Create string template and replace with contents of array

Say I have a string like so:

const m = 'Map<?,?>';

and an array like so:

const types = ['string', 'boolean'];

and I am looking to generate:

const r = 'Map<string,boolean>'

how can I do that? Note that I don't need to use a question mark in the template string, I can use a different character, maybe like this:

const m = 'Map<%s,%s>';

Also please note that I cannot use ES6 template strings of this nature ${} because my template is not wrapped in a function.

I believe this will work in Node.js:

util.format(m, ...types);

but I guess what I am looking for is a way to use a question mark character (?) instead of using %s.

asked Nov 21 '18 at 22:35

MrCholo

1,408933

add a comment |

Say I have a string like so:

const m = 'Map<?,?>';

and an array like so:

const types = ['string', 'boolean'];

and I am looking to generate:

const r = 'Map<string,boolean>'

how can I do that? Note that I don't need to use a question mark in the template string, I can use a different character, maybe like this:

const m = 'Map<%s,%s>';

Also please note that I cannot use ES6 template strings of this nature ${} because my template is not wrapped in a function.

I believe this will work in Node.js:

util.format(m, ...types);

but I guess what I am looking for is a way to use a question mark character (?) instead of using %s.

asked Nov 21 '18 at 22:35

MrCholo

1,408933

add a comment |

Say I have a string like so:

const m = 'Map<?,?>';

and an array like so:

const types = ['string', 'boolean'];

and I am looking to generate:

const r = 'Map<string,boolean>'

how can I do that? Note that I don't need to use a question mark in the template string, I can use a different character, maybe like this:

const m = 'Map<%s,%s>';

Also please note that I cannot use ES6 template strings of this nature ${} because my template is not wrapped in a function.

I believe this will work in Node.js:

util.format(m, ...types);

but I guess what I am looking for is a way to use a question mark character (?) instead of using %s.

asked Nov 21 '18 at 22:35

MrCholo

1,408933

Say I have a string like so:

const m = 'Map<?,?>';

and an array like so:

const types = ['string', 'boolean'];

and I am looking to generate:

const r = 'Map<string,boolean>'

how can I do that? Note that I don't need to use a question mark in the template string, I can use a different character, maybe like this:

const m = 'Map<%s,%s>';

Also please note that I cannot use ES6 template strings of this nature ${} because my template is not wrapped in a function.

I believe this will work in Node.js:

util.format(m, ...types);

but I guess what I am looking for is a way to use a question mark character (?) instead of using %s.

javascript node.js

asked Nov 21 '18 at 22:35

MrCholo

1,408933

asked Nov 21 '18 at 22:35

MrCholo

1,408933

asked Nov 21 '18 at 22:35

MrCholo

1,408933

asked Nov 21 '18 at 22:35

MrCholo

1,408933

asked Nov 21 '18 at 22:35

MrCholo

1,408933

add a comment |

3 Answers
3

active

oldest

votes

The smallest way is probably using a combination of reduce and replace.

Replace only replaces the first found position and reduce makes it easy to iterate over the given replacements.

const m = 'Map<?,?>';



const types = ['string', 'boolean'];



const n = types.reduce((n, t) => n.replace('?', t), m);

console.log(n);

Or to formalize it into a function:

function format(str, key, replacements){

    return replacements.reduce((s, t) => s.replace(k, t), str);

}

edited Nov 21 '18 at 22:51

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

probably a good idea to make a note on the comma operator, or explain that in the comments here. Oh just kidding, that's not a comma operator.

– MrCholo
Nov 21 '18 at 22:42

What do you mean by comma operator?

– Sebastian Speitel
Nov 21 '18 at 22:42

I misread your code, but this is the comma op: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– MrCholo
Nov 21 '18 at 22:43

Is anything else unclear?

– Sebastian Speitel
Nov 21 '18 at 22:44

1

This is probably the best answer possible. I'm just miffed that I never thought of iterating over a string like this, and especially in a way that works with a const string. Nice answer!

– zfrisch
Nov 21 '18 at 22:52

|
show 10 more comments

You can also use replace() directly with a little fancy closure work to capture the index (which sadly isn't passed to the replace callback function):

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



let rep = m.replace(/?/g, ((i) => () => types[i++])(0)) // i is simple a count of matches starting at 0

console.log(rep)

If the immediately-executed function is too weird, you can wrap the whole thing as a simple function that's a little easier on the eyes:

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



function replaceMatch(str, arr){

    let i = 0

    return str.replace(/?/g, () => arr[i++])

}



console.log(replaceMatch(m, types))

edited Nov 21 '18 at 23:15

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

1

Interesting answer! That is one weird looking closure

– zfrisch
Nov 21 '18 at 23:01

add a comment |

You only need one ?

const types = ['string', 'boolean'];



console.log('Map<?>'.split('?').join(types))

const types = ['string', 'boolean'];



console.log('Map<?>'.replace('?', types))

or if you wished to keep ?,?

const types = ['string', 'boolean'];



console.log('Map<?,?>'.replace('?,?', types))

edited Nov 22 '18 at 10:27

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

The smallest way is probably using a combination of reduce and replace.

Replace only replaces the first found position and reduce makes it easy to iterate over the given replacements.

const m = 'Map<?,?>';



const types = ['string', 'boolean'];



const n = types.reduce((n, t) => n.replace('?', t), m);

console.log(n);

Or to formalize it into a function:

function format(str, key, replacements){

    return replacements.reduce((s, t) => s.replace(k, t), str);

}

edited Nov 21 '18 at 22:51

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

probably a good idea to make a note on the comma operator, or explain that in the comments here. Oh just kidding, that's not a comma operator.

– MrCholo
Nov 21 '18 at 22:42

What do you mean by comma operator?

– Sebastian Speitel
Nov 21 '18 at 22:42

I misread your code, but this is the comma op: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– MrCholo
Nov 21 '18 at 22:43

Is anything else unclear?

– Sebastian Speitel
Nov 21 '18 at 22:44

1

This is probably the best answer possible. I'm just miffed that I never thought of iterating over a string like this, and especially in a way that works with a const string. Nice answer!

– zfrisch
Nov 21 '18 at 22:52

|
show 10 more comments

The smallest way is probably using a combination of reduce and replace.

Replace only replaces the first found position and reduce makes it easy to iterate over the given replacements.

const m = 'Map<?,?>';



const types = ['string', 'boolean'];



const n = types.reduce((n, t) => n.replace('?', t), m);

console.log(n);

Or to formalize it into a function:

function format(str, key, replacements){

    return replacements.reduce((s, t) => s.replace(k, t), str);

}

edited Nov 21 '18 at 22:51

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

probably a good idea to make a note on the comma operator, or explain that in the comments here. Oh just kidding, that's not a comma operator.

– MrCholo
Nov 21 '18 at 22:42

What do you mean by comma operator?

– Sebastian Speitel
Nov 21 '18 at 22:42

I misread your code, but this is the comma op: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– MrCholo
Nov 21 '18 at 22:43

Is anything else unclear?

– Sebastian Speitel
Nov 21 '18 at 22:44

1

This is probably the best answer possible. I'm just miffed that I never thought of iterating over a string like this, and especially in a way that works with a const string. Nice answer!

– zfrisch
Nov 21 '18 at 22:52

|
show 10 more comments

The smallest way is probably using a combination of reduce and replace.

Replace only replaces the first found position and reduce makes it easy to iterate over the given replacements.

const m = 'Map<?,?>';



const types = ['string', 'boolean'];



const n = types.reduce((n, t) => n.replace('?', t), m);

console.log(n);

Or to formalize it into a function:

function format(str, key, replacements){

    return replacements.reduce((s, t) => s.replace(k, t), str);

}

edited Nov 21 '18 at 22:51

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

The smallest way is probably using a combination of reduce and replace.

Replace only replaces the first found position and reduce makes it easy to iterate over the given replacements.

const m = 'Map<?,?>';



const types = ['string', 'boolean'];



const n = types.reduce((n, t) => n.replace('?', t), m);

console.log(n);

Or to formalize it into a function:

function format(str, key, replacements){

    return replacements.reduce((s, t) => s.replace(k, t), str);

}

const m = 'Map<?,?>';



const types = ['string', 'boolean'];



const n = types.reduce((n, t) => n.replace('?', t), m);

console.log(n);

const m = 'Map<?,?>';



const types = ['string', 'boolean'];



const n = types.reduce((n, t) => n.replace('?', t), m);

console.log(n);

edited Nov 21 '18 at 22:51

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

edited Nov 21 '18 at 22:51

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

answered Nov 21 '18 at 22:40

Sebastian Speitel

4,5102525

probably a good idea to make a note on the comma operator, or explain that in the comments here. Oh just kidding, that's not a comma operator.

– MrCholo
Nov 21 '18 at 22:42

What do you mean by comma operator?

– Sebastian Speitel
Nov 21 '18 at 22:42

I misread your code, but this is the comma op: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– MrCholo
Nov 21 '18 at 22:43

Is anything else unclear?

– Sebastian Speitel
Nov 21 '18 at 22:44

1

This is probably the best answer possible. I'm just miffed that I never thought of iterating over a string like this, and especially in a way that works with a const string. Nice answer!

– zfrisch
Nov 21 '18 at 22:52

|
show 10 more comments

probably a good idea to make a note on the comma operator, or explain that in the comments here. Oh just kidding, that's not a comma operator.

– MrCholo
Nov 21 '18 at 22:42

What do you mean by comma operator?

– Sebastian Speitel
Nov 21 '18 at 22:42

I misread your code, but this is the comma op: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– MrCholo
Nov 21 '18 at 22:43

Is anything else unclear?

– Sebastian Speitel
Nov 21 '18 at 22:44

1

This is probably the best answer possible. I'm just miffed that I never thought of iterating over a string like this, and especially in a way that works with a const string. Nice answer!

– zfrisch
Nov 21 '18 at 22:52

probably a good idea to make a note on the comma operator, or explain that in the comments here. Oh just kidding, that's not a comma operator.

– MrCholo
Nov 21 '18 at 22:42

What do you mean by comma operator?

– Sebastian Speitel
Nov 21 '18 at 22:42

I misread your code, but this is the comma op: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– MrCholo
Nov 21 '18 at 22:43

Is anything else unclear?

– Sebastian Speitel
Nov 21 '18 at 22:44

This is probably the best answer possible. I'm just miffed that I never thought of iterating over a string like this, and especially in a way that works with a const string. Nice answer!

– zfrisch
Nov 21 '18 at 22:52

|
show 10 more comments

You can also use replace() directly with a little fancy closure work to capture the index (which sadly isn't passed to the replace callback function):

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



let rep = m.replace(/?/g, ((i) => () => types[i++])(0)) // i is simple a count of matches starting at 0

console.log(rep)

If the immediately-executed function is too weird, you can wrap the whole thing as a simple function that's a little easier on the eyes:

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



function replaceMatch(str, arr){

    let i = 0

    return str.replace(/?/g, () => arr[i++])

}



console.log(replaceMatch(m, types))

edited Nov 21 '18 at 23:15

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

1

Interesting answer! That is one weird looking closure

– zfrisch
Nov 21 '18 at 23:01

add a comment |

You can also use replace() directly with a little fancy closure work to capture the index (which sadly isn't passed to the replace callback function):

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



let rep = m.replace(/?/g, ((i) => () => types[i++])(0)) // i is simple a count of matches starting at 0

console.log(rep)

If the immediately-executed function is too weird, you can wrap the whole thing as a simple function that's a little easier on the eyes:

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



function replaceMatch(str, arr){

    let i = 0

    return str.replace(/?/g, () => arr[i++])

}



console.log(replaceMatch(m, types))

edited Nov 21 '18 at 23:15

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

1

Interesting answer! That is one weird looking closure

– zfrisch
Nov 21 '18 at 23:01

add a comment |

You can also use replace() directly with a little fancy closure work to capture the index (which sadly isn't passed to the replace callback function):

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



let rep = m.replace(/?/g, ((i) => () => types[i++])(0)) // i is simple a count of matches starting at 0

console.log(rep)

If the immediately-executed function is too weird, you can wrap the whole thing as a simple function that's a little easier on the eyes:

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



function replaceMatch(str, arr){

    let i = 0

    return str.replace(/?/g, () => arr[i++])

}



console.log(replaceMatch(m, types))

edited Nov 21 '18 at 23:15

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

You can also use replace() directly with a little fancy closure work to capture the index (which sadly isn't passed to the replace callback function):

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



let rep = m.replace(/?/g, ((i) => () => types[i++])(0)) // i is simple a count of matches starting at 0

console.log(rep)

If the immediately-executed function is too weird, you can wrap the whole thing as a simple function that's a little easier on the eyes:

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



function replaceMatch(str, arr){

    let i = 0

    return str.replace(/?/g, () => arr[i++])

}



console.log(replaceMatch(m, types))

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



let rep = m.replace(/?/g, ((i) => () => types[i++])(0)) // i is simple a count of matches starting at 0

console.log(rep)

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



let rep = m.replace(/?/g, ((i) => () => types[i++])(0)) // i is simple a count of matches starting at 0

console.log(rep)

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



function replaceMatch(str, arr){

    let i = 0

    return str.replace(/?/g, () => arr[i++])

}



console.log(replaceMatch(m, types))

const m = 'Map<?,?>';

const types = ['string', 'boolean'];



function replaceMatch(str, arr){

    let i = 0

    return str.replace(/?/g, () => arr[i++])

}



console.log(replaceMatch(m, types))

edited Nov 21 '18 at 23:15

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

edited Nov 21 '18 at 23:15

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

answered Nov 21 '18 at 22:53

Mark Meyer

38.6k33159

1

Interesting answer! That is one weird looking closure

– zfrisch
Nov 21 '18 at 23:01

add a comment |

1

Interesting answer! That is one weird looking closure

– zfrisch
Nov 21 '18 at 23:01

Interesting answer! That is one weird looking closure

– zfrisch
Nov 21 '18 at 23:01

add a comment |

You only need one ?

const types = ['string', 'boolean'];



console.log('Map<?>'.split('?').join(types))

const types = ['string', 'boolean'];



console.log('Map<?>'.replace('?', types))

or if you wished to keep ?,?

const types = ['string', 'boolean'];



console.log('Map<?,?>'.replace('?,?', types))

edited Nov 22 '18 at 10:27

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

add a comment |

You only need one ?

const types = ['string', 'boolean'];



console.log('Map<?>'.split('?').join(types))

const types = ['string', 'boolean'];



console.log('Map<?>'.replace('?', types))

or if you wished to keep ?,?

const types = ['string', 'boolean'];



console.log('Map<?,?>'.replace('?,?', types))

edited Nov 22 '18 at 10:27

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

add a comment |

You only need one ?

const types = ['string', 'boolean'];



console.log('Map<?>'.split('?').join(types))

const types = ['string', 'boolean'];



console.log('Map<?>'.replace('?', types))

or if you wished to keep ?,?

const types = ['string', 'boolean'];



console.log('Map<?,?>'.replace('?,?', types))

edited Nov 22 '18 at 10:27

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

You only need one ?

const types = ['string', 'boolean'];



console.log('Map<?>'.split('?').join(types))

const types = ['string', 'boolean'];



console.log('Map<?>'.replace('?', types))

or if you wished to keep ?,?

const types = ['string', 'boolean'];



console.log('Map<?,?>'.replace('?,?', types))

const types = ['string', 'boolean'];



console.log('Map<?>'.split('?').join(types))

const types = ['string', 'boolean'];



console.log('Map<?>'.split('?').join(types))

const types = ['string', 'boolean'];



console.log('Map<?>'.replace('?', types))

const types = ['string', 'boolean'];



console.log('Map<?>'.replace('?', types))

const types = ['string', 'boolean'];



console.log('Map<?,?>'.replace('?,?', types))

const types = ['string', 'boolean'];



console.log('Map<?,?>'.replace('?,?', types))

edited Nov 22 '18 at 10:27

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

edited Nov 22 '18 at 10:27

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

answered Nov 21 '18 at 22:42

kemicofa

10.2k43982

add a comment |

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