + number;

  }

  return number + ""; // always return a string

}

you could bake this into a library if you want to conserve namespace or whatever. Like with jQuery's extend.

edited Dec 6 '17 at 15:21

community wiki

9 revs, 5 users 66%
Peter Bailey

edited Dec 6 '17 at 15:21

community wiki

9 revs, 5 users 66%
Peter Bailey

community wiki

9 revs, 5 users 66%
Peter Bailey

community wiki

9 revs, 5 users 66%
Peter Bailey

1

Now you just need to take care of numbers like 50.1234 and you've got a readable version of my solution below! I did, however, assume that we were just left padding, not overall padding.
– coderjoe
Aug 12 '09 at 20:16

5

Not a fan of creating a new array each time you want to pad a value. Readers should be aware of the the potential memory and GC impact if they're doing large #'s of these (e.g. on a node.js server that's doing 1000's of these operations per second. @profitehlolz's answer would seem to be the better solution.)
– broofa
Sep 24 '12 at 12:21

6

This does not work for negative values: zeroFill(-10, 7) -> "0000-10"
– digitalbath
Nov 7 '12 at 15:20

2

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:35

8

As of ECMAScript 2017, there's a build-in function padStart.
– Tomas Nikodym
Sep 11 '16 at 21:23

|
show 9 more comments

1

Now you just need to take care of numbers like 50.1234 and you've got a readable version of my solution below! I did, however, assume that we were just left padding, not overall padding.
– coderjoe
Aug 12 '09 at 20:16

5

Not a fan of creating a new array each time you want to pad a value. Readers should be aware of the the potential memory and GC impact if they're doing large #'s of these (e.g. on a node.js server that's doing 1000's of these operations per second. @profitehlolz's answer would seem to be the better solution.)
– broofa
Sep 24 '12 at 12:21

6

This does not work for negative values: zeroFill(-10, 7) -> "0000-10"
– digitalbath
Nov 7 '12 at 15:20

2

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:35

8

As of ECMAScript 2017, there's a build-in function padStart.
– Tomas Nikodym
Sep 11 '16 at 21:23

Now you just need to take care of numbers like 50.1234 and you've got a readable version of my solution below! I did, however, assume that we were just left padding, not overall padding.
– coderjoe
Aug 12 '09 at 20:16

Not a fan of creating a new array each time you want to pad a value. Readers should be aware of the the potential memory and GC impact if they're doing large #'s of these (e.g. on a node.js server that's doing 1000's of these operations per second. @profitehlolz's answer would seem to be the better solution.)
– broofa
Sep 24 '12 at 12:21

This does not work for negative values: zeroFill(-10, 7) -> "0000-10"
– digitalbath
Nov 7 '12 at 15:20

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:35

As of ECMAScript 2017, there's a build-in function padStart.
– Tomas Nikodym
Sep 11 '16 at 21:23

|
show 9 more comments

up vote
277
down vote

Simple way. You could add string multiplication for the pad and turn it into a function.

var pad = "000000";

var n = '5';

var result = (pad+n).slice(-pad.length);

As a function,

function paddy(num, padlen, padchar) {

    var pad_char = typeof padchar !== 'undefined' ? padchar : '0';

    var pad = new Array(1 + padlen).join(pad_char);

    return (pad + num).slice(-pad.length);

}

var fu = paddy(14, 5); // 00014

var bar = paddy(2, 4, '#'); // ###2

edited Mar 23 at 14:49

community wiki

4 revs, 2 users 83%
profitehlolz

84

This is a very nice solution, best I've seen. For clarity, here's a simpler version I'm using to pad the minutes in time formatting: ('0'+minutes).slice(-2)
– Luke Ehresman
Sep 14 '12 at 0:22

14

This is 5 times slower than the implementation with a while loop: gist.github.com/4382935
– andrewrk
Dec 26 '12 at 20:37

36

This has a FATAL FLAW with larger than expected numbers -- which is an extremely common occurrence. For example, paddy (888, 2) yields 88 and not 888 as required. This answer also does not handle negative numbers.
– Brock Adams
Jun 15 '13 at 22:20

7

@BrockAdams, zerofill is typically used for fixed width number/formatting cases -- therefore I think it'd actually be less likely (or even a non-issue) if given a 3 digit number when trying to do 2 digit zerofill.
– Seaux
Dec 8 '13 at 23:15

7

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

|
show 2 more comments

up vote
277
down vote

Simple way. You could add string multiplication for the pad and turn it into a function.

var pad = "000000";

var n = '5';

var result = (pad+n).slice(-pad.length);

As a function,

function paddy(num, padlen, padchar) {

    var pad_char = typeof padchar !== 'undefined' ? padchar : '0';

    var pad = new Array(1 + padlen).join(pad_char);

    return (pad + num).slice(-pad.length);

}

var fu = paddy(14, 5); // 00014

var bar = paddy(2, 4, '#'); // ###2

edited Mar 23 at 14:49

community wiki

4 revs, 2 users 83%
profitehlolz

84

This is a very nice solution, best I've seen. For clarity, here's a simpler version I'm using to pad the minutes in time formatting: ('0'+minutes).slice(-2)
– Luke Ehresman
Sep 14 '12 at 0:22

14

This is 5 times slower than the implementation with a while loop: gist.github.com/4382935
– andrewrk
Dec 26 '12 at 20:37

36

This has a FATAL FLAW with larger than expected numbers -- which is an extremely common occurrence. For example, paddy (888, 2) yields 88 and not 888 as required. This answer also does not handle negative numbers.
– Brock Adams
Jun 15 '13 at 22:20

7

@BrockAdams, zerofill is typically used for fixed width number/formatting cases -- therefore I think it'd actually be less likely (or even a non-issue) if given a 3 digit number when trying to do 2 digit zerofill.
– Seaux
Dec 8 '13 at 23:15

7

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

|
show 2 more comments

up vote
277
down vote

Simple way. You could add string multiplication for the pad and turn it into a function.

var pad = "000000";

var n = '5';

var result = (pad+n).slice(-pad.length);

As a function,

function paddy(num, padlen, padchar) {

    var pad_char = typeof padchar !== 'undefined' ? padchar : '0';

    var pad = new Array(1 + padlen).join(pad_char);

    return (pad + num).slice(-pad.length);

}

var fu = paddy(14, 5); // 00014

var bar = paddy(2, 4, '#'); // ###2

edited Mar 23 at 14:49

community wiki

4 revs, 2 users 83%
profitehlolz

Simple way. You could add string multiplication for the pad and turn it into a function.

var pad = "000000";

var n = '5';

var result = (pad+n).slice(-pad.length);

As a function,

function paddy(num, padlen, padchar) {

    var pad_char = typeof padchar !== 'undefined' ? padchar : '0';

    var pad = new Array(1 + padlen).join(pad_char);

    return (pad + num).slice(-pad.length);

}

var fu = paddy(14, 5); // 00014

var bar = paddy(2, 4, '#'); // ###2

edited Mar 23 at 14:49

community wiki

4 revs, 2 users 83%
profitehlolz

edited Mar 23 at 14:49

community wiki

4 revs, 2 users 83%
profitehlolz

community wiki

4 revs, 2 users 83%
profitehlolz

community wiki

4 revs, 2 users 83%
profitehlolz

84

This is a very nice solution, best I've seen. For clarity, here's a simpler version I'm using to pad the minutes in time formatting: ('0'+minutes).slice(-2)
– Luke Ehresman
Sep 14 '12 at 0:22

14

This is 5 times slower than the implementation with a while loop: gist.github.com/4382935
– andrewrk
Dec 26 '12 at 20:37

36

This has a FATAL FLAW with larger than expected numbers -- which is an extremely common occurrence. For example, paddy (888, 2) yields 88 and not 888 as required. This answer also does not handle negative numbers.
– Brock Adams
Jun 15 '13 at 22:20

7

@BrockAdams, zerofill is typically used for fixed width number/formatting cases -- therefore I think it'd actually be less likely (or even a non-issue) if given a 3 digit number when trying to do 2 digit zerofill.
– Seaux
Dec 8 '13 at 23:15

7

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

|
show 2 more comments

84

This is a very nice solution, best I've seen. For clarity, here's a simpler version I'm using to pad the minutes in time formatting: ('0'+minutes).slice(-2)
– Luke Ehresman
Sep 14 '12 at 0:22

14

This is 5 times slower than the implementation with a while loop: gist.github.com/4382935
– andrewrk
Dec 26 '12 at 20:37

36

This has a FATAL FLAW with larger than expected numbers -- which is an extremely common occurrence. For example, paddy (888, 2) yields 88 and not 888 as required. This answer also does not handle negative numbers.
– Brock Adams
Jun 15 '13 at 22:20

7

@BrockAdams, zerofill is typically used for fixed width number/formatting cases -- therefore I think it'd actually be less likely (or even a non-issue) if given a 3 digit number when trying to do 2 digit zerofill.
– Seaux
Dec 8 '13 at 23:15

7

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

This is a very nice solution, best I've seen. For clarity, here's a simpler version I'm using to pad the minutes in time formatting: ('0'+minutes).slice(-2)
– Luke Ehresman
Sep 14 '12 at 0:22

This is 5 times slower than the implementation with a while loop: gist.github.com/4382935
– andrewrk
Dec 26 '12 at 20:37

This has a FATAL FLAW with larger than expected numbers -- which is an extremely common occurrence. For example, paddy (888, 2) yields 88 and not 888 as required. This answer also does not handle negative numbers.
– Brock Adams
Jun 15 '13 at 22:20

@BrockAdams, zerofill is typically used for fixed width number/formatting cases -- therefore I think it'd actually be less likely (or even a non-issue) if given a 3 digit number when trying to do 2 digit zerofill.
– Seaux
Dec 8 '13 at 23:15

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

|
show 2 more comments

up vote
228
down vote

I can't believe all the complex answers on here... Just use this:

var zerofilled = ('0000'+n).slice(-4);

edited Sep 11 at 7:40

community wiki

2 revs, 2 users 67%
Seaux

17

Good except for negative numbers and numbers longer than 4 digits.
– wberry
Sep 19 '14 at 22:40

7

@wberry this was not intended to be production ready code, but more a simple explanation of the basics to solving the question at hand. Aka it's very easy to determine if a number is positive or negative and add the appropriate sign if needed, so i didn't want to cluster my easy example with that. Also, it's just as easy to make a function taking a digit length as an arg and using that instead of my hardcoded values.
– Seaux
Sep 21 '14 at 4:13

4

This seems like a good solution for simple cases. Easier to understand.
– Artur Carvalho
Jan 28 '15 at 12:23

2

@OmShankar by "complex answers", I wasn't referring to explained or detailed answers (which are obviously encouraged on this site even though my answer isn't), but rather answers that contained unnecessary code complexity.
– Seaux
Aug 10 '15 at 23:10

3

I like this answer. In my own specific case, I know that the numbers are always positive and never more than 3 digits, and this is often the case when you're padding with leading 0's and your input is well known. Nice!
– Patrick Chu
Dec 16 '16 at 22:12

|
show 7 more comments

up vote
228
down vote

I can't believe all the complex answers on here... Just use this:

var zerofilled = ('0000'+n).slice(-4);

edited Sep 11 at 7:40

community wiki

2 revs, 2 users 67%
Seaux

17

Good except for negative numbers and numbers longer than 4 digits.
– wberry
Sep 19 '14 at 22:40

7

@wberry this was not intended to be production ready code, but more a simple explanation of the basics to solving the question at hand. Aka it's very easy to determine if a number is positive or negative and add the appropriate sign if needed, so i didn't want to cluster my easy example with that. Also, it's just as easy to make a function taking a digit length as an arg and using that instead of my hardcoded values.
– Seaux
Sep 21 '14 at 4:13

4

This seems like a good solution for simple cases. Easier to understand.
– Artur Carvalho
Jan 28 '15 at 12:23

2

@OmShankar by "complex answers", I wasn't referring to explained or detailed answers (which are obviously encouraged on this site even though my answer isn't), but rather answers that contained unnecessary code complexity.
– Seaux
Aug 10 '15 at 23:10

3

I like this answer. In my own specific case, I know that the numbers are always positive and never more than 3 digits, and this is often the case when you're padding with leading 0's and your input is well known. Nice!
– Patrick Chu
Dec 16 '16 at 22:12

|
show 7 more comments

up vote
228
down vote

I can't believe all the complex answers on here... Just use this:

var zerofilled = ('0000'+n).slice(-4);

edited Sep 11 at 7:40

community wiki

2 revs, 2 users 67%
Seaux

I can't believe all the complex answers on here... Just use this:

var zerofilled = ('0000'+n).slice(-4);

edited Sep 11 at 7:40

community wiki

2 revs, 2 users 67%
Seaux

edited Sep 11 at 7:40

community wiki

2 revs, 2 users 67%
Seaux

community wiki

2 revs, 2 users 67%
Seaux

community wiki

2 revs, 2 users 67%
Seaux

17

Good except for negative numbers and numbers longer than 4 digits.
– wberry
Sep 19 '14 at 22:40

7

@wberry this was not intended to be production ready code, but more a simple explanation of the basics to solving the question at hand. Aka it's very easy to determine if a number is positive or negative and add the appropriate sign if needed, so i didn't want to cluster my easy example with that. Also, it's just as easy to make a function taking a digit length as an arg and using that instead of my hardcoded values.
– Seaux
Sep 21 '14 at 4:13

4

This seems like a good solution for simple cases. Easier to understand.
– Artur Carvalho
Jan 28 '15 at 12:23

2

@OmShankar by "complex answers", I wasn't referring to explained or detailed answers (which are obviously encouraged on this site even though my answer isn't), but rather answers that contained unnecessary code complexity.
– Seaux
Aug 10 '15 at 23:10

3

I like this answer. In my own specific case, I know that the numbers are always positive and never more than 3 digits, and this is often the case when you're padding with leading 0's and your input is well known. Nice!
– Patrick Chu
Dec 16 '16 at 22:12

|
show 7 more comments

17

Good except for negative numbers and numbers longer than 4 digits.
– wberry
Sep 19 '14 at 22:40

7

@wberry this was not intended to be production ready code, but more a simple explanation of the basics to solving the question at hand. Aka it's very easy to determine if a number is positive or negative and add the appropriate sign if needed, so i didn't want to cluster my easy example with that. Also, it's just as easy to make a function taking a digit length as an arg and using that instead of my hardcoded values.
– Seaux
Sep 21 '14 at 4:13

4

This seems like a good solution for simple cases. Easier to understand.
– Artur Carvalho
Jan 28 '15 at 12:23

2

@OmShankar by "complex answers", I wasn't referring to explained or detailed answers (which are obviously encouraged on this site even though my answer isn't), but rather answers that contained unnecessary code complexity.
– Seaux
Aug 10 '15 at 23:10

3

I like this answer. In my own specific case, I know that the numbers are always positive and never more than 3 digits, and this is often the case when you're padding with leading 0's and your input is well known. Nice!
– Patrick Chu
Dec 16 '16 at 22:12

Good except for negative numbers and numbers longer than 4 digits.
– wberry
Sep 19 '14 at 22:40

@wberry this was not intended to be production ready code, but more a simple explanation of the basics to solving the question at hand. Aka it's very easy to determine if a number is positive or negative and add the appropriate sign if needed, so i didn't want to cluster my easy example with that. Also, it's just as easy to make a function taking a digit length as an arg and using that instead of my hardcoded values.
– Seaux
Sep 21 '14 at 4:13

This seems like a good solution for simple cases. Easier to understand.
– Artur Carvalho
Jan 28 '15 at 12:23

@OmShankar by "complex answers", I wasn't referring to explained or detailed answers (which are obviously encouraged on this site even though my answer isn't), but rather answers that contained unnecessary code complexity.
– Seaux
Aug 10 '15 at 23:10

I like this answer. In my own specific case, I know that the numbers are always positive and never more than 3 digits, and this is often the case when you're padding with leading 0's and your input is well known. Nice!
– Patrick Chu
Dec 16 '16 at 22:12

|
show 7 more comments

up vote
104
down vote

I actually had to come up with something like this recently.
I figured there had to be a way to do it without using loops.

This is what I came up with.

function zeroPad(num, numZeros) {

    var n = Math.abs(num);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( num < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Then just use it providing a number to zero pad:

> zeroPad(50,4);

"0050"

If the number is larger than the padding, the number will expand beyond the padding:

> zeroPad(51234, 3);

"51234"

Decimals are fine too!

> zeroPad(51.1234, 4);

"0051.1234"

If you don't mind polluting the global namespace you can add it to Number directly:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

And if you'd rather have decimals take up space in the padding:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - n.toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Cheers!

XDR came up with a logarithmic variation that seems to perform better.

WARNING: This function fails if num equals zero (e.g. zeropad(0, 2))

function zeroPad (num, numZeros) {

    var an = Math.abs (num);

    var digitCount = 1 + Math.floor (Math.log (an) / Math.LN10);

    if (digitCount >= numZeros) {

        return num;

    }

    var zeroString = Math.pow (10, numZeros - digitCount).toString ().substr (1);

    return num < 0 ? '-' + zeroString + an : zeroString + an;

}

Speaking of performance, tomsmeding compared the top 3 answers (4 with the log variation). Guess which one majorly outperformed the other two? :)

edited May 23 '17 at 12:34

community wiki

6 revs, 3 users 78%
coderjoe

9

This is good, I like how it's readable and robust. The only thing I would change is the name of the numZeros parameter, since it's misleading. It's not the number of zeros you want to add, it's the minimum length the number should be. A better name would be numLength or even length.
– Senseful
Jan 2 '12 at 4:23

12

+1. Unlike the higher voted answers, this one handles negative numbers and does not return gross errors on overflow!!?!
– Brock Adams
Jun 15 '13 at 22:39

10

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

4

I improved the performance of this answer by over 100% by using logarithms. Please see the logarithmic test case at http://jsperf.com/left-zero-pad/10
– XDR
Aug 18 '14 at 4:30

2

The original solution is better, logarithmic variation doesn't work if num can be zero...
– Ondra C.
Mar 17 '15 at 10:03

|
show 7 more comments

up vote
104
down vote

I actually had to come up with something like this recently.
I figured there had to be a way to do it without using loops.

This is what I came up with.

function zeroPad(num, numZeros) {

    var n = Math.abs(num);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( num < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Then just use it providing a number to zero pad:

> zeroPad(50,4);

"0050"

If the number is larger than the padding, the number will expand beyond the padding:

> zeroPad(51234, 3);

"51234"

Decimals are fine too!

> zeroPad(51.1234, 4);

"0051.1234"

If you don't mind polluting the global namespace you can add it to Number directly:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

And if you'd rather have decimals take up space in the padding:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - n.toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Cheers!

XDR came up with a logarithmic variation that seems to perform better.

WARNING: This function fails if num equals zero (e.g. zeropad(0, 2))

function zeroPad (num, numZeros) {

    var an = Math.abs (num);

    var digitCount = 1 + Math.floor (Math.log (an) / Math.LN10);

    if (digitCount >= numZeros) {

        return num;

    }

    var zeroString = Math.pow (10, numZeros - digitCount).toString ().substr (1);

    return num < 0 ? '-' + zeroString + an : zeroString + an;

}

Speaking of performance, tomsmeding compared the top 3 answers (4 with the log variation). Guess which one majorly outperformed the other two? :)

edited May 23 '17 at 12:34

community wiki

6 revs, 3 users 78%
coderjoe

9

This is good, I like how it's readable and robust. The only thing I would change is the name of the numZeros parameter, since it's misleading. It's not the number of zeros you want to add, it's the minimum length the number should be. A better name would be numLength or even length.
– Senseful
Jan 2 '12 at 4:23

12

+1. Unlike the higher voted answers, this one handles negative numbers and does not return gross errors on overflow!!?!
– Brock Adams
Jun 15 '13 at 22:39

10

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

4

I improved the performance of this answer by over 100% by using logarithms. Please see the logarithmic test case at http://jsperf.com/left-zero-pad/10
– XDR
Aug 18 '14 at 4:30

2

The original solution is better, logarithmic variation doesn't work if num can be zero...
– Ondra C.
Mar 17 '15 at 10:03

|
show 7 more comments

up vote
104
down vote

I actually had to come up with something like this recently.
I figured there had to be a way to do it without using loops.

This is what I came up with.

function zeroPad(num, numZeros) {

    var n = Math.abs(num);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( num < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Then just use it providing a number to zero pad:

> zeroPad(50,4);

"0050"

If the number is larger than the padding, the number will expand beyond the padding:

> zeroPad(51234, 3);

"51234"

Decimals are fine too!

> zeroPad(51.1234, 4);

"0051.1234"

If you don't mind polluting the global namespace you can add it to Number directly:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

And if you'd rather have decimals take up space in the padding:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - n.toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Cheers!

XDR came up with a logarithmic variation that seems to perform better.

WARNING: This function fails if num equals zero (e.g. zeropad(0, 2))

function zeroPad (num, numZeros) {

    var an = Math.abs (num);

    var digitCount = 1 + Math.floor (Math.log (an) / Math.LN10);

    if (digitCount >= numZeros) {

        return num;

    }

    var zeroString = Math.pow (10, numZeros - digitCount).toString ().substr (1);

    return num < 0 ? '-' + zeroString + an : zeroString + an;

}

Speaking of performance, tomsmeding compared the top 3 answers (4 with the log variation). Guess which one majorly outperformed the other two? :)

edited May 23 '17 at 12:34

community wiki

6 revs, 3 users 78%
coderjoe

I actually had to come up with something like this recently.
I figured there had to be a way to do it without using loops.

This is what I came up with.

function zeroPad(num, numZeros) {

    var n = Math.abs(num);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( num < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Then just use it providing a number to zero pad:

> zeroPad(50,4);

"0050"

If the number is larger than the padding, the number will expand beyond the padding:

> zeroPad(51234, 3);

"51234"

Decimals are fine too!

> zeroPad(51.1234, 4);

"0051.1234"

If you don't mind polluting the global namespace you can add it to Number directly:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

And if you'd rather have decimals take up space in the padding:

Number.prototype.leftZeroPad = function(numZeros) {

    var n = Math.abs(this);

    var zeros = Math.max(0, numZeros - n.toString().length );

    var zeroString = Math.pow(10,zeros).toString().substr(1);

    if( this < 0 ) {

        zeroString = '-' + zeroString;

    }



    return zeroString+n;

}

Cheers!

XDR came up with a logarithmic variation that seems to perform better.

WARNING: This function fails if num equals zero (e.g. zeropad(0, 2))

function zeroPad (num, numZeros) {

    var an = Math.abs (num);

    var digitCount = 1 + Math.floor (Math.log (an) / Math.LN10);

    if (digitCount >= numZeros) {

        return num;

    }

    var zeroString = Math.pow (10, numZeros - digitCount).toString ().substr (1);

    return num < 0 ? '-' + zeroString + an : zeroString + an;

}

Speaking of performance, tomsmeding compared the top 3 answers (4 with the log variation). Guess which one majorly outperformed the other two? :)

edited May 23 '17 at 12:34

community wiki

6 revs, 3 users 78%
coderjoe

edited May 23 '17 at 12:34

community wiki

6 revs, 3 users 78%
coderjoe

community wiki

6 revs, 3 users 78%
coderjoe

community wiki

6 revs, 3 users 78%
coderjoe

9

This is good, I like how it's readable and robust. The only thing I would change is the name of the numZeros parameter, since it's misleading. It's not the number of zeros you want to add, it's the minimum length the number should be. A better name would be numLength or even length.
– Senseful
Jan 2 '12 at 4:23

12

+1. Unlike the higher voted answers, this one handles negative numbers and does not return gross errors on overflow!!?!
– Brock Adams
Jun 15 '13 at 22:39

10

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

4

I improved the performance of this answer by over 100% by using logarithms. Please see the logarithmic test case at http://jsperf.com/left-zero-pad/10
– XDR
Aug 18 '14 at 4:30

2

The original solution is better, logarithmic variation doesn't work if num can be zero...
– Ondra C.
Mar 17 '15 at 10:03

|
show 7 more comments

9

This is good, I like how it's readable and robust. The only thing I would change is the name of the numZeros parameter, since it's misleading. It's not the number of zeros you want to add, it's the minimum length the number should be. A better name would be numLength or even length.
– Senseful
Jan 2 '12 at 4:23

12

+1. Unlike the higher voted answers, this one handles negative numbers and does not return gross errors on overflow!!?!
– Brock Adams
Jun 15 '13 at 22:39

10

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

4

I improved the performance of this answer by over 100% by using logarithms. Please see the logarithmic test case at http://jsperf.com/left-zero-pad/10
– XDR
Aug 18 '14 at 4:30

2

The original solution is better, logarithmic variation doesn't work if num can be zero...
– Ondra C.
Mar 17 '15 at 10:03

This is good, I like how it's readable and robust. The only thing I would change is the name of the numZeros parameter, since it's misleading. It's not the number of zeros you want to add, it's the minimum length the number should be. A better name would be numLength or even length.
– Senseful
Jan 2 '12 at 4:23

+1. Unlike the higher voted answers, this one handles negative numbers and does not return gross errors on overflow!!?!
– Brock Adams
Jun 15 '13 at 22:39

Performance measure amongst three top answers here: jsperf.com/left-zero-pad
– tomsmeding
Feb 21 '14 at 16:36

I improved the performance of this answer by over 100% by using logarithms. Please see the logarithmic test case at http://jsperf.com/left-zero-pad/10
– XDR
Aug 18 '14 at 4:30

The original solution is better, logarithmic variation doesn't work if num can be zero...
– Ondra C.
Mar 17 '15 at 10:03

|
show 7 more comments

up vote
55
down vote

Here's what I used to pad a number up to 7 characters.

("0000000" + number).slice(-7)

This approach will probably suffice for most people.

Edit: If you want to make it more generic you can do this:

("0".repeat(padding) + number).slice(-padding)

edited Jun 27 '17 at 9:46

community wiki

6 revs
chetbox

This fails, badly, on large numbers and negative numbers. number = 123456789, with the code, above, for example.
– Brock Adams
Jun 15 '13 at 22:49

This solution is the best possible ever for certains scenarios! I've got no clue why we didn't come up with this before. The scenario is: you've got a number which is always positive [as Brock mentioned] and you know that it won't take more characters than your limit. In our case we have a score which we store in Amazon SDB. As you know SDB can't compare numbers, so all scores have to be zero padded. This is extremely easy and effective!
– Krystian
Oct 10 '13 at 11:27

2

Apologies, I should have mentioned that this won't work for negative numbers. I think this deals with the more common positive number case, though. It does what I need, at least!
– chetbox
Oct 16 '13 at 11:33

Needed this again and realised you don't need new String. You can just use string literals.
– chetbox
Mar 4 '16 at 15:56

2

(new Array(padding + 1).join("0") can be replaced with "0".repeat(padding)
– Pavel
Jun 22 '17 at 16:17

|
show 3 more comments

up vote
55
down vote

Here's what I used to pad a number up to 7 characters.

("0000000" + number).slice(-7)

This approach will probably suffice for most people.

Edit: If you want to make it more generic you can do this:

("0".repeat(padding) + number).slice(-padding)

edited Jun 27 '17 at 9:46

community wiki

6 revs
chetbox

This fails, badly, on large numbers and negative numbers. number = 123456789, with the code, above, for example.
– Brock Adams
Jun 15 '13 at 22:49

This solution is the best possible ever for certains scenarios! I've got no clue why we didn't come up with this before. The scenario is: you've got a number which is always positive [as Brock mentioned] and you know that it won't take more characters than your limit. In our case we have a score which we store in Amazon SDB. As you know SDB can't compare numbers, so all scores have to be zero padded. This is extremely easy and effective!
– Krystian
Oct 10 '13 at 11:27

2

Apologies, I should have mentioned that this won't work for negative numbers. I think this deals with the more common positive number case, though. It does what I need, at least!
– chetbox
Oct 16 '13 at 11:33

Needed this again and realised you don't need new String. You can just use string literals.
– chetbox
Mar 4 '16 at 15:56

2

(new Array(padding + 1).join("0") can be replaced with "0".repeat(padding)
– Pavel
Jun 22 '17 at 16:17

|
show 3 more comments

up vote
55
down vote

Here's what I used to pad a number up to 7 characters.

("0000000" + number).slice(-7)

This approach will probably suffice for most people.

Edit: If you want to make it more generic you can do this:

("0".repeat(padding) + number).slice(-padding)

edited Jun 27 '17 at 9:46

community wiki

6 revs
chetbox

Here's what I used to pad a number up to 7 characters.

("0000000" + number).slice(-7)

This approach will probably suffice for most people.

Edit: If you want to make it more generic you can do this:

("0".repeat(padding) + number).slice(-padding)

edited Jun 27 '17 at 9:46

community wiki

6 revs
chetbox

edited Jun 27 '17 at 9:46

community wiki

6 revs
chetbox

community wiki

6 revs
chetbox

community wiki

6 revs
chetbox

This fails, badly, on large numbers and negative numbers. number = 123456789, with the code, above, for example.
– Brock Adams
Jun 15 '13 at 22:49

This solution is the best possible ever for certains scenarios! I've got no clue why we didn't come up with this before. The scenario is: you've got a number which is always positive [as Brock mentioned] and you know that it won't take more characters than your limit. In our case we have a score which we store in Amazon SDB. As you know SDB can't compare numbers, so all scores have to be zero padded. This is extremely easy and effective!
– Krystian
Oct 10 '13 at 11:27

2

Apologies, I should have mentioned that this won't work for negative numbers. I think this deals with the more common positive number case, though. It does what I need, at least!
– chetbox
Oct 16 '13 at 11:33

Needed this again and realised you don't need new String. You can just use string literals.
– chetbox
Mar 4 '16 at 15:56

2

(new Array(padding + 1).join("0") can be replaced with "0".repeat(padding)
– Pavel
Jun 22 '17 at 16:17

|
show 3 more comments

This fails, badly, on large numbers and negative numbers. number = 123456789, with the code, above, for example.
– Brock Adams
Jun 15 '13 at 22:49

This solution is the best possible ever for certains scenarios! I've got no clue why we didn't come up with this before. The scenario is: you've got a number which is always positive [as Brock mentioned] and you know that it won't take more characters than your limit. In our case we have a score which we store in Amazon SDB. As you know SDB can't compare numbers, so all scores have to be zero padded. This is extremely easy and effective!
– Krystian
Oct 10 '13 at 11:27

2

Apologies, I should have mentioned that this won't work for negative numbers. I think this deals with the more common positive number case, though. It does what I need, at least!
– chetbox
Oct 16 '13 at 11:33

Needed this again and realised you don't need new String. You can just use string literals.
– chetbox
Mar 4 '16 at 15:56

2

(new Array(padding + 1).join("0") can be replaced with "0".repeat(padding)
– Pavel
Jun 22 '17 at 16:17

This fails, badly, on large numbers and negative numbers. number = 123456789, with the code, above, for example.
– Brock Adams
Jun 15 '13 at 22:49

This solution is the best possible ever for certains scenarios! I've got no clue why we didn't come up with this before. The scenario is: you've got a number which is always positive [as Brock mentioned] and you know that it won't take more characters than your limit. In our case we have a score which we store in Amazon SDB. As you know SDB can't compare numbers, so all scores have to be zero padded. This is extremely easy and effective!
– Krystian
Oct 10 '13 at 11:27

Apologies, I should have mentioned that this won't work for negative numbers. I think this deals with the more common positive number case, though. It does what I need, at least!
– chetbox
Oct 16 '13 at 11:33

Needed this again and realised you don't need new String. You can just use string literals.
– chetbox
Mar 4 '16 at 15:56

(new Array(padding + 1).join("0") can be replaced with "0".repeat(padding)
– Pavel
Jun 22 '17 at 16:17

|
show 3 more comments

up vote
23
down vote

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumIntegerDigits: 3,

    useGrouping: false

});

This will produce the value of "005" for text. You can also use the toLocaleString function of Number to pad zeros to the right side of the decimal point.

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumFractionDigits: 2,

    useGrouping: false

});

This will produce the value of "5.00" for text. Change useGrouping to true to use comma separators for thousands.

Complete Example

edited May 28 '15 at 21:01

community wiki

2 revs, 2 users 95%
Daniel Barbalace

Wow, this is a super clean solution and works in node.js.
– jishi
Feb 6 '16 at 22:21

I chuckled when I read the intro -- "needless complicated suggestions". Software development, as in all engineering disciplines, involves weighing trade-offs. I tried this "standard way" myself -- and timed it. It definitely works, but I would never choose to use it for any kind of repetitive application due to how slow it is compared to many of the other "home rolled" solutions.
– bearvarine
Sep 16 '17 at 20:47

True, a lot of built-in JavaScript functions perform poorly when looping over lots of data, but I would argue you should not use JavaScript for that, even server-side like Node.js. If you have lots of data to process server-side, you should use a better platform like .NET or Java. If you are processing client-side data for display to the end user, you should only process the data for what you are currently rendering to the end user's screen. For example, render only the visible rows of a table and don't process data for other roads.
– Daniel Barbalace
Jul 23 at 21:03

add a comment |

up vote
23
down vote

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumIntegerDigits: 3,

    useGrouping: false

});

This will produce the value of "005" for text. You can also use the toLocaleString function of Number to pad zeros to the right side of the decimal point.

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumFractionDigits: 2,

    useGrouping: false

});

This will produce the value of "5.00" for text. Change useGrouping to true to use comma separators for thousands.

Complete Example

edited May 28 '15 at 21:01

community wiki

2 revs, 2 users 95%
Daniel Barbalace

Wow, this is a super clean solution and works in node.js.
– jishi
Feb 6 '16 at 22:21

I chuckled when I read the intro -- "needless complicated suggestions". Software development, as in all engineering disciplines, involves weighing trade-offs. I tried this "standard way" myself -- and timed it. It definitely works, but I would never choose to use it for any kind of repetitive application due to how slow it is compared to many of the other "home rolled" solutions.
– bearvarine
Sep 16 '17 at 20:47

True, a lot of built-in JavaScript functions perform poorly when looping over lots of data, but I would argue you should not use JavaScript for that, even server-side like Node.js. If you have lots of data to process server-side, you should use a better platform like .NET or Java. If you are processing client-side data for display to the end user, you should only process the data for what you are currently rendering to the end user's screen. For example, render only the visible rows of a table and don't process data for other roads.
– Daniel Barbalace
Jul 23 at 21:03

add a comment |

up vote
23
down vote

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumIntegerDigits: 3,

    useGrouping: false

});

This will produce the value of "005" for text. You can also use the toLocaleString function of Number to pad zeros to the right side of the decimal point.

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumFractionDigits: 2,

    useGrouping: false

});

This will produce the value of "5.00" for text. Change useGrouping to true to use comma separators for thousands.

Complete Example

edited May 28 '15 at 21:01

community wiki

2 revs, 2 users 95%
Daniel Barbalace

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumIntegerDigits: 3,

    useGrouping: false

});

This will produce the value of "005" for text. You can also use the toLocaleString function of Number to pad zeros to the right side of the decimal point.

var amount = 5;



var text = amount.toLocaleString('en-US',

{

    style: 'decimal',

    minimumFractionDigits: 2,

    useGrouping: false

});

This will produce the value of "5.00" for text. Change useGrouping to true to use comma separators for thousands.

Complete Example

edited May 28 '15 at 21:01

community wiki

2 revs, 2 users 95%
Daniel Barbalace

edited May 28 '15 at 21:01

community wiki

2 revs, 2 users 95%
Daniel Barbalace

community wiki

2 revs, 2 users 95%
Daniel Barbalace

community wiki

2 revs, 2 users 95%
Daniel Barbalace

Wow, this is a super clean solution and works in node.js.
– jishi
Feb 6 '16 at 22:21

I chuckled when I read the intro -- "needless complicated suggestions". Software development, as in all engineering disciplines, involves weighing trade-offs. I tried this "standard way" myself -- and timed it. It definitely works, but I would never choose to use it for any kind of repetitive application due to how slow it is compared to many of the other "home rolled" solutions.
– bearvarine
Sep 16 '17 at 20:47

True, a lot of built-in JavaScript functions perform poorly when looping over lots of data, but I would argue you should not use JavaScript for that, even server-side like Node.js. If you have lots of data to process server-side, you should use a better platform like .NET or Java. If you are processing client-side data for display to the end user, you should only process the data for what you are currently rendering to the end user's screen. For example, render only the visible rows of a table and don't process data for other roads.
– Daniel Barbalace
Jul 23 at 21:03

add a comment |

Wow, this is a super clean solution and works in node.js.
– jishi
Feb 6 '16 at 22:21

I chuckled when I read the intro -- "needless complicated suggestions". Software development, as in all engineering disciplines, involves weighing trade-offs. I tried this "standard way" myself -- and timed it. It definitely works, but I would never choose to use it for any kind of repetitive application due to how slow it is compared to many of the other "home rolled" solutions.
– bearvarine
Sep 16 '17 at 20:47

True, a lot of built-in JavaScript functions perform poorly when looping over lots of data, but I would argue you should not use JavaScript for that, even server-side like Node.js. If you have lots of data to process server-side, you should use a better platform like .NET or Java. If you are processing client-side data for display to the end user, you should only process the data for what you are currently rendering to the end user's screen. For example, render only the visible rows of a table and don't process data for other roads.
– Daniel Barbalace
Jul 23 at 21:03

Wow, this is a super clean solution and works in node.js.
– jishi
Feb 6 '16 at 22:21

I chuckled when I read the intro -- "needless complicated suggestions". Software development, as in all engineering disciplines, involves weighing trade-offs. I tried this "standard way" myself -- and timed it. It definitely works, but I would never choose to use it for any kind of repetitive application due to how slow it is compared to many of the other "home rolled" solutions.
– bearvarine
Sep 16 '17 at 20:47

True, a lot of built-in JavaScript functions perform poorly when looping over lots of data, but I would argue you should not use JavaScript for that, even server-side like Node.js. If you have lots of data to process server-side, you should use a better platform like .NET or Java. If you are processing client-side data for display to the end user, you should only process the data for what you are currently rendering to the end user's screen. For example, render only the visible rows of a table and don't process data for other roads.
– Daniel Barbalace
Jul 23 at 21:03

add a comment |

up vote
20
down vote

If the fill number is known in advance not to exceed a certain value, there's another way to do this with no loops:

var fillZeroes = "00000000000000000000";  // max number of zero fill ever asked for in global



function zeroFill(number, width) {

    // make sure it's a string

    var input = number + "";  

    var prefix = "";

    if (input.charAt(0) === '-') {

        prefix = "-";

        input = input.slice(1);

        --width;

    }

    var fillAmt = Math.max(width - input.length, 0);

    return prefix + fillZeroes.slice(0, fillAmt) + input;

}

Test cases here: http://jsfiddle.net/jfriend00/N87mZ/

edited Jun 16 '13 at 1:08

community wiki

jfriend00

@BrockAdams - fixed for both cases you mention. If the number is already as wide as the fill width, then no fill is added.
– jfriend00
Jun 16 '13 at 1:08

Thanks; +1. The negative numbers are slightly off from the usual convention, though. In your test case, -88 should yield "-00088", for example.
– Brock Adams
Jun 16 '13 at 1:57

1

@BrockAdams - I wasn't sure whether calling zeroFill(-88, 5) should produce -00088 or -0088? I guess it depends upon whether you want the width argument to be the entire width of the number or just the number of digits (not including the negative sign). An implementer can easily switch the behavior to not include the negative sign by just removing the --width line of code.
– jfriend00
Jun 16 '13 at 3:07

add a comment |

up vote
20
down vote

If the fill number is known in advance not to exceed a certain value, there's another way to do this with no loops:

var fillZeroes = "00000000000000000000";  // max number of zero fill ever asked for in global



function zeroFill(number, width) {

    // make sure it's a string

    var input = number + "";  

    var prefix = "";

    if (input.charAt(0) === '-') {

        prefix = "-";

        input = input.slice(1);

        --width;

    }

    var fillAmt = Math.max(width - input.length, 0);

    return prefix + fillZeroes.slice(0, fillAmt) + input;

}

Test cases here: http://jsfiddle.net/jfriend00/N87mZ/

edited Jun 16 '13 at 1:08

community wiki

jfriend00

@BrockAdams - fixed for both cases you mention. If the number is already as wide as the fill width, then no fill is added.
– jfriend00
Jun 16 '13 at 1:08

Thanks; +1. The negative numbers are slightly off from the usual convention, though. In your test case, -88 should yield "-00088", for example.
– Brock Adams
Jun 16 '13 at 1:57

1

@BrockAdams - I wasn't sure whether calling zeroFill(-88, 5) should produce -00088 or -0088? I guess it depends upon whether you want the width argument to be the entire width of the number or just the number of digits (not including the negative sign). An implementer can easily switch the behavior to not include the negative sign by just removing the --width line of code.
– jfriend00
Jun 16 '13 at 3:07

add a comment |

up vote
20
down vote

If the fill number is known in advance not to exceed a certain value, there's another way to do this with no loops:

var fillZeroes = "00000000000000000000";  // max number of zero fill ever asked for in global



function zeroFill(number, width) {

    // make sure it's a string

    var input = number + "";  

    var prefix = "";

    if (input.charAt(0) === '-') {

        prefix = "-";

        input = input.slice(1);

        --width;

    }

    var fillAmt = Math.max(width - input.length, 0);

    return prefix + fillZeroes.slice(0, fillAmt) + input;

}

Test cases here: http://jsfiddle.net/jfriend00/N87mZ/

edited Jun 16 '13 at 1:08

community wiki

jfriend00

If the fill number is known in advance not to exceed a certain value, there's another way to do this with no loops:

var fillZeroes = "00000000000000000000";  // max number of zero fill ever asked for in global



function zeroFill(number, width) {

    // make sure it's a string

    var input = number + "";  

    var prefix = "";

    if (input.charAt(0) === '-') {

        prefix = "-";

        input = input.slice(1);

        --width;

    }

    var fillAmt = Math.max(width - input.length, 0);

    return prefix + fillZeroes.slice(0, fillAmt) + input;

}

Test cases here: http://jsfiddle.net/jfriend00/N87mZ/

edited Jun 16 '13 at 1:08

community wiki

jfriend00

edited Jun 16 '13 at 1:08

community wiki

jfriend00

community wiki

jfriend00

community wiki

jfriend00

@BrockAdams - fixed for both cases you mention. If the number is already as wide as the fill width, then no fill is added.
– jfriend00
Jun 16 '13 at 1:08

Thanks; +1. The negative numbers are slightly off from the usual convention, though. In your test case, -88 should yield "-00088", for example.
– Brock Adams
Jun 16 '13 at 1:57

1

@BrockAdams - I wasn't sure whether calling zeroFill(-88, 5) should produce -00088 or -0088? I guess it depends upon whether you want the width argument to be the entire width of the number or just the number of digits (not including the negative sign). An implementer can easily switch the behavior to not include the negative sign by just removing the --width line of code.
– jfriend00
Jun 16 '13 at 3:07

add a comment |

@BrockAdams - fixed for both cases you mention. If the number is already as wide as the fill width, then no fill is added.
– jfriend00
Jun 16 '13 at 1:08

Thanks; +1. The negative numbers are slightly off from the usual convention, though. In your test case, -88 should yield "-00088", for example.
– Brock Adams
Jun 16 '13 at 1:57

1

@BrockAdams - I wasn't sure whether calling zeroFill(-88, 5) should produce -00088 or -0088? I guess it depends upon whether you want the width argument to be the entire width of the number or just the number of digits (not including the negative sign). An implementer can easily switch the behavior to not include the negative sign by just removing the --width line of code.
– jfriend00
Jun 16 '13 at 3:07

@BrockAdams - fixed for both cases you mention. If the number is already as wide as the fill width, then no fill is added.
– jfriend00
Jun 16 '13 at 1:08

Thanks; +1. The negative numbers are slightly off from the usual convention, though. In your test case, -88 should yield "-00088", for example.
– Brock Adams
Jun 16 '13 at 1:57

@BrockAdams - I wasn't sure whether calling zeroFill(-88, 5) should produce -00088 or -0088? I guess it depends upon whether you want the width argument to be the entire width of the number or just the number of digits (not including the negative sign). An implementer can easily switch the behavior to not include the negative sign by just removing the --width line of code.
– jfriend00
Jun 16 '13 at 3:07

add a comment |

up vote
18
down vote

In a proposed (stage 3) ES2017 method .padStart() you can simply now do (when implemented/supported):

string.padStart(maxLength, "0"); //max length is the max string length, not max # of fills

edited May 16 '16 at 14:25

community wiki

2 revs
Sterling Archer

I've been looking for this one for some time now. Thank you for giving some visibility. I hope in 2030 we have it working.
– dinigo
Jun 20 '17 at 10:55

1

Its works now @dinigo ^^ it is live
– Paulo Roberto
Nov 22 '17 at 12:47

yep, I've used it a couple of times
– dinigo
Nov 22 '17 at 13:02

add a comment |

up vote
18
down vote

In a proposed (stage 3) ES2017 method .padStart() you can simply now do (when implemented/supported):

string.padStart(maxLength, "0"); //max length is the max string length, not max # of fills

edited May 16 '16 at 14:25

community wiki

2 revs
Sterling Archer

I've been looking for this one for some time now. Thank you for giving some visibility. I hope in 2030 we have it working.
– dinigo
Jun 20 '17 at 10:55

1

Its works now @dinigo ^^ it is live
– Paulo Roberto
Nov 22 '17 at 12:47

yep, I've used it a couple of times
– dinigo
Nov 22 '17 at 13:02

add a comment |

up vote
18
down vote

In a proposed (stage 3) ES2017 method .padStart() you can simply now do (when implemented/supported):

string.padStart(maxLength, "0"); //max length is the max string length, not max # of fills

edited May 16 '16 at 14:25

community wiki

2 revs
Sterling Archer

In a proposed (stage 3) ES2017 method .padStart() you can simply now do (when implemented/supported):

string.padStart(maxLength, "0"); //max length is the max string length, not max # of fills

edited May 16 '16 at 14:25

community wiki

2 revs
Sterling Archer

edited May 16 '16 at 14:25

community wiki

2 revs
Sterling Archer

community wiki

2 revs
Sterling Archer

community wiki

2 revs
Sterling Archer

I've been looking for this one for some time now. Thank you for giving some visibility. I hope in 2030 we have it working.
– dinigo
Jun 20 '17 at 10:55

1

Its works now @dinigo ^^ it is live
– Paulo Roberto
Nov 22 '17 at 12:47

yep, I've used it a couple of times
– dinigo
Nov 22 '17 at 13:02

add a comment |

I've been looking for this one for some time now. Thank you for giving some visibility. I hope in 2030 we have it working.
– dinigo
Jun 20 '17 at 10:55

1

Its works now @dinigo ^^ it is live
– Paulo Roberto
Nov 22 '17 at 12:47

yep, I've used it a couple of times
– dinigo
Nov 22 '17 at 13:02

I've been looking for this one for some time now. Thank you for giving some visibility. I hope in 2030 we have it working.
– dinigo
Jun 20 '17 at 10:55

Its works now @dinigo ^^ it is live
– Paulo Roberto
Nov 22 '17 at 12:47

yep, I've used it a couple of times
– dinigo
Nov 22 '17 at 13:02

add a comment |

up vote
17
down vote

The quick and dirty way:

y = (new Array(count + 1 - x.toString().length)).join('0') + x;

For x = 5 and count = 6 you'll have y = "000005"

edited May 24 '12 at 22:47

community wiki

Johann Philipp Strathausen

1

I got y="0005" with the above, y = (new Array(count + 1 - x.toString().length)).join('0') + x; is what gave me y="000005"
– forforf
May 23 '12 at 15:30

@forforf thanks, corrected it now!
– Johann Philipp Strathausen
May 24 '12 at 22:48

Thanks. I always like the shortest code solution.
– Orr Siloni
Nov 12 '12 at 14:08

3

@OrrSiloni: the shortest code solution is actually ('0000'+n).slice(-4)
– Seaux
Dec 8 '13 at 23:17

add a comment |

up vote
17
down vote

The quick and dirty way:

y = (new Array(count + 1 - x.toString().length)).join('0') + x;

For x = 5 and count = 6 you'll have y = "000005"

edited May 24 '12 at 22:47

community wiki

Johann Philipp Strathausen

1

I got y="0005" with the above, y = (new Array(count + 1 - x.toString().length)).join('0') + x; is what gave me y="000005"
– forforf
May 23 '12 at 15:30

@forforf thanks, corrected it now!
– Johann Philipp Strathausen
May 24 '12 at 22:48

Thanks. I always like the shortest code solution.
– Orr Siloni
Nov 12 '12 at 14:08

3

@OrrSiloni: the shortest code solution is actually ('0000'+n).slice(-4)
– Seaux
Dec 8 '13 at 23:17

add a comment |

up vote
17
down vote

The quick and dirty way:

y = (new Array(count + 1 - x.toString().length)).join('0') + x;

For x = 5 and count = 6 you'll have y = "000005"

edited May 24 '12 at 22:47

community wiki

Johann Philipp Strathausen

The quick and dirty way:

y = (new Array(count + 1 - x.toString().length)).join('0') + x;

For x = 5 and count = 6 you'll have y = "000005"

edited May 24 '12 at 22:47

community wiki

Johann Philipp Strathausen

edited May 24 '12 at 22:47

community wiki

Johann Philipp Strathausen

community wiki

Johann Philipp Strathausen

community wiki

Johann Philipp Strathausen

1

I got y="0005" with the above, y = (new Array(count + 1 - x.toString().length)).join('0') + x; is what gave me y="000005"
– forforf
May 23 '12 at 15:30

@forforf thanks, corrected it now!
– Johann Philipp Strathausen
May 24 '12 at 22:48

Thanks. I always like the shortest code solution.
– Orr Siloni
Nov 12 '12 at 14:08

3

@OrrSiloni: the shortest code solution is actually ('0000'+n).slice(-4)
– Seaux
Dec 8 '13 at 23:17

add a comment |

1

I got y="0005" with the above, y = (new Array(count + 1 - x.toString().length)).join('0') + x; is what gave me y="000005"
– forforf
May 23 '12 at 15:30

@forforf thanks, corrected it now!
– Johann Philipp Strathausen
May 24 '12 at 22:48

Thanks. I always like the shortest code solution.
– Orr Siloni
Nov 12 '12 at 14:08

3

@OrrSiloni: the shortest code solution is actually ('0000'+n).slice(-4)
– Seaux
Dec 8 '13 at 23:17

I got y="0005" with the above, y = (new Array(count + 1 - x.toString().length)).join('0') + x; is what gave me y="000005"
– forforf
May 23 '12 at 15:30

@forforf thanks, corrected it now!
– Johann Philipp Strathausen
May 24 '12 at 22:48

Thanks. I always like the shortest code solution.
– Orr Siloni
Nov 12 '12 at 14:08

@OrrSiloni: the shortest code solution is actually ('0000'+n).slice(-4)
– Seaux
Dec 8 '13 at 23:17

add a comment |

up vote
12
down vote

Here's a quick function I came up with to do the job. If anyone has a simpler approach, feel free to share!

function zerofill(number, length) {

    // Setup

    var result = number.toString();

    var pad = length - result.length;



    while(pad > 0) {

        result = '0' + result;

        pad--;

    }



    return result;

}

answered Aug 12 '09 at 16:54

community wiki

Wilco

I don't think there will be a "simpler" solution - it will always be a function of some sort. We could have a algorithm discussion, but at the end of the day, you'll still end up with a function that zerofills a number.
– Peter Bailey
Aug 12 '09 at 17:54

My general advice is to use "for" loops as they are generally more stable and faster in ECMAScript languages (ActionScript 1-3, and JavaScript)
– cwallenpoole
Aug 12 '09 at 18:03

Or, skip iteration altogether ;)
– Peter Bailey
Aug 12 '09 at 20:05

Simpler function? Probably not. One that doesn't loop? That is possible... though the algorithm isn't entirely trivial.
– coderjoe
Aug 12 '09 at 20:09

1

Wow, pretty much all the above comments are incorrect. @profitehlolz's solution is simpler and suitable for inlining (doesn't need to be a function). Meanwhile, for .vs. while performance is not different enough to be interesting: jsperf.com/fors-vs-while/34
– broofa
Sep 24 '12 at 12:07

|
show 3 more comments

up vote
12
down vote

Here's a quick function I came up with to do the job. If anyone has a simpler approach, feel free to share!

function zerofill(number, length) {

    // Setup

    var result = number.toString();

    var pad = length - result.length;



    while(pad > 0) {

        result = '0' + result;

        pad--;

    }



    return result;

}

answered Aug 12 '09 at 16:54

community wiki

Wilco

I don't think there will be a "simpler" solution - it will always be a function of some sort. We could have a algorithm discussion, but at the end of the day, you'll still end up with a function that zerofills a number.
– Peter Bailey
Aug 12 '09 at 17:54

My general advice is to use "for" loops as they are generally more stable and faster in ECMAScript languages (ActionScript 1-3, and JavaScript)
– cwallenpoole
Aug 12 '09 at 18:03

Or, skip iteration altogether ;)
– Peter Bailey
Aug 12 '09 at 20:05

Simpler function? Probably not. One that doesn't loop? That is possible... though the algorithm isn't entirely trivial.
– coderjoe
Aug 12 '09 at 20:09

1

Wow, pretty much all the above comments are incorrect. @profitehlolz's solution is simpler and suitable for inlining (doesn't need to be a function). Meanwhile, for .vs. while performance is not different enough to be interesting: jsperf.com/fors-vs-while/34
– broofa
Sep 24 '12 at 12:07

|
show 3 more comments

up vote
12
down vote

Here's a quick function I came up with to do the job. If anyone has a simpler approach, feel free to share!

function zerofill(number, length) {

    // Setup

    var result = number.toString();

    var pad = length - result.length;



    while(pad > 0) {

        result = '0' + result;

        pad--;

    }



    return result;

}

answered Aug 12 '09 at 16:54

community wiki

Wilco

Here's a quick function I came up with to do the job. If anyone has a simpler approach, feel free to share!

function zerofill(number, length) {

    // Setup

    var result = number.toString();

    var pad = length - result.length;



    while(pad > 0) {

        result = '0' + result;

        pad--;

    }



    return result;

}

answered Aug 12 '09 at 16:54

community wiki

Wilco

answered Aug 12 '09 at 16:54

community wiki

Wilco

community wiki

Wilco

community wiki

Wilco

I don't think there will be a "simpler" solution - it will always be a function of some sort. We could have a algorithm discussion, but at the end of the day, you'll still end up with a function that zerofills a number.
– Peter Bailey
Aug 12 '09 at 17:54

My general advice is to use "for" loops as they are generally more stable and faster in ECMAScript languages (ActionScript 1-3, and JavaScript)
– cwallenpoole
Aug 12 '09 at 18:03

Or, skip iteration altogether ;)
– Peter Bailey
Aug 12 '09 at 20:05

Simpler function? Probably not. One that doesn't loop? That is possible... though the algorithm isn't entirely trivial.
– coderjoe
Aug 12 '09 at 20:09

1

Wow, pretty much all the above comments are incorrect. @profitehlolz's solution is simpler and suitable for inlining (doesn't need to be a function). Meanwhile, for .vs. while performance is not different enough to be interesting: jsperf.com/fors-vs-while/34
– broofa
Sep 24 '12 at 12:07

|
show 3 more comments

I don't think there will be a "simpler" solution - it will always be a function of some sort. We could have a algorithm discussion, but at the end of the day, you'll still end up with a function that zerofills a number.
– Peter Bailey
Aug 12 '09 at 17:54

My general advice is to use "for" loops as they are generally more stable and faster in ECMAScript languages (ActionScript 1-3, and JavaScript)
– cwallenpoole
Aug 12 '09 at 18:03

Or, skip iteration altogether ;)
– Peter Bailey
Aug 12 '09 at 20:05

Simpler function? Probably not. One that doesn't loop? That is possible... though the algorithm isn't entirely trivial.
– coderjoe
Aug 12 '09 at 20:09

1

Wow, pretty much all the above comments are incorrect. @profitehlolz's solution is simpler and suitable for inlining (doesn't need to be a function). Meanwhile, for .vs. while performance is not different enough to be interesting: jsperf.com/fors-vs-while/34
– broofa
Sep 24 '12 at 12:07

I don't think there will be a "simpler" solution - it will always be a function of some sort. We could have a algorithm discussion, but at the end of the day, you'll still end up with a function that zerofills a number.
– Peter Bailey
Aug 12 '09 at 17:54

My general advice is to use "for" loops as they are generally more stable and faster in ECMAScript languages (ActionScript 1-3, and JavaScript)
– cwallenpoole
Aug 12 '09 at 18:03

Or, skip iteration altogether ;)
– Peter Bailey
Aug 12 '09 at 20:05

Simpler function? Probably not. One that doesn't loop? That is possible... though the algorithm isn't entirely trivial.
– coderjoe
Aug 12 '09 at 20:09

Wow, pretty much all the above comments are incorrect. @profitehlolz's solution is simpler and suitable for inlining (doesn't need to be a function). Meanwhile, for .vs. while performance is not different enough to be interesting: jsperf.com/fors-vs-while/34
– broofa
Sep 24 '12 at 12:07

|
show 3 more comments

up vote
11
down vote

Late to the party here, but I often use this construct for doing ad-hoc padding of some value n, known to be a positive, decimal:

(offset + n + '').substr(1);

Where offset is 10^^digits.

E.g. Padding to 5 digits, where n = 123:

(1e5 + 123 + '').substr(1); // => 00123

The hexidecimal version of this is slightly more verbose:

(0x100000 + 0x123).toString(16).substr(1); // => 00123

Note 1: I like @profitehlolz's solution as well, which is the string version of this, using slice()'s nifty negative-index feature.

answered Sep 24 '12 at 12:25

community wiki

broofa

Very elegant and neat!
– Marc
Feb 13 '13 at 9:42

add a comment |

up vote
11
down vote

Late to the party here, but I often use this construct for doing ad-hoc padding of some value n, known to be a positive, decimal:

(offset + n + '').substr(1);

Where offset is 10^^digits.

E.g. Padding to 5 digits, where n = 123:

(1e5 + 123 + '').substr(1); // => 00123

The hexidecimal version of this is slightly more verbose:

(0x100000 + 0x123).toString(16).substr(1); // => 00123

Note 1: I like @profitehlolz's solution as well, which is the string version of this, using slice()'s nifty negative-index feature.

answered Sep 24 '12 at 12:25

community wiki

broofa

Very elegant and neat!
– Marc
Feb 13 '13 at 9:42

add a comment |

up vote
11
down vote

Late to the party here, but I often use this construct for doing ad-hoc padding of some value n, known to be a positive, decimal:

(offset + n + '').substr(1);

Where offset is 10^^digits.

E.g. Padding to 5 digits, where n = 123:

(1e5 + 123 + '').substr(1); // => 00123

The hexidecimal version of this is slightly more verbose:

(0x100000 + 0x123).toString(16).substr(1); // => 00123

Note 1: I like @profitehlolz's solution as well, which is the string version of this, using slice()'s nifty negative-index feature.

answered Sep 24 '12 at 12:25

community wiki

broofa

Late to the party here, but I often use this construct for doing ad-hoc padding of some value n, known to be a positive, decimal:

(offset + n + '').substr(1);

Where offset is 10^^digits.

E.g. Padding to 5 digits, where n = 123:

(1e5 + 123 + '').substr(1); // => 00123

The hexidecimal version of this is slightly more verbose:

(0x100000 + 0x123).toString(16).substr(1); // => 00123

Note 1: I like @profitehlolz's solution as well, which is the string version of this, using slice()'s nifty negative-index feature.

answered Sep 24 '12 at 12:25

community wiki

broofa

answered Sep 24 '12 at 12:25

community wiki

broofa

community wiki

broofa

community wiki

broofa

Very elegant and neat!
– Marc
Feb 13 '13 at 9:42

add a comment |

Very elegant and neat!
– Marc
Feb 13 '13 at 9:42

Very elegant and neat!
– Marc
Feb 13 '13 at 9:42

add a comment |

up vote
11
down vote

I really don't know why, but no one did it in the most obvious way. Here it's my implementation.

Function:

/** Pad a number with 0 on the left */

function zeroPad(number, digits) {

    var num = number+"";

    while(num.length < digits){

        num='0'+num;

    }

    return num;

}

Prototype:

Number.prototype.zeroPad=function(digits){

    var num=this+"";

    while(num.length < digits){

        num='0'+num;

    }

    return(num);

};

answered Feb 5 '13 at 14:45

community wiki

Vitim.us

add a comment |

up vote
11
down vote

I really don't know why, but no one did it in the most obvious way. Here it's my implementation.

Function:

/** Pad a number with 0 on the left */

function zeroPad(number, digits) {

    var num = number+"";

    while(num.length < digits){

        num='0'+num;

    }

    return num;

}

Prototype:

Number.prototype.zeroPad=function(digits){

    var num=this+"";

    while(num.length < digits){

        num='0'+num;

    }

    return(num);

};

answered Feb 5 '13 at 14:45

community wiki

Vitim.us

add a comment |

up vote
11
down vote

I really don't know why, but no one did it in the most obvious way. Here it's my implementation.

Function:

/** Pad a number with 0 on the left */

function zeroPad(number, digits) {

    var num = number+"";

    while(num.length < digits){

        num='0'+num;

    }

    return num;

}

Prototype:

Number.prototype.zeroPad=function(digits){

    var num=this+"";

    while(num.length < digits){

        num='0'+num;

    }

    return(num);

};

answered Feb 5 '13 at 14:45

community wiki

Vitim.us

I really don't know why, but no one did it in the most obvious way. Here it's my implementation.

Function:

/** Pad a number with 0 on the left */

function zeroPad(number, digits) {

    var num = number+"";

    while(num.length < digits){

        num='0'+num;

    }

    return num;

}

Prototype:

Number.prototype.zeroPad=function(digits){

    var num=this+"";

    while(num.length < digits){

        num='0'+num;

    }

    return(num);

};

answered Feb 5 '13 at 14:45

community wiki

Vitim.us

answered Feb 5 '13 at 14:45

community wiki

Vitim.us

community wiki

Vitim.us

community wiki

Vitim.us

add a comment |

up vote
9
down vote

I use this snipet to get a 5 digits representation

(value+100000).toString().slice(-5) // "00123" with value=123

answered Sep 3 '13 at 11:08

community wiki

jasto salto

this is a smart way to add leading zero . I have done something similar to add leading zero for fixed length binary integer
– Chris.Huang
May 11 '15 at 2:14

add a comment |

up vote
9
down vote

I use this snipet to get a 5 digits representation

(value+100000).toString().slice(-5) // "00123" with value=123

answered Sep 3 '13 at 11:08

community wiki

jasto salto

this is a smart way to add leading zero . I have done something similar to add leading zero for fixed length binary integer
– Chris.Huang
May 11 '15 at 2:14

add a comment |

up vote
9
down vote

I use this snipet to get a 5 digits representation

(value+100000).toString().slice(-5) // "00123" with value=123

answered Sep 3 '13 at 11:08

community wiki

jasto salto

I use this snipet to get a 5 digits representation

(value+100000).toString().slice(-5) // "00123" with value=123

answered Sep 3 '13 at 11:08

community wiki

jasto salto

answered Sep 3 '13 at 11:08

community wiki

jasto salto

community wiki

jasto salto

community wiki

jasto salto

this is a smart way to add leading zero . I have done something similar to add leading zero for fixed length binary integer
– Chris.Huang
May 11 '15 at 2:14

add a comment |

this is a smart way to add leading zero . I have done something similar to add leading zero for fixed length binary integer
– Chris.Huang
May 11 '15 at 2:14

this is a smart way to add leading zero . I have done something similar to add leading zero for fixed length binary integer
– Chris.Huang
May 11 '15 at 2:14

add a comment |

up vote
8
down vote

The power of Math!

x = integer to pad

y = number of zeroes to pad

function zeroPad(x, y)

{

   y = Math.max(y-1,0);

   var n = (x / Math.pow(10,y)).toFixed(y);

   return n.replace('.','');  

}

answered Mar 21 '13 at 1:40

community wiki

tir

Wow, that is a very clever way to solve this problem. I already had my own solution for this that is exactly functionally equivalent, and even relatively short, but this is concise and probably quite fast.
– pseudosavant
Jan 9 '14 at 21:10

add a comment |

up vote
8
down vote

The power of Math!

x = integer to pad

y = number of zeroes to pad

function zeroPad(x, y)

{

   y = Math.max(y-1,0);

   var n = (x / Math.pow(10,y)).toFixed(y);

   return n.replace('.','');  

}

answered Mar 21 '13 at 1:40

community wiki

tir

Wow, that is a very clever way to solve this problem. I already had my own solution for this that is exactly functionally equivalent, and even relatively short, but this is concise and probably quite fast.
– pseudosavant
Jan 9 '14 at 21:10

add a comment |

up vote
8
down vote

The power of Math!

x = integer to pad

y = number of zeroes to pad

function zeroPad(x, y)

{

   y = Math.max(y-1,0);

   var n = (x / Math.pow(10,y)).toFixed(y);

   return n.replace('.','');  

}

answered Mar 21 '13 at 1:40

community wiki

tir

The power of Math!

x = integer to pad

y = number of zeroes to pad

function zeroPad(x, y)

{

   y = Math.max(y-1,0);

   var n = (x / Math.pow(10,y)).toFixed(y);

   return n.replace('.','');  

}

answered Mar 21 '13 at 1:40

community wiki

tir

answered Mar 21 '13 at 1:40

community wiki

tir

community wiki

tir

community wiki

tir

Wow, that is a very clever way to solve this problem. I already had my own solution for this that is exactly functionally equivalent, and even relatively short, but this is concise and probably quite fast.
– pseudosavant
Jan 9 '14 at 21:10

add a comment |

Wow, that is a very clever way to solve this problem. I already had my own solution for this that is exactly functionally equivalent, and even relatively short, but this is concise and probably quite fast.
– pseudosavant
Jan 9 '14 at 21:10

Wow, that is a very clever way to solve this problem. I already had my own solution for this that is exactly functionally equivalent, and even relatively short, but this is concise and probably quite fast.
– pseudosavant
Jan 9 '14 at 21:10

add a comment |

up vote
7
down vote

function zPad(n, l, r){

    return(a=String(n).match(/(^-?)(d*).?(d*)/))?a[1]+(Array(l).join(0)+a[2]).slice(-Math.max(l,a[2].length))+('undefined'!==typeof r?(0<r?'.':'')+(a[3]+Array(r+1).join(0)).slice(0,r):a[3]?'.'+a[3]:''):0

}

           zPad(6, 2) === '06'

          zPad(-6, 2) === '-06'

       zPad(600.2, 2) === '600.2'

        zPad(-600, 2) === '-600'

         zPad(6.2, 3) === '006.2'

        zPad(-6.2, 3) === '-006.2'

      zPad(6.2, 3, 0) === '006'

        zPad(6, 2, 3) === '06.000'

    zPad(600.2, 2, 3) === '600.200'

zPad(-600.1499, 2, 3) === '-600.149'

edited Feb 4 '14 at 22:52

community wiki

4 revs
Jack Allan

add a comment |

up vote
7
down vote

function zPad(n, l, r){

    return(a=String(n).match(/(^-?)(d*).?(d*)/))?a[1]+(Array(l).join(0)+a[2]).slice(-Math.max(l,a[2].length))+('undefined'!==typeof r?(0<r?'.':'')+(a[3]+Array(r+1).join(0)).slice(0,r):a[3]?'.'+a[3]:''):0

}

           zPad(6, 2) === '06'

          zPad(-6, 2) === '-06'

       zPad(600.2, 2) === '600.2'

        zPad(-600, 2) === '-600'

         zPad(6.2, 3) === '006.2'

        zPad(-6.2, 3) === '-006.2'

      zPad(6.2, 3, 0) === '006'

        zPad(6, 2, 3) === '06.000'

    zPad(600.2, 2, 3) === '600.200'

zPad(-600.1499, 2, 3) === '-600.149'

edited Feb 4 '14 at 22:52

community wiki

4 revs
Jack Allan

add a comment |

up vote
7
down vote

function zPad(n, l, r){

    return(a=String(n).match(/(^-?)(d*).?(d*)/))?a[1]+(Array(l).join(0)+a[2]).slice(-Math.max(l,a[2].length))+('undefined'!==typeof r?(0<r?'.':'')+(a[3]+Array(r+1).join(0)).slice(0,r):a[3]?'.'+a[3]:''):0

}

           zPad(6, 2) === '06'

          zPad(-6, 2) === '-06'

       zPad(600.2, 2) === '600.2'

        zPad(-600, 2) === '-600'

         zPad(6.2, 3) === '006.2'

        zPad(-6.2, 3) === '-006.2'

      zPad(6.2, 3, 0) === '006'

        zPad(6, 2, 3) === '06.000'

    zPad(600.2, 2, 3) === '600.200'

zPad(-600.1499, 2, 3) === '-600.149'

edited Feb 4 '14 at 22:52

community wiki

4 revs
Jack Allan

function zPad(n, l, r){

    return(a=String(n).match(/(^-?)(d*).?(d*)/))?a[1]+(Array(l).join(0)+a[2]).slice(-Math.max(l,a[2].length))+('undefined'!==typeof r?(0<r?'.':'')+(a[3]+Array(r+1).join(0)).slice(0,r):a[3]?'.'+a[3]:''):0

}

           zPad(6, 2) === '06'

          zPad(-6, 2) === '-06'

       zPad(600.2, 2) === '600.2'

        zPad(-600, 2) === '-600'

         zPad(6.2, 3) === '006.2'

        zPad(-6.2, 3) === '-006.2'

      zPad(6.2, 3, 0) === '006'

        zPad(6, 2, 3) === '06.000'

    zPad(600.2, 2, 3) === '600.200'

zPad(-600.1499, 2, 3) === '-600.149'

edited Feb 4 '14 at 22:52

community wiki

4 revs
Jack Allan

edited Feb 4 '14 at 22:52

community wiki

4 revs
Jack Allan

community wiki

4 revs
Jack Allan

community wiki

4 revs
Jack Allan

add a comment |

up vote
7
down vote

Don't reinvent the wheel, use underscore string:

jsFiddle

var numToPad = '5';



alert(_.str.pad(numToPad, 6, '0')); // yields: '000005'

answered Sep 5 '14 at 12:34

community wiki

Benny Bottema

1

Pure JavaScript! Short and simple, and a jsFiddle example! EXCELLENT! :D
– tfont
Oct 10 '14 at 23:41

add a comment |

up vote
7
down vote

Don't reinvent the wheel, use underscore string:

jsFiddle

var numToPad = '5';



alert(_.str.pad(numToPad, 6, '0')); // yields: '000005'

answered Sep 5 '14 at 12:34

community wiki

Benny Bottema

1

Pure JavaScript! Short and simple, and a jsFiddle example! EXCELLENT! :D
– tfont
Oct 10 '14 at 23:41

add a comment |

up vote
7
down vote

Don't reinvent the wheel, use underscore string:

jsFiddle

var numToPad = '5';



alert(_.str.pad(numToPad, 6, '0')); // yields: '000005'

answered Sep 5 '14 at 12:34

community wiki

Benny Bottema

Don't reinvent the wheel, use underscore string:

jsFiddle

var numToPad = '5';



alert(_.str.pad(numToPad, 6, '0')); // yields: '000005'

answered Sep 5 '14 at 12:34

community wiki

Benny Bottema

answered Sep 5 '14 at 12:34

community wiki

Benny Bottema

community wiki

Benny Bottema

community wiki

Benny Bottema

1

Pure JavaScript! Short and simple, and a jsFiddle example! EXCELLENT! :D
– tfont
Oct 10 '14 at 23:41

add a comment |

1

Pure JavaScript! Short and simple, and a jsFiddle example! EXCELLENT! :D
– tfont
Oct 10 '14 at 23:41

Pure JavaScript! Short and simple, and a jsFiddle example! EXCELLENT! :D
– tfont
Oct 10 '14 at 23:41

add a comment |

up vote
7
down vote

This is the ES6 solution.

function pad(num, len) {

  return '0'.repeat(len - num.toString().length) + num;

}

alert(pad(1234,6));

answered May 22 '15 at 3:38

community wiki

Lewis

Clearly the most elegant solution I would just modify it to avoid 2 string conversions: const numStr = String(num) and return '0'.repeat(len - numStr.length) + numStr;
– ngryman
Sep 26 '16 at 21:52

add a comment |

up vote
7
down vote

This is the ES6 solution.

function pad(num, len) {

  return '0'.repeat(len - num.toString().length) + num;

}

alert(pad(1234,6));

answered May 22 '15 at 3:38

community wiki

Lewis

Clearly the most elegant solution I would just modify it to avoid 2 string conversions: const numStr = String(num) and return '0'.repeat(len - numStr.length) + numStr;
– ngryman
Sep 26 '16 at 21:52

add a comment |

up vote
7
down vote

This is the ES6 solution.

function pad(num, len) {

  return '0'.repeat(len - num.toString().length) + num;

}

alert(pad(1234,6));

answered May 22 '15 at 3:38

community wiki

Lewis

This is the ES6 solution.

function pad(num, len) {

  return '0'.repeat(len - num.toString().length) + num;

}

alert(pad(1234,6));

function pad(num, len) {

  return '0'.repeat(len - num.toString().length) + num;

}

alert(pad(1234,6));

function pad(num, len) {

  return '0'.repeat(len - num.toString().length) + num;

}

alert(pad(1234,6));

answered May 22 '15 at 3:38

community wiki

Lewis

answered May 22 '15 at 3:38

community wiki

Lewis

community wiki

Lewis

community wiki

Lewis

Clearly the most elegant solution I would just modify it to avoid 2 string conversions: const numStr = String(num) and return '0'.repeat(len - numStr.length) + numStr;
– ngryman
Sep 26 '16 at 21:52

add a comment |

Clearly the most elegant solution I would just modify it to avoid 2 string conversions: const numStr = String(num) and return '0'.repeat(len - numStr.length) + numStr;
– ngryman
Sep 26 '16 at 21:52

Clearly the most elegant solution I would just modify it to avoid 2 string conversions: const numStr = String(num) and return '0'.repeat(len - numStr.length) + numStr;
– ngryman
Sep 26 '16 at 21:52

add a comment |

up vote
7
down vote

I didn't see anyone point out the fact that when you use String.prototype.substr() with a negative number it counts from the right.

A one liner solution to the OP's question, a 6-digit zerofilled representation of the number 5, is:

console.log(("00000000" + 5).substr(-6));

Generalizing we'll get:

function pad(num, len) { return ("00000000" + num).substr(-len) };



console.log(pad(5, 6));

console.log(pad(45, 6));

console.log(pad(345, 6));

console.log(pad(2345, 6));

console.log(pad(12345, 6));

edited Sep 21 '17 at 22:16

community wiki

2 revs
Stephen Quan

nice one liner solution
– rave
Sep 21 '17 at 21:06

add a comment |

up vote
7
down vote

I didn't see anyone point out the fact that when you use String.prototype.substr() with a negative number it counts from the right.

A one liner solution to the OP's question, a 6-digit zerofilled representation of the number 5, is:

console.log(("00000000" + 5).substr(-6));

Generalizing we'll get:

function pad(num, len) { return ("00000000" + num).substr(-len) };



console.log(pad(5, 6));

console.log(pad(45, 6));

console.log(pad(345, 6));

console.log(pad(2345, 6));

console.log(pad(12345, 6));

edited Sep 21 '17 at 22:16

community wiki

2 revs
Stephen Quan

nice one liner solution
– rave
Sep 21 '17 at 21:06

add a comment |

up vote
7
down vote

I didn't see anyone point out the fact that when you use String.prototype.substr() with a negative number it counts from the right.

A one liner solution to the OP's question, a 6-digit zerofilled representation of the number 5, is:

console.log(("00000000" + 5).substr(-6));

Generalizing we'll get:

function pad(num, len) { return ("00000000" + num).substr(-len) };



console.log(pad(5, 6));

console.log(pad(45, 6));

console.log(pad(345, 6));

console.log(pad(2345, 6));

console.log(pad(12345, 6));

edited Sep 21 '17 at 22:16

community wiki

2 revs
Stephen Quan

I didn't see anyone point out the fact that when you use String.prototype.substr() with a negative number it counts from the right.

A one liner solution to the OP's question, a 6-digit zerofilled representation of the number 5, is:

console.log(("00000000" + 5).substr(-6));

Generalizing we'll get:

function pad(num, len) { return ("00000000" + num).substr(-len) };



console.log(pad(5, 6));

console.log(pad(45, 6));

console.log(pad(345, 6));

console.log(pad(2345, 6));

console.log(pad(12345, 6));

console.log(("00000000" + 5).substr(-6));

console.log(("00000000" + 5).substr(-6));

function pad(num, len) { return ("00000000" + num).substr(-len) };



console.log(pad(5, 6));

console.log(pad(45, 6));

console.log(pad(345, 6));

console.log(pad(2345, 6));

console.log(pad(12345, 6));

function pad(num, len) { return ("00000000" + num).substr(-len) };



console.log(pad(5, 6));

console.log(pad(45, 6));

console.log(pad(345, 6));

console.log(pad(2345, 6));

console.log(pad(12345, 6));

edited Sep 21 '17 at 22:16

community wiki

2 revs
Stephen Quan

edited Sep 21 '17 at 22:16

community wiki

2 revs
Stephen Quan

community wiki

2 revs
Stephen Quan

community wiki

2 revs
Stephen Quan

nice one liner solution
– rave
Sep 21 '17 at 21:06

add a comment |

nice one liner solution
– rave
Sep 21 '17 at 21:06

nice one liner solution
– rave
Sep 21 '17 at 21:06

add a comment |

up vote
6
down vote

After a, long, long time of testing 15 different functions/methods found in this questions answers, I now know which is the best (the most versatile and quickest).

The code I used can be found here:
https://gist.github.com/NextToNothing/6325915

Feel free to modify and test the code yourself.

In order to get the most versatile method, you have to use a loop. This is because with very large numbers others are likely to fail, whereas, this will succeed.

So, which loop to use? Well, that would be a while loop. A for loop is still fast, but a while loop is just slightly quicker(a couple of ms) - and cleaner.

Answers like those by Wilco, Aleksandar Toplek or Vitim.us will do the job perfectly.

My function is:

function pad(str, max, padder) {

  padder = typeof padder === "undefined" ? "0" : padder;

  return str.toString().length < max ? pad(padder.toString() + str, max, padder) : str;

}

You can use my function with, or without, setting the padding variable. So like this:

pad(1, 3); // Returns '001'

// - Or -

pad(1, 3, "x"); // Returns 'xx1'

Personally, after my tests, I would use a method with a while loop, like Aleksandar Toplek or Vitim.us. However, I would modify it slightly so that you are able to set the padding string.

So, I would use this code:

function padLeft(str, len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    str = str + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

padLeft(1, 3); // Returns '001'

// - Or -

padLeft(1, 3, "x"); // Returns 'xx1'

You could also use it as a prototype function, by using this code:

Number.prototype.padLeft = function(len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    var str = this + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

var num = 1;



num.padLeft(3); // Returns '001'

// - Or -

num.padLeft(3, "x"); // Returns 'xx1'

answered Aug 24 '13 at 4:12

community wiki

Jack B

I put this in a jsfiddle to make it quick for others to test, adding your version of the code: jsfiddle.net/kevinmicke/vnvghw7y/2 Your version is always very competitive, and sometimes the fastest. Thanks for the fairly exhaustive testing.
– kevinmicke
Oct 5 '15 at 19:17

add a comment |

up vote
6
down vote

After a, long, long time of testing 15 different functions/methods found in this questions answers, I now know which is the best (the most versatile and quickest).

The code I used can be found here:
https://gist.github.com/NextToNothing/6325915

Feel free to modify and test the code yourself.

In order to get the most versatile method, you have to use a loop. This is because with very large numbers others are likely to fail, whereas, this will succeed.

So, which loop to use? Well, that would be a while loop. A for loop is still fast, but a while loop is just slightly quicker(a couple of ms) - and cleaner.

Answers like those by Wilco, Aleksandar Toplek or Vitim.us will do the job perfectly.

My function is:

function pad(str, max, padder) {

  padder = typeof padder === "undefined" ? "0" : padder;

  return str.toString().length < max ? pad(padder.toString() + str, max, padder) : str;

}

You can use my function with, or without, setting the padding variable. So like this:

pad(1, 3); // Returns '001'

// - Or -

pad(1, 3, "x"); // Returns 'xx1'

Personally, after my tests, I would use a method with a while loop, like Aleksandar Toplek or Vitim.us. However, I would modify it slightly so that you are able to set the padding string.

So, I would use this code:

function padLeft(str, len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    str = str + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

padLeft(1, 3); // Returns '001'

// - Or -

padLeft(1, 3, "x"); // Returns 'xx1'

You could also use it as a prototype function, by using this code:

Number.prototype.padLeft = function(len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    var str = this + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

var num = 1;



num.padLeft(3); // Returns '001'

// - Or -

num.padLeft(3, "x"); // Returns 'xx1'

answered Aug 24 '13 at 4:12

community wiki

Jack B

I put this in a jsfiddle to make it quick for others to test, adding your version of the code: jsfiddle.net/kevinmicke/vnvghw7y/2 Your version is always very competitive, and sometimes the fastest. Thanks for the fairly exhaustive testing.
– kevinmicke
Oct 5 '15 at 19:17

add a comment |

up vote
6
down vote

After a, long, long time of testing 15 different functions/methods found in this questions answers, I now know which is the best (the most versatile and quickest).

The code I used can be found here:
https://gist.github.com/NextToNothing/6325915

Feel free to modify and test the code yourself.

In order to get the most versatile method, you have to use a loop. This is because with very large numbers others are likely to fail, whereas, this will succeed.

So, which loop to use? Well, that would be a while loop. A for loop is still fast, but a while loop is just slightly quicker(a couple of ms) - and cleaner.

Answers like those by Wilco, Aleksandar Toplek or Vitim.us will do the job perfectly.

My function is:

function pad(str, max, padder) {

  padder = typeof padder === "undefined" ? "0" : padder;

  return str.toString().length < max ? pad(padder.toString() + str, max, padder) : str;

}

You can use my function with, or without, setting the padding variable. So like this:

pad(1, 3); // Returns '001'

// - Or -

pad(1, 3, "x"); // Returns 'xx1'

Personally, after my tests, I would use a method with a while loop, like Aleksandar Toplek or Vitim.us. However, I would modify it slightly so that you are able to set the padding string.

So, I would use this code:

function padLeft(str, len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    str = str + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

padLeft(1, 3); // Returns '001'

// - Or -

padLeft(1, 3, "x"); // Returns 'xx1'

You could also use it as a prototype function, by using this code:

Number.prototype.padLeft = function(len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    var str = this + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

var num = 1;



num.padLeft(3); // Returns '001'

// - Or -

num.padLeft(3, "x"); // Returns 'xx1'

answered Aug 24 '13 at 4:12

community wiki

Jack B

After a, long, long time of testing 15 different functions/methods found in this questions answers, I now know which is the best (the most versatile and quickest).

The code I used can be found here:
https://gist.github.com/NextToNothing/6325915

Feel free to modify and test the code yourself.

In order to get the most versatile method, you have to use a loop. This is because with very large numbers others are likely to fail, whereas, this will succeed.

So, which loop to use? Well, that would be a while loop. A for loop is still fast, but a while loop is just slightly quicker(a couple of ms) - and cleaner.

Answers like those by Wilco, Aleksandar Toplek or Vitim.us will do the job perfectly.

My function is:

function pad(str, max, padder) {

  padder = typeof padder === "undefined" ? "0" : padder;

  return str.toString().length < max ? pad(padder.toString() + str, max, padder) : str;

}

You can use my function with, or without, setting the padding variable. So like this:

pad(1, 3); // Returns '001'

// - Or -

pad(1, 3, "x"); // Returns 'xx1'

Personally, after my tests, I would use a method with a while loop, like Aleksandar Toplek or Vitim.us. However, I would modify it slightly so that you are able to set the padding string.

So, I would use this code:

function padLeft(str, len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    str = str + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

padLeft(1, 3); // Returns '001'

// - Or -

padLeft(1, 3, "x"); // Returns 'xx1'

You could also use it as a prototype function, by using this code:

Number.prototype.padLeft = function(len, pad) {

    pad = typeof pad === "undefined" ? "0" : pad + "";

    var str = this + "";

    while(str.length < len) {

        str = pad + str;

    }

    return str;

}



// Usage

var num = 1;



num.padLeft(3); // Returns '001'

// - Or -

num.padLeft(3, "x"); // Returns 'xx1'

answered Aug 24 '13 at 4:12

community wiki

Jack B

answered Aug 24 '13 at 4:12

community wiki

Jack B

community wiki

Jack B

community wiki

Jack B

I put this in a jsfiddle to make it quick for others to test, adding your version of the code: jsfiddle.net/kevinmicke/vnvghw7y/2 Your version is always very competitive, and sometimes the fastest. Thanks for the fairly exhaustive testing.
– kevinmicke
Oct 5 '15 at 19:17

add a comment |

I put this in a jsfiddle to make it quick for others to test, adding your version of the code: jsfiddle.net/kevinmicke/vnvghw7y/2 Your version is always very competitive, and sometimes the fastest. Thanks for the fairly exhaustive testing.
– kevinmicke
Oct 5 '15 at 19:17

I put this in a jsfiddle to make it quick for others to test, adding your version of the code: jsfiddle.net/kevinmicke/vnvghw7y/2 Your version is always very competitive, and sometimes the fastest. Thanks for the fairly exhaustive testing.
– kevinmicke
Oct 5 '15 at 19:17

add a comment |

up vote
6
down vote

ECMAScript 2017:
use padStart or padEnd

'abc'.padStart(10);         // "       abc"

'abc'.padStart(10, "foo");  // "foofoofabc"

'abc'.padStart(6,"123465"); // "123abc"

More info:

https://github.com/tc39/proposal-string-pad-start-end

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Nov 28 '16 at 8:49

community wiki

juFo

Voted up because it works in my scenario and would be nice to have in the specs. (Although it doesn't really answer the question, as it is not about "0"s :-) )
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:28

add a comment |

up vote
6
down vote

ECMAScript 2017:
use padStart or padEnd

'abc'.padStart(10);         // "       abc"

'abc'.padStart(10, "foo");  // "foofoofabc"

'abc'.padStart(6,"123465"); // "123abc"

More info:

https://github.com/tc39/proposal-string-pad-start-end

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Nov 28 '16 at 8:49

community wiki

juFo

Voted up because it works in my scenario and would be nice to have in the specs. (Although it doesn't really answer the question, as it is not about "0"s :-) )
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:28

add a comment |

up vote
6
down vote

ECMAScript 2017:
use padStart or padEnd

'abc'.padStart(10);         // "       abc"

'abc'.padStart(10, "foo");  // "foofoofabc"

'abc'.padStart(6,"123465"); // "123abc"

More info:

https://github.com/tc39/proposal-string-pad-start-end

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Nov 28 '16 at 8:49

community wiki

juFo

ECMAScript 2017:
use padStart or padEnd

'abc'.padStart(10);         // "       abc"

'abc'.padStart(10, "foo");  // "foofoofabc"

'abc'.padStart(6,"123465"); // "123abc"

More info:

https://github.com/tc39/proposal-string-pad-start-end

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Nov 28 '16 at 8:49

community wiki

juFo

answered Nov 28 '16 at 8:49

community wiki

juFo

community wiki

juFo

community wiki

juFo

Voted up because it works in my scenario and would be nice to have in the specs. (Although it doesn't really answer the question, as it is not about "0"s :-) )
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:28

add a comment |

Voted up because it works in my scenario and would be nice to have in the specs. (Although it doesn't really answer the question, as it is not about "0"s :-) )
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:28

Voted up because it works in my scenario and would be nice to have in the specs. (Although it doesn't really answer the question, as it is not about "0"s :-) )
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:28

add a comment |

up vote
4
down vote

The latest way to do this is much simpler:

var number = 2

number.toLocaleString(undefined, {minimumIntegerDigits:2})

output: "02"

answered Feb 29 '16 at 4:00

community wiki

MrE

(8).toLocaleString(undefined, {minimumIntegerDigits: 5}) returns "00.008" for me, based on my locale.
– le_m
Mar 18 '17 at 21:58

add a comment |

up vote
4
down vote

The latest way to do this is much simpler:

var number = 2

number.toLocaleString(undefined, {minimumIntegerDigits:2})

output: "02"

answered Feb 29 '16 at 4:00

community wiki

MrE

(8).toLocaleString(undefined, {minimumIntegerDigits: 5}) returns "00.008" for me, based on my locale.
– le_m
Mar 18 '17 at 21:58

add a comment |

up vote
4
down vote

The latest way to do this is much simpler:

var number = 2

number.toLocaleString(undefined, {minimumIntegerDigits:2})

output: "02"

answered Feb 29 '16 at 4:00

community wiki

MrE

The latest way to do this is much simpler:

var number = 2

number.toLocaleString(undefined, {minimumIntegerDigits:2})

output: "02"

answered Feb 29 '16 at 4:00

community wiki

MrE

answered Feb 29 '16 at 4:00

community wiki

MrE

community wiki

MrE

community wiki

MrE

(8).toLocaleString(undefined, {minimumIntegerDigits: 5}) returns "00.008" for me, based on my locale.
– le_m
Mar 18 '17 at 21:58

add a comment |

(8).toLocaleString(undefined, {minimumIntegerDigits: 5}) returns "00.008" for me, based on my locale.
– le_m
Mar 18 '17 at 21:58

(8).toLocaleString(undefined, {minimumIntegerDigits: 5}) returns "00.008" for me, based on my locale.
– le_m
Mar 18 '17 at 21:58

add a comment |

up vote
4
down vote

Just an another solution, but I think it's more legible.

function zeroFill(text, size)

{

  while (text.length < size){

  	text = "0" + text;

  }

  

  return text;

}

edited Jul 22 '16 at 14:47

community wiki

3 revs
Arthur

Same as which other?
– Fabio Turati
Jul 22 '16 at 14:14

add a comment |

up vote
4
down vote

Just an another solution, but I think it's more legible.

function zeroFill(text, size)

{

  while (text.length < size){

  	text = "0" + text;

  }

  

  return text;

}

edited Jul 22 '16 at 14:47

community wiki

3 revs
Arthur

Same as which other?
– Fabio Turati
Jul 22 '16 at 14:14

add a comment |

up vote
4
down vote

Just an another solution, but I think it's more legible.

function zeroFill(text, size)

{

  while (text.length < size){

  	text = "0" + text;

  }

  

  return text;

}

edited Jul 22 '16 at 14:47

community wiki

3 revs
Arthur

Just an another solution, but I think it's more legible.

function zeroFill(text, size)

{

  while (text.length < size){

  	text = "0" + text;

  }

  

  return text;

}

function zeroFill(text, size)

{

  while (text.length < size){

  	text = "0" + text;

  }

  

  return text;

}

function zeroFill(text, size)

{

  while (text.length < size){

  	text = "0" + text;

  }

  

  return text;

}

edited Jul 22 '16 at 14:47

community wiki

3 revs
Arthur

edited Jul 22 '16 at 14:47

community wiki

3 revs
Arthur

community wiki

3 revs
Arthur

community wiki

3 revs
Arthur

Same as which other?
– Fabio Turati
Jul 22 '16 at 14:14

add a comment |

Same as which other?
– Fabio Turati
Jul 22 '16 at 14:14

Same as which other?
– Fabio Turati
Jul 22 '16 at 14:14

add a comment |

up vote
4
down vote

In all modern browsers you can use

numberStr.padStart(numberLength, "0");

function zeroFill(num, numLength) {

  var numberStr = num.toString();



  return numberStr.padStart(numLength, "0");

}



var numbers = [0, 1, 12, 123, 1234, 12345];



numbers.forEach(

  function(num) {

    var numString = num.toString();

    

    var paddedNum = zeroFill(numString, 5);



    console.log(paddedNum);

  }

);

Here is the MDN reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Aug 21 '17 at 22:07

community wiki

Lifehack

add a comment |

up vote
4
down vote

In all modern browsers you can use

numberStr.padStart(numberLength, "0");

function zeroFill(num, numLength) {

  var numberStr = num.toString();



  return numberStr.padStart(numLength, "0");

}



var numbers = [0, 1, 12, 123, 1234, 12345];



numbers.forEach(

  function(num) {

    var numString = num.toString();

    

    var paddedNum = zeroFill(numString, 5);



    console.log(paddedNum);

  }

);

Here is the MDN reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Aug 21 '17 at 22:07

community wiki

Lifehack

add a comment |

up vote
4
down vote

In all modern browsers you can use

numberStr.padStart(numberLength, "0");

function zeroFill(num, numLength) {

  var numberStr = num.toString();



  return numberStr.padStart(numLength, "0");

}



var numbers = [0, 1, 12, 123, 1234, 12345];



numbers.forEach(

  function(num) {

    var numString = num.toString();

    

    var paddedNum = zeroFill(numString, 5);



    console.log(paddedNum);

  }

);

Here is the MDN reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

answered Aug 21 '17 at 22:07

community wiki

Lifehack

In all modern browsers you can use

numberStr.padStart(numberLength, "0");

function zeroFill(num, numLength) {

  var numberStr = num.toString();



  return numberStr.padStart(numLength, "0");

}



var numbers = [0, 1, 12, 123, 1234, 12345];



numbers.forEach(

  function(num) {

    var numString = num.toString();

    

    var paddedNum = zeroFill(numString, 5);



    console.log(paddedNum);

  }

);

Here is the MDN reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

function zeroFill(num, numLength) {

  var numberStr = num.toString();



  return numberStr.padStart(numLength, "0");

}



var numbers = [0, 1, 12, 123, 1234, 12345];



numbers.forEach(

  function(num) {

    var numString = num.toString();

    

    var paddedNum = zeroFill(numString, 5);



    console.log(paddedNum);

  }

);

function zeroFill(num, numLength) {

  var numberStr = num.toString();



  return numberStr.padStart(numLength, "0");

}



var numbers = [0, 1, 12, 123, 1234, 12345];



numbers.forEach(

  function(num) {

    var numString = num.toString();

    

    var paddedNum = zeroFill(numString, 5);



    console.log(paddedNum);

  }

);

answered Aug 21 '17 at 22:07

community wiki

Lifehack

answered Aug 21 '17 at 22:07

community wiki

Lifehack

community wiki

Lifehack

community wiki

Lifehack

add a comment |

up vote
3
down vote

This one is less native, but may be the fastest...

zeroPad = function (num, count) {

    var pad = (num + '').length - count;

    while(--pad > -1) {

        num = '0' + num;

    }

    return num;

};

answered Sep 6 '09 at 23:05

community wiki

Jonathan Neal

i think you want to do pad = count - (num + '').length. negative numbers aren't handled well, but apart from that, it's not bad. +1
– nickf
Sep 6 '09 at 23:29

add a comment |

up vote
3
down vote

This one is less native, but may be the fastest...

zeroPad = function (num, count) {

    var pad = (num + '').length - count;

    while(--pad > -1) {

        num = '0' + num;

    }

    return num;

};

answered Sep 6 '09 at 23:05

community wiki

Jonathan Neal

i think you want to do pad = count - (num + '').length. negative numbers aren't handled well, but apart from that, it's not bad. +1
– nickf
Sep 6 '09 at 23:29

add a comment |

up vote
3
down vote

This one is less native, but may be the fastest...

zeroPad = function (num, count) {

    var pad = (num + '').length - count;

    while(--pad > -1) {

        num = '0' + num;

    }

    return num;

};

answered Sep 6 '09 at 23:05

community wiki

Jonathan Neal

This one is less native, but may be the fastest...

zeroPad = function (num, count) {

    var pad = (num + '').length - count;

    while(--pad > -1) {

        num = '0' + num;

    }

    return num;

};

answered Sep 6 '09 at 23:05

community wiki

Jonathan Neal

answered Sep 6 '09 at 23:05

community wiki

Jonathan Neal

community wiki

Jonathan Neal

community wiki

Jonathan Neal

i think you want to do pad = count - (num + '').length. negative numbers aren't handled well, but apart from that, it's not bad. +1
– nickf
Sep 6 '09 at 23:29

add a comment |

i think you want to do pad = count - (num + '').length. negative numbers aren't handled well, but apart from that, it's not bad. +1
– nickf
Sep 6 '09 at 23:29

i think you want to do pad = count - (num + '').length. negative numbers aren't handled well, but apart from that, it's not bad. +1
– nickf
Sep 6 '09 at 23:29

add a comment |

up vote
3
down vote

My solution

Number.prototype.PadLeft = function (length, digit) {

    var str = '' + this;

    while (str.length < length) {

        str = (digit || '0') + str;

    }

    return str;

};

Usage

var a = 567.25;

a.PadLeft(10); // 0000567.25



var b = 567.25;

b.PadLeft(20, '2'); // 22222222222222567.25

answered Dec 27 '12 at 17:15

community wiki

Aleksandar Toplek

add a comment |

up vote
3
down vote

My solution

Number.prototype.PadLeft = function (length, digit) {

    var str = '' + this;

    while (str.length < length) {

        str = (digit || '0') + str;

    }

    return str;

};

Usage

var a = 567.25;

a.PadLeft(10); // 0000567.25



var b = 567.25;

b.PadLeft(20, '2'); // 22222222222222567.25

answered Dec 27 '12 at 17:15

community wiki

Aleksandar Toplek

add a comment |

up vote
3
down vote

My solution

Number.prototype.PadLeft = function (length, digit) {

    var str = '' + this;

    while (str.length < length) {

        str = (digit || '0') + str;

    }

    return str;

};

Usage

var a = 567.25;

a.PadLeft(10); // 0000567.25



var b = 567.25;

b.PadLeft(20, '2'); // 22222222222222567.25

answered Dec 27 '12 at 17:15

community wiki

Aleksandar Toplek

My solution

Number.prototype.PadLeft = function (length, digit) {

    var str = '' + this;

    while (str.length < length) {

        str = (digit || '0') + str;

    }

    return str;

};

Usage

var a = 567.25;

a.PadLeft(10); // 0000567.25



var b = 567.25;

b.PadLeft(20, '2'); // 22222222222222567.25

answered Dec 27 '12 at 17:15

community wiki

Aleksandar Toplek

answered Dec 27 '12 at 17:15

community wiki

Aleksandar Toplek

community wiki

Aleksandar Toplek

community wiki

Aleksandar Toplek

add a comment |

up vote
3
down vote

Not that this question needs more answers, but I thought I would add the simple lodash version of this.

_.padLeft(number, 6, '0')

answered Jul 13 '15 at 15:21

community wiki

Art

1

Better: var zfill = _.partialRight(_.padStart, '0');
– Droogans
Mar 14 '16 at 18:03

2

Some people will probably protest "I don't want to use lodash" but it's hardly a valid argument anymore since you can install only this function: npm install --save lodash.padleft, then import padLeft from 'lodash.padleft' and omit the _.
– Andy
Oct 14 '16 at 20:44

add a comment |

up vote
3
down vote

Not that this question needs more answers, but I thought I would add the simple lodash version of this.

_.padLeft(number, 6, '0')

answered Jul 13 '15 at 15:21

community wiki

Art

1

Better: var zfill = _.partialRight(_.padStart, '0');
– Droogans
Mar 14 '16 at 18:03

2

Some people will probably protest "I don't want to use lodash" but it's hardly a valid argument anymore since you can install only this function: npm install --save lodash.padleft, then import padLeft from 'lodash.padleft' and omit the _.
– Andy
Oct 14 '16 at 20:44

add a comment |

up vote
3
down vote

Not that this question needs more answers, but I thought I would add the simple lodash version of this.

_.padLeft(number, 6, '0')

answered Jul 13 '15 at 15:21

community wiki

Art

Not that this question needs more answers, but I thought I would add the simple lodash version of this.

_.padLeft(number, 6, '0')

answered Jul 13 '15 at 15:21

community wiki

Art

answered Jul 13 '15 at 15:21

community wiki

Art

community wiki

Art

community wiki

Art

1

Better: var zfill = _.partialRight(_.padStart, '0');
– Droogans
Mar 14 '16 at 18:03

2

Some people will probably protest "I don't want to use lodash" but it's hardly a valid argument anymore since you can install only this function: npm install --save lodash.padleft, then import padLeft from 'lodash.padleft' and omit the _.
– Andy
Oct 14 '16 at 20:44

add a comment |

1

Better: var zfill = _.partialRight(_.padStart, '0');
– Droogans
Mar 14 '16 at 18:03

2

Some people will probably protest "I don't want to use lodash" but it's hardly a valid argument anymore since you can install only this function: npm install --save lodash.padleft, then import padLeft from 'lodash.padleft' and omit the _.
– Andy
Oct 14 '16 at 20:44

Better: var zfill = _.partialRight(_.padStart, '0');
– Droogans
Mar 14 '16 at 18:03

Some people will probably protest "I don't want to use lodash" but it's hardly a valid argument anymore since you can install only this function: npm install --save lodash.padleft, then import padLeft from 'lodash.padleft' and omit the _.
– Andy
Oct 14 '16 at 20:44

add a comment |

up vote
3
down vote

Why not use recursion?

function padZero(s, n) {

    s = s.toString(); // in case someone passes a number

    return s.length >= n ? s : padZero('0' + s, n);

}

edited Oct 18 '16 at 12:12

community wiki

2 revs
lex82

padZero(223, 3) fails with '0223'
– Serginho
Oct 18 '16 at 10:55

Well, I assumed that this function is called with a string as the first parameter. However, I fixed it.
– lex82
Oct 18 '16 at 12:12

I love recursions :-)
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:33

add a comment |

up vote
3
down vote

Why not use recursion?

function padZero(s, n) {

    s = s.toString(); // in case someone passes a number

    return s.length >= n ? s : padZero('0' + s, n);

}

edited Oct 18 '16 at 12:12

community wiki

2 revs
lex82

padZero(223, 3) fails with '0223'
– Serginho
Oct 18 '16 at 10:55

Well, I assumed that this function is called with a string as the first parameter. However, I fixed it.
– lex82
Oct 18 '16 at 12:12

I love recursions :-)
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:33

add a comment |

up vote
3
down vote

Why not use recursion?

function padZero(s, n) {

    s = s.toString(); // in case someone passes a number

    return s.length >= n ? s : padZero('0' + s, n);

}

edited Oct 18 '16 at 12:12

community wiki

2 revs
lex82

Why not use recursion?

function padZero(s, n) {

    s = s.toString(); // in case someone passes a number

    return s.length >= n ? s : padZero('0' + s, n);

}

edited Oct 18 '16 at 12:12

community wiki

2 revs
lex82

edited Oct 18 '16 at 12:12

community wiki

2 revs
lex82

community wiki

2 revs
lex82

community wiki

2 revs
lex82

padZero(223, 3) fails with '0223'
– Serginho
Oct 18 '16 at 10:55

Well, I assumed that this function is called with a string as the first parameter. However, I fixed it.
– lex82
Oct 18 '16 at 12:12

I love recursions :-)
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:33

add a comment |

padZero(223, 3) fails with '0223'
– Serginho
Oct 18 '16 at 10:55

Well, I assumed that this function is called with a string as the first parameter. However, I fixed it.
– lex82
Oct 18 '16 at 12:12

I love recursions :-)
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:33

padZero(223, 3) fails with '0223'
– Serginho
Oct 18 '16 at 10:55

Well, I assumed that this function is called with a string as the first parameter. However, I fixed it.
– lex82
Oct 18 '16 at 12:12

I love recursions :-)
– Xan-Kun Clark-Davis
Jun 6 '17 at 9:33

add a comment |

up vote
2
down vote

Some monkeypatching also works

String.prototype.padLeft = function (n, c) {

  if (isNaN(n))

    return null;

  c = c || "0";

  return (new Array(n).join(c).substring(0, this.length-n)) + this; 

};

var paddedValue = "123".padLeft(6); // returns "000123"

var otherPadded = "TEXT".padLeft(8, " "); // returns "    TEXT"

answered Sep 7 '09 at 0:07

community wiki

Rodrigo

1

-1 monkeypatching the toplevel namespacing in javascript is bad practice.
– Jeremy Wall
Sep 7 '09 at 0:37

2

True for Object and Array objects, but String is not bad
– Rodrigo
Sep 11 '09 at 16:56

add a comment |

up vote
2
down vote

Some monkeypatching also works

String.prototype.padLeft = function (n, c) {

  if (isNaN(n))

    return null;

  c = c || "0";

  return (new Array(n).join(c).substring(0, this.length-n)) + this; 

};

var paddedValue = "123".padLeft(6); // returns "000123"

var otherPadded = "TEXT".padLeft(8, " "); // returns "    TEXT"

answered Sep 7 '09 at 0:07

community wiki

Rodrigo

1

-1 monkeypatching the toplevel namespacing in javascript is bad practice.
– Jeremy Wall
Sep 7 '09 at 0:37

2

True for Object and Array objects, but String is not bad
– Rodrigo
Sep 11 '09 at 16:56

add a comment |

up vote
2
down vote

Some monkeypatching also works

String.prototype.padLeft = function (n, c) {

  if (isNaN(n))

    return null;

  c = c || "0";

  return (new Array(n).join(c).substring(0, this.length-n)) + this; 

};

var paddedValue = "123".padLeft(6); // returns "000123"

var otherPadded = "TEXT".padLeft(8, " "); // returns "    TEXT"

answered Sep 7 '09 at 0:07

community wiki

Rodrigo

Some monkeypatching also works

String.prototype.padLeft = function (n, c) {

  if (isNaN(n))

    return null;

  c = c || "0";

  return (new Array(n).join(c).substring(0, this.length-n)) + this; 

};

var paddedValue = "123".padLeft(6); // returns "000123"

var otherPadded = "TEXT".padLeft(8, " "); // returns "    TEXT"

answered Sep 7 '09 at 0:07

community wiki

Rodrigo

answered Sep 7 '09 at 0:07

community wiki

Rodrigo

community wiki

Rodrigo

community wiki

Rodrigo

1

-1 monkeypatching the toplevel namespacing in javascript is bad practice.
– Jeremy Wall
Sep 7 '09 at 0:37

2

True for Object and Array objects, but String is not bad
– Rodrigo
Sep 11 '09 at 16:56

add a comment |

1

-1 monkeypatching the toplevel namespacing in javascript is bad practice.
– Jeremy Wall
Sep 7 '09 at 0:37

2

True for Object and Array objects, but String is not bad
– Rodrigo
Sep 11 '09 at 16:56

-1 monkeypatching the toplevel namespacing in javascript is bad practice.
– Jeremy Wall
Sep 7 '09 at 0:37

True for Object and Array objects, but String is not bad
– Rodrigo
Sep 11 '09 at 16:56

add a comment |

up vote
2
down vote

function pad(toPad, padChar, length){

    return (String(toPad).length < length)

        ? new Array(length - String(toPad).length + 1).join(padChar) + String(toPad)

        : toPad;

}

pad(5, 0, 6) = 000005

pad('10', 0, 2) = 10 // don't pad if not necessary

pad('S', 'O', 2) = SO

...etc.

Cheers

edited May 23 '12 at 3:50

community wiki

Madbreaks

This doesn't work. pad(100, '0', 4) => 000100
– drewish
May 22 '12 at 20:23

I think it should be more like: var s = String(toPad); return (s.length < length) ? new Array(length - s.length + 1).join('0') + s : s; if you fix it I'll remove my down vote.
– drewish
May 22 '12 at 20:24

@drewish Thanks for catching that, I agree with your edit (except for the hardcoded 0 join char :) -- Answer updated.
– Madbreaks
May 22 '12 at 21:25

Ah yeah I'd hard coded it in my usage.
– drewish
May 23 '12 at 2:04

I think it'd be better to store the string into a temporary variable. I did some benchmarking of that here: jsperf.com/padding-with-memoization it's also got benchmarks for the use of the new in "new Array" the results for new are all over the place but memiozation seems like the way to go.
– drewish
May 23 '12 at 16:28

add a comment |

up vote
2
down vote

function pad(toPad, padChar, length){

    return (String(toPad).length < length)

        ? new Array(length - String(toPad).length + 1).join(padChar) + String(toPad)

        : toPad;

}

pad(5, 0, 6) = 000005

pad('10', 0, 2) = 10 // don't pad if not necessary

pad('S', 'O', 2) = SO

...etc.

Cheers

edited May 23 '12 at 3:50

community wiki

Madbreaks

This doesn't work. pad(100, '0', 4) => 000100
– drewish
May 22 '12 at 20:23

I think it should be more like: var s = String(toPad); return (s.length < length) ? new Array(length - s.length + 1).join('0') + s : s; if you fix it I'll remove my down vote.
– drewish
May 22 '12 at 20:24

@drewish Thanks for catching that, I agree with your edit (except for the hardcoded 0 join char :) -- Answer updated.
– Madbreaks
May 22 '12 at 21:25

Ah yeah I'd hard coded it in my usage.
– drewish
May 23 '12 at 2:04

I think it'd be better to store the string into a temporary variable. I did some benchmarking of that here: jsperf.com/padding-with-memoization it's also got benchmarks for the use of the new in "new Array" the results for new are all over the place but memiozation seems like the way to go.
– drewish
May 23 '12 at 16:28

add a comment |

up vote
2
down vote

function pad(toPad, padChar, length){

    return (String(toPad).length < length)

        ? new Array(length - String(toPad).length + 1).join(padChar) + String(toPad)

        : toPad;

}

pad(5, 0, 6) = 000005

pad('10', 0, 2) = 10 // don't pad if not necessary

pad('S', 'O', 2) = SO

...etc.

Cheers

edited May 23 '12 at 3:50

community wiki

Madbreaks

function pad(toPad, padChar, length){

    return (String(toPad).length < length)

        ? new Array(length - String(toPad).length + 1).join(padChar) + String(toPad)

        : toPad;

}

pad(5, 0, 6) = 000005

pad('10', 0, 2) = 10 // don't pad if not necessary

pad('S', 'O', 2) = SO

...etc.

Cheers

edited May 23 '12 at 3:50

community wiki

Madbreaks

edited May 23 '12 at 3:50

community wiki

Madbreaks

community wiki

Madbreaks

community wiki

Madbreaks

This doesn't work. pad(100, '0', 4) => 000100
– drewish
May 22 '12 at 20:23

I think it should be more like: var s = String(toPad); return (s.length < length) ? new Array(length - s.length + 1).join('0') + s : s; if you fix it I'll remove my down vote.
– drewish
May 22 '12 at 20:24

@drewish Thanks for catching that, I agree with your edit (except for the hardcoded 0 join char :) -- Answer updated.
– Madbreaks
May 22 '12 at 21:25

Ah yeah I'd hard coded it in my usage.
– drewish
May 23 '12 at 2:04

I think it'd be better to store the string into a temporary variable. I did some benchmarking of that here: jsperf.com/padding-with-memoization it's also got benchmarks for the use of the new in "new Array" the results for new are all over the place but memiozation seems like the way to go.
– drewish
May 23 '12 at 16:28

add a comment |

This doesn't work. pad(100, '0', 4) => 000100
– drewish
May 22 '12 at 20:23

I think it should be more like: var s = String(toPad); return (s.length < length) ? new Array(length - s.length + 1).join('0') + s : s; if you fix it I'll remove my down vote.
– drewish
May 22 '12 at 20:24

@drewish Thanks for catching that, I agree with your edit (except for the hardcoded 0 join char :) -- Answer updated.
– Madbreaks
May 22 '12 at 21:25

Ah yeah I'd hard coded it in my usage.
– drewish
May 23 '12 at 2:04

I think it'd be better to store the string into a temporary variable. I did some benchmarking of that here: jsperf.com/padding-with-memoization it's also got benchmarks for the use of the new in "new Array" the results for new are all over the place but memiozation seems like the way to go.
– drewish
May 23 '12 at 16:28

This doesn't work. pad(100, '0', 4) => 000100
– drewish
May 22 '12 at 20:23

I think it should be more like: var s = String(toPad); return (s.length < length) ? new Array(length - s.length + 1).join('0') + s : s; if you fix it I'll remove my down vote.
– drewish
May 22 '12 at 20:24

@drewish Thanks for catching that, I agree with your edit (except for the hardcoded 0 join char :) -- Answer updated.
– Madbreaks
May 22 '12 at 21:25

Ah yeah I'd hard coded it in my usage.
– drewish
May 23 '12 at 2:04

I think it'd be better to store the string into a temporary variable. I did some benchmarking of that here: jsperf.com/padding-with-memoization it's also got benchmarks for the use of the new in "new Array" the results for new are all over the place but memiozation seems like the way to go.
– drewish
May 23 '12 at 16:28

add a comment |

up vote
2
down vote

The simplest, most straight-forward solution you will find.

function zerofill(number,length) {

    var output = number.toString();

    while(output.length < length) {

      output = '0' + output;

    }

    return output;

}

answered Jan 7 '15 at 9:22

community wiki

d4nyll

add a comment |

up vote
2
down vote

The simplest, most straight-forward solution you will find.

function zerofill(number,length) {

    var output = number.toString();

    while(output.length < length) {

      output = '0' + output;

    }

    return output;

}

answered Jan 7 '15 at 9:22

community wiki

d4nyll

add a comment |

up vote
2
down vote

The simplest, most straight-forward solution you will find.

function zerofill(number,length) {

    var output = number.toString();

    while(output.length < length) {

      output = '0' + output;

    }

    return output;

}

answered Jan 7 '15 at 9:22

community wiki

d4nyll

The simplest, most straight-forward solution you will find.

function zerofill(number,length) {

    var output = number.toString();

    while(output.length < length) {

      output = '0' + output;

    }

    return output;

}

answered Jan 7 '15 at 9:22

community wiki

d4nyll

answered Jan 7 '15 at 9:22

community wiki

d4nyll

community wiki

d4nyll

community wiki

d4nyll

add a comment |

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