How to increment the last number in a string; bash

7

I have a string that looks something like
/foo/bar/baz59_ 5stuff.thing

I would like to increase the last number (5 in the example) by one, if it's greater than another variable. A 9 would be increased to 10. Note that the last number could be multiple digits also; also that "stuff.thing" could be anything; other than a number; so it can't be hard coded.

The above example would result in /foo/bar/baz59_ 6stuff.thing

I've found multiple questions (and answers) that would extract the last number from the string, and obviously that could then be used in a comparison.
The issue I'm having is how to ensure that when I do the replace, I only replace the last number (since obviously I can't just replace "5" for "6"). Can anyone make any suggestions?

awk/sed/bash/grep are all viable.

bash perl awk sed

share|improve this question

edited Jan 25 at 16:12

UKMonkey

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If the last number is 9, what's the desired outcome? What if it's 59?
  
  – James Brown
  Jan 25 at 14:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  if the last number is 9; then the desired is 10. 59; 60. 99->100 etc (question updated for posterity)
  
  – UKMonkey
  Jan 25 at 14:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's always preceded by an underscore this last number of yours?
  
  – gboffi
  Jan 25 at 14:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @gboffi no; the previous character is not fixed.
  
  – UKMonkey
  Jan 25 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if it's greater than a script argument you mean the whole string, or the number? if the number, do you mean increment the last number in the string that is greater than an argument, or increment the last number in the string if it is greater than an argument?
  
  – ysth
  Jan 25 at 15:31
  

  
  
  
  

 | 
show 2 more comments

7

I have a string that looks something like
/foo/bar/baz59_ 5stuff.thing

I would like to increase the last number (5 in the example) by one, if it's greater than another variable. A 9 would be increased to 10. Note that the last number could be multiple digits also; also that "stuff.thing" could be anything; other than a number; so it can't be hard coded.

The above example would result in /foo/bar/baz59_ 6stuff.thing

I've found multiple questions (and answers) that would extract the last number from the string, and obviously that could then be used in a comparison.
The issue I'm having is how to ensure that when I do the replace, I only replace the last number (since obviously I can't just replace "5" for "6"). Can anyone make any suggestions?

awk/sed/bash/grep are all viable.

bash perl awk sed

share|improve this question

edited Jan 25 at 16:12

UKMonkey

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If the last number is 9, what's the desired outcome? What if it's 59?
  
  – James Brown
  Jan 25 at 14:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  if the last number is 9; then the desired is 10. 59; 60. 99->100 etc (question updated for posterity)
  
  – UKMonkey
  Jan 25 at 14:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's always preceded by an underscore this last number of yours?
  
  – gboffi
  Jan 25 at 14:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @gboffi no; the previous character is not fixed.
  
  – UKMonkey
  Jan 25 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if it's greater than a script argument you mean the whole string, or the number? if the number, do you mean increment the last number in the string that is greater than an argument, or increment the last number in the string if it is greater than an argument?
  
  – ysth
  Jan 25 at 15:31
  

  
  
  
  

 | 
show 2 more comments

7

7

7

1

I have a string that looks something like
/foo/bar/baz59_ 5stuff.thing

I would like to increase the last number (5 in the example) by one, if it's greater than another variable. A 9 would be increased to 10. Note that the last number could be multiple digits also; also that "stuff.thing" could be anything; other than a number; so it can't be hard coded.

The above example would result in /foo/bar/baz59_ 6stuff.thing

I've found multiple questions (and answers) that would extract the last number from the string, and obviously that could then be used in a comparison.
The issue I'm having is how to ensure that when I do the replace, I only replace the last number (since obviously I can't just replace "5" for "6"). Can anyone make any suggestions?

awk/sed/bash/grep are all viable.

bash perl awk sed

share|improve this question

edited Jan 25 at 16:12

UKMonkey

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

I have a string that looks something like
/foo/bar/baz59_ 5stuff.thing

I would like to increase the last number (5 in the example) by one, if it's greater than another variable. A 9 would be increased to 10. Note that the last number could be multiple digits also; also that "stuff.thing" could be anything; other than a number; so it can't be hard coded.

The above example would result in /foo/bar/baz59_ 6stuff.thing

I've found multiple questions (and answers) that would extract the last number from the string, and obviously that could then be used in a comparison.
The issue I'm having is how to ensure that when I do the replace, I only replace the last number (since obviously I can't just replace "5" for "6"). Can anyone make any suggestions?

awk/sed/bash/grep are all viable.

bash perl awk sed

bash perl awk sed

share|improve this question

edited Jan 25 at 16:12

UKMonkey

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

share|improve this question

edited Jan 25 at 16:12

UKMonkey

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

share|improve this question

share|improve this question

edited Jan 25 at 16:12

UKMonkey

edited Jan 25 at 16:12

UKMonkey

edited Jan 25 at 16:12

UKMonkey

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

asked Jan 25 at 14:34

UKMonkeyUKMonkey

5,78431227

5,78431227

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If the last number is 9, what's the desired outcome? What if it's 59?
  
  – James Brown
  Jan 25 at 14:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  if the last number is 9; then the desired is 10. 59; 60. 99->100 etc (question updated for posterity)
  
  – UKMonkey
  Jan 25 at 14:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's always preceded by an underscore this last number of yours?
  
  – gboffi
  Jan 25 at 14:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @gboffi no; the previous character is not fixed.
  
  – UKMonkey
  Jan 25 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if it's greater than a script argument you mean the whole string, or the number? if the number, do you mean increment the last number in the string that is greater than an argument, or increment the last number in the string if it is greater than an argument?
  
  – ysth
  Jan 25 at 15:31
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If the last number is 9, what's the desired outcome? What if it's 59?
  
  – James Brown
  Jan 25 at 14:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  if the last number is 9; then the desired is 10. 59; 60. 99->100 etc (question updated for posterity)
  
  – UKMonkey
  Jan 25 at 14:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's always preceded by an underscore this last number of yours?
  
  – gboffi
  Jan 25 at 14:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @gboffi no; the previous character is not fixed.
  
  – UKMonkey
  Jan 25 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if it's greater than a script argument you mean the whole string, or the number? if the number, do you mean increment the last number in the string that is greater than an argument, or increment the last number in the string if it is greater than an argument?
  
  – ysth
  Jan 25 at 15:31
  

  
  
  
  

1

1

If the last number is 9, what's the desired outcome? What if it's 59?

– James Brown
Jan 25 at 14:38

If the last number is 9, what's the desired outcome? What if it's 59?

– James Brown
Jan 25 at 14:38

1

1

if the last number is 9; then the desired is 10. 59; 60. 99->100 etc (question updated for posterity)

– UKMonkey
Jan 25 at 14:42

if the last number is 9; then the desired is 10. 59; 60. 99->100 etc (question updated for posterity)

– UKMonkey
Jan 25 at 14:42

It's always preceded by an underscore this last number of yours?

– gboffi
Jan 25 at 14:58

It's always preceded by an underscore this last number of yours?

– gboffi
Jan 25 at 14:58

@gboffi no; the previous character is not fixed.

– UKMonkey
Jan 25 at 15:01

@gboffi no; the previous character is not fixed.

– UKMonkey
Jan 25 at 15:01

if it's greater than a script argument you mean the whole string, or the number? if the number, do you mean increment the last number in the string that is greater than an argument, or increment the last number in the string if it is greater than an argument?

– ysth
Jan 25 at 15:31

if it's greater than a script argument you mean the whole string, or the number? if the number, do you mean increment the last number in the string that is greater than an argument, or increment the last number in the string if it is greater than an argument?

– ysth
Jan 25 at 15:31

 | 
show 2 more comments

6 Answers
6

active

oldest

votes

7

Updated Answer

Thanks to @EdMorton for pointing out the further requirement of the number exceeding a threshold. That can be done like this:

perl -spe 's/(d+)(?!.*d+)/$1>$thresh? $1+1 : $1/e' <<<  "abc123_456.txt" -- -thresh=500

Original Answer

You can evaluate/calculate a replacement with /e in Perl regexes. Here I just add 1 to the captured string of digits but you can do more complicated stuff:

perl -pe 's/(d+)(?!.*d+)/$1+1/e' <<< "abc123_456.txt"

abc123_457.txt

The (?!.*d+) is (hopefully) a negative look-ahead for any more digits.

The $1 represents any sequence of digits captured in the capture group (d+).

Note that this would need modification to handle decimal numbers, negative numbers and scientific notation - but that is possible.

share|improve this answer

edited Jan 25 at 17:15

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems to work perfectly; and even handles a string containing quotes
  
  – UKMonkey
  Jan 25 at 15:06
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  or s/.*(?<!d)K(d+)/$1+1/sea
  
  – ysth
  Jan 25 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @EdMorton Well spotted! I hadn't seen that, but have added it in now.
  
  – Mark Setchell
  Jan 25 at 17:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  or s/(d+)(?=D*$)/$1+1/ea
  
  – Miller
  Jan 25 at 20:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you all for your alternative suggestions.
  
  – Mark Setchell
  Jan 25 at 20:55
  

  
  
  
  

add a comment | 

4

Using bash regular expression matching:

$ f="/foo/bar/baz59_ 99stuff.thing"

$ [[ $f =~ ([0-9]+)([^0-9]+)$ ]]

OK, what do we have now?

$ declare -p BASH_REMATCH

declare -ar BASH_REMATCH=([0]="99stuff.thing" [1]="99" [2]="stuff.thing")

So we can construct the new filename

if [[ $f =~ ([0-9]+)([^0-9]+)$ ]]; then

    prefix=${f%${BASH_REMATCH[0]}}           # remove "99stuff.thing" from $f

    number=$(( 10#${BASH_REMATCH[1]} + 1 ))  # use "10#" to force base10

    new=${prefix}${number}${BASH_REMATCH[2]}

    echo $new

fi

# => /foo/bar/baz59_ 100stuff.thing

share|improve this answer

edited Jan 25 at 15:22

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

add a comment | 

4

With GNU awk for the 3rd arg to match():

$ awk -v t=3 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 6stuff.thing

Just set t to whatever your threshold value is for incrementing, e.g.:

$ awk -v t=7 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 5stuff.thing

share|improve this answer

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

add a comment | 

1

if it's greater than a script argument.

If I get it correctly(I am assuming you are passing an argument through a script and if its value is greater than string's 2nd field digit then increase 1 into that 2nd field's digit), could you please try following once.

cat script.ksh

value=$1

echo "/foo/bar/baz59_ 5stuff.thing" | 

awk -v arg="$value" '

  match($2,/[0-9]+/){

    val=substr($2,RSTART,RLENGTH)

    val=val<arg?val+1:val

    $2=val substr($2,RSTART+RLENGTH)

}

1'

Here is an example when I run script.ksh it gives following output.

/script.ksh 7

/foo/bar/baz59_ 6stuff.thing

share|improve this answer

edited Jan 25 at 15:10

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, could you please check this one also once and let me know if this works for you?
  
  – RavinderSingh13
  Jan 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  works fine! thanks
  
  – UKMonkey
  Jan 25 at 15:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, cool glad that it helped you, why I have specifically asked is no other solution have taken argument from script and compared it, so I wanted to make sure I understood question correctly, if you tested by passing argument then cool :)
  
  – RavinderSingh13
  Jan 25 at 15:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  it was the principle behind the replacement that I cared about - the full implementation such as how to plug in the string etc is a worry I didn't want to infect the question with.
  
  – UKMonkey
  Jan 25 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, well(if I get it correctly) if that is complete requirement then you need not to worry now, you have complete solution now :)
  
  – RavinderSingh13
  Jan 25 at 15:29
  
  
  

  
  
  
  

add a comment | 

1

Here is a shorter gnu awk approach:

cat incr.awk

{

   n = split($0, a, /[0-9]+/, b)

   for(i=1; i<n; i++)

      s = s a[i] b[i] + (b[i] < max && i == n-1 ? 1 : 0)

   print s a[i]

}

Then use it as:

awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 5stuff.thing'

/foo/bar/baz59_ 6stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 79stuff.thing'

/foo/bar/baz59_ 80stuff.thing





awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 90stuff.thing'

/foo/bar/baz59_ 90stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 80stuff.thing'

/foo/bar/baz59_ 80stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/stuff.thing'

/foo/bar/stuff.thing

share|improve this answer

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

add a comment | 

1

An awk:

$ echo /foo/bar/baz59_ 99stuff.thing |

awk '

/[0-9]/ {

    rstart=1                                    # keep track of the start

    while(match(substr($0,rstart),/[0-9]+/)) {  # while numbers last

        rstart+=RSTART+RLENGTH-1                # increase rstart

        rlength=RLENGTH                         # remember length too

    }

    v=substr($0,rstart-rlength,rlength)+1                    # increase last number

    print substr($0,1,rstart-rlength-1) v substr($0,rstart)  # print in parts

    next

}1'                                             # in case there was no number

/foo/bar/baz59_ 100stuff.thing

Edit:

Whoops, I missed the argument requirement (increase the last number - - by a one, if it's greater than a script argument):

$ echo /foo/bar/baz59_ 99stuff.thing |

awk -v arg=100 '

/[0-9]/ {

    rstart=1

    while(match(substr($0,rstart),/[0-9]+/)) {

        rstart+=RSTART+RLENGTH-1

        rlength=RLENGTH

    }

    v=substr($0,rstart-rlength,rlength)

    if(0+v>arg) {                           # test if v greater that argument

        print substr($0,1,rstart-rlength-1) v+1 substr($0,rstart)

        next

    }

}1'

Output now:

/foo/bar/baz59_ 99stuff.thing

share|improve this answer

edited Jan 25 at 15:56

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated with the script argument requirement. Missed it at first.
  
  – James Brown
  Jan 25 at 15:58
  

  
  
  
  

add a comment | 

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

7

Updated Answer

Thanks to @EdMorton for pointing out the further requirement of the number exceeding a threshold. That can be done like this:

perl -spe 's/(d+)(?!.*d+)/$1>$thresh? $1+1 : $1/e' <<<  "abc123_456.txt" -- -thresh=500

Original Answer

You can evaluate/calculate a replacement with /e in Perl regexes. Here I just add 1 to the captured string of digits but you can do more complicated stuff:

perl -pe 's/(d+)(?!.*d+)/$1+1/e' <<< "abc123_456.txt"

abc123_457.txt

The (?!.*d+) is (hopefully) a negative look-ahead for any more digits.

The $1 represents any sequence of digits captured in the capture group (d+).

Note that this would need modification to handle decimal numbers, negative numbers and scientific notation - but that is possible.

share|improve this answer

edited Jan 25 at 17:15

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems to work perfectly; and even handles a string containing quotes
  
  – UKMonkey
  Jan 25 at 15:06
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  or s/.*(?<!d)K(d+)/$1+1/sea
  
  – ysth
  Jan 25 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @EdMorton Well spotted! I hadn't seen that, but have added it in now.
  
  – Mark Setchell
  Jan 25 at 17:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  or s/(d+)(?=D*$)/$1+1/ea
  
  – Miller
  Jan 25 at 20:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you all for your alternative suggestions.
  
  – Mark Setchell
  Jan 25 at 20:55
  

  
  
  
  

add a comment | 

7

Updated Answer

Thanks to @EdMorton for pointing out the further requirement of the number exceeding a threshold. That can be done like this:

perl -spe 's/(d+)(?!.*d+)/$1>$thresh? $1+1 : $1/e' <<<  "abc123_456.txt" -- -thresh=500

Original Answer

You can evaluate/calculate a replacement with /e in Perl regexes. Here I just add 1 to the captured string of digits but you can do more complicated stuff:

perl -pe 's/(d+)(?!.*d+)/$1+1/e' <<< "abc123_456.txt"

abc123_457.txt

The (?!.*d+) is (hopefully) a negative look-ahead for any more digits.

The $1 represents any sequence of digits captured in the capture group (d+).

Note that this would need modification to handle decimal numbers, negative numbers and scientific notation - but that is possible.

share|improve this answer

edited Jan 25 at 17:15

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems to work perfectly; and even handles a string containing quotes
  
  – UKMonkey
  Jan 25 at 15:06
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  or s/.*(?<!d)K(d+)/$1+1/sea
  
  – ysth
  Jan 25 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @EdMorton Well spotted! I hadn't seen that, but have added it in now.
  
  – Mark Setchell
  Jan 25 at 17:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  or s/(d+)(?=D*$)/$1+1/ea
  
  – Miller
  Jan 25 at 20:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you all for your alternative suggestions.
  
  – Mark Setchell
  Jan 25 at 20:55
  

  
  
  
  

add a comment | 

7

7

7

Updated Answer

Thanks to @EdMorton for pointing out the further requirement of the number exceeding a threshold. That can be done like this:

perl -spe 's/(d+)(?!.*d+)/$1>$thresh? $1+1 : $1/e' <<<  "abc123_456.txt" -- -thresh=500

Original Answer

You can evaluate/calculate a replacement with /e in Perl regexes. Here I just add 1 to the captured string of digits but you can do more complicated stuff:

perl -pe 's/(d+)(?!.*d+)/$1+1/e' <<< "abc123_456.txt"

abc123_457.txt

The (?!.*d+) is (hopefully) a negative look-ahead for any more digits.

The $1 represents any sequence of digits captured in the capture group (d+).

Note that this would need modification to handle decimal numbers, negative numbers and scientific notation - but that is possible.

share|improve this answer

edited Jan 25 at 17:15

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

Updated Answer

Thanks to @EdMorton for pointing out the further requirement of the number exceeding a threshold. That can be done like this:

perl -spe 's/(d+)(?!.*d+)/$1>$thresh? $1+1 : $1/e' <<<  "abc123_456.txt" -- -thresh=500

Original Answer

You can evaluate/calculate a replacement with /e in Perl regexes. Here I just add 1 to the captured string of digits but you can do more complicated stuff:

perl -pe 's/(d+)(?!.*d+)/$1+1/e' <<< "abc123_456.txt"

abc123_457.txt

The (?!.*d+) is (hopefully) a negative look-ahead for any more digits.

The $1 represents any sequence of digits captured in the capture group (d+).

Note that this would need modification to handle decimal numbers, negative numbers and scientific notation - but that is possible.

share|improve this answer

edited Jan 25 at 17:15

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

share|improve this answer

share|improve this answer

edited Jan 25 at 17:15

edited Jan 25 at 17:15

edited Jan 25 at 17:15

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

answered Jan 25 at 14:58

Mark SetchellMark Setchell

91.6k781184

91.6k781184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems to work perfectly; and even handles a string containing quotes
  
  – UKMonkey
  Jan 25 at 15:06
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  or s/.*(?<!d)K(d+)/$1+1/sea
  
  – ysth
  Jan 25 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @EdMorton Well spotted! I hadn't seen that, but have added it in now.
  
  – Mark Setchell
  Jan 25 at 17:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  or s/(d+)(?=D*$)/$1+1/ea
  
  – Miller
  Jan 25 at 20:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you all for your alternative suggestions.
  
  – Mark Setchell
  Jan 25 at 20:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems to work perfectly; and even handles a string containing quotes
  
  – UKMonkey
  Jan 25 at 15:06
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  or s/.*(?<!d)K(d+)/$1+1/sea
  
  – ysth
  Jan 25 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @EdMorton Well spotted! I hadn't seen that, but have added it in now.
  
  – Mark Setchell
  Jan 25 at 17:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  or s/(d+)(?=D*$)/$1+1/ea
  
  – Miller
  Jan 25 at 20:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you all for your alternative suggestions.
  
  – Mark Setchell
  Jan 25 at 20:55
  

  
  
  
  

This seems to work perfectly; and even handles a string containing quotes

– UKMonkey
Jan 25 at 15:06

This seems to work perfectly; and even handles a string containing quotes

– UKMonkey
Jan 25 at 15:06

5

5

or s/.*(?<!d)K(d+)/$1+1/sea

– ysth
Jan 25 at 15:27

or s/.*(?<!d)K(d+)/$1+1/sea

– ysth
Jan 25 at 15:27

2

2

@EdMorton Well spotted! I hadn't seen that, but have added it in now.

– Mark Setchell
Jan 25 at 17:16

@EdMorton Well spotted! I hadn't seen that, but have added it in now.

– Mark Setchell
Jan 25 at 17:16

1

1

or s/(d+)(?=D*$)/$1+1/ea

– Miller
Jan 25 at 20:17

or s/(d+)(?=D*$)/$1+1/ea

– Miller
Jan 25 at 20:17

Thank you all for your alternative suggestions.

– Mark Setchell
Jan 25 at 20:55

Thank you all for your alternative suggestions.

– Mark Setchell
Jan 25 at 20:55

add a comment | 

4

Using bash regular expression matching:

$ f="/foo/bar/baz59_ 99stuff.thing"

$ [[ $f =~ ([0-9]+)([^0-9]+)$ ]]

OK, what do we have now?

$ declare -p BASH_REMATCH

declare -ar BASH_REMATCH=([0]="99stuff.thing" [1]="99" [2]="stuff.thing")

So we can construct the new filename

if [[ $f =~ ([0-9]+)([^0-9]+)$ ]]; then

    prefix=${f%${BASH_REMATCH[0]}}           # remove "99stuff.thing" from $f

    number=$(( 10#${BASH_REMATCH[1]} + 1 ))  # use "10#" to force base10

    new=${prefix}${number}${BASH_REMATCH[2]}

    echo $new

fi

# => /foo/bar/baz59_ 100stuff.thing

share|improve this answer

edited Jan 25 at 15:22

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

add a comment | 

4

Using bash regular expression matching:

$ f="/foo/bar/baz59_ 99stuff.thing"

$ [[ $f =~ ([0-9]+)([^0-9]+)$ ]]

OK, what do we have now?

$ declare -p BASH_REMATCH

declare -ar BASH_REMATCH=([0]="99stuff.thing" [1]="99" [2]="stuff.thing")

So we can construct the new filename

if [[ $f =~ ([0-9]+)([^0-9]+)$ ]]; then

    prefix=${f%${BASH_REMATCH[0]}}           # remove "99stuff.thing" from $f

    number=$(( 10#${BASH_REMATCH[1]} + 1 ))  # use "10#" to force base10

    new=${prefix}${number}${BASH_REMATCH[2]}

    echo $new

fi

# => /foo/bar/baz59_ 100stuff.thing

share|improve this answer

edited Jan 25 at 15:22

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

add a comment | 

4

4

4

Using bash regular expression matching:

$ f="/foo/bar/baz59_ 99stuff.thing"

$ [[ $f =~ ([0-9]+)([^0-9]+)$ ]]

OK, what do we have now?

$ declare -p BASH_REMATCH

declare -ar BASH_REMATCH=([0]="99stuff.thing" [1]="99" [2]="stuff.thing")

So we can construct the new filename

if [[ $f =~ ([0-9]+)([^0-9]+)$ ]]; then

    prefix=${f%${BASH_REMATCH[0]}}           # remove "99stuff.thing" from $f

    number=$(( 10#${BASH_REMATCH[1]} + 1 ))  # use "10#" to force base10

    new=${prefix}${number}${BASH_REMATCH[2]}

    echo $new

fi

# => /foo/bar/baz59_ 100stuff.thing

share|improve this answer

edited Jan 25 at 15:22

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

Using bash regular expression matching:

$ f="/foo/bar/baz59_ 99stuff.thing"

$ [[ $f =~ ([0-9]+)([^0-9]+)$ ]]

OK, what do we have now?

$ declare -p BASH_REMATCH

declare -ar BASH_REMATCH=([0]="99stuff.thing" [1]="99" [2]="stuff.thing")

So we can construct the new filename

if [[ $f =~ ([0-9]+)([^0-9]+)$ ]]; then

    prefix=${f%${BASH_REMATCH[0]}}           # remove "99stuff.thing" from $f

    number=$(( 10#${BASH_REMATCH[1]} + 1 ))  # use "10#" to force base10

    new=${prefix}${number}${BASH_REMATCH[2]}

    echo $new

fi

# => /foo/bar/baz59_ 100stuff.thing

share|improve this answer

edited Jan 25 at 15:22

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

share|improve this answer

share|improve this answer

edited Jan 25 at 15:22

edited Jan 25 at 15:22

edited Jan 25 at 15:22

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

answered Jan 25 at 15:10

glenn jackmanglenn jackman

170k26147240

170k26147240

add a comment | 

add a comment | 

4

With GNU awk for the 3rd arg to match():

$ awk -v t=3 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 6stuff.thing

Just set t to whatever your threshold value is for incrementing, e.g.:

$ awk -v t=7 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 5stuff.thing

share|improve this answer

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

add a comment | 

4

With GNU awk for the 3rd arg to match():

$ awk -v t=3 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 6stuff.thing

Just set t to whatever your threshold value is for incrementing, e.g.:

$ awk -v t=7 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 5stuff.thing

share|improve this answer

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

add a comment | 

4

4

4

With GNU awk for the 3rd arg to match():

$ awk -v t=3 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 6stuff.thing

Just set t to whatever your threshold value is for incrementing, e.g.:

$ awk -v t=7 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 5stuff.thing

share|improve this answer

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

With GNU awk for the 3rd arg to match():

$ awk -v t=3 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 6stuff.thing

Just set t to whatever your threshold value is for incrementing, e.g.:

$ awk -v t=7 'match($0,/(.*)([0-9]+)([^0-9]*)$/,a) && a[2]>t{a[2]++; $0=a[1] a[2] a[3]} 1' file

/foo/bar/baz59_ 5stuff.thing

share|improve this answer

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

share|improve this answer

share|improve this answer

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

answered Jan 25 at 16:58

Ed MortonEd Morton

112k1245103

112k1245103

add a comment | 

add a comment | 

1

if it's greater than a script argument.

If I get it correctly(I am assuming you are passing an argument through a script and if its value is greater than string's 2nd field digit then increase 1 into that 2nd field's digit), could you please try following once.

cat script.ksh

value=$1

echo "/foo/bar/baz59_ 5stuff.thing" | 

awk -v arg="$value" '

  match($2,/[0-9]+/){

    val=substr($2,RSTART,RLENGTH)

    val=val<arg?val+1:val

    $2=val substr($2,RSTART+RLENGTH)

}

1'

Here is an example when I run script.ksh it gives following output.

/script.ksh 7

/foo/bar/baz59_ 6stuff.thing

share|improve this answer

edited Jan 25 at 15:10

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, could you please check this one also once and let me know if this works for you?
  
  – RavinderSingh13
  Jan 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  works fine! thanks
  
  – UKMonkey
  Jan 25 at 15:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, cool glad that it helped you, why I have specifically asked is no other solution have taken argument from script and compared it, so I wanted to make sure I understood question correctly, if you tested by passing argument then cool :)
  
  – RavinderSingh13
  Jan 25 at 15:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  it was the principle behind the replacement that I cared about - the full implementation such as how to plug in the string etc is a worry I didn't want to infect the question with.
  
  – UKMonkey
  Jan 25 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, well(if I get it correctly) if that is complete requirement then you need not to worry now, you have complete solution now :)
  
  – RavinderSingh13
  Jan 25 at 15:29
  
  
  

  
  
  
  

add a comment | 

1

if it's greater than a script argument.

If I get it correctly(I am assuming you are passing an argument through a script and if its value is greater than string's 2nd field digit then increase 1 into that 2nd field's digit), could you please try following once.

cat script.ksh

value=$1

echo "/foo/bar/baz59_ 5stuff.thing" | 

awk -v arg="$value" '

  match($2,/[0-9]+/){

    val=substr($2,RSTART,RLENGTH)

    val=val<arg?val+1:val

    $2=val substr($2,RSTART+RLENGTH)

}

1'

Here is an example when I run script.ksh it gives following output.

/script.ksh 7

/foo/bar/baz59_ 6stuff.thing

share|improve this answer

edited Jan 25 at 15:10

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, could you please check this one also once and let me know if this works for you?
  
  – RavinderSingh13
  Jan 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  works fine! thanks
  
  – UKMonkey
  Jan 25 at 15:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, cool glad that it helped you, why I have specifically asked is no other solution have taken argument from script and compared it, so I wanted to make sure I understood question correctly, if you tested by passing argument then cool :)
  
  – RavinderSingh13
  Jan 25 at 15:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  it was the principle behind the replacement that I cared about - the full implementation such as how to plug in the string etc is a worry I didn't want to infect the question with.
  
  – UKMonkey
  Jan 25 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, well(if I get it correctly) if that is complete requirement then you need not to worry now, you have complete solution now :)
  
  – RavinderSingh13
  Jan 25 at 15:29
  
  
  

  
  
  
  

add a comment | 

1

1

1

if it's greater than a script argument.

If I get it correctly(I am assuming you are passing an argument through a script and if its value is greater than string's 2nd field digit then increase 1 into that 2nd field's digit), could you please try following once.

cat script.ksh

value=$1

echo "/foo/bar/baz59_ 5stuff.thing" | 

awk -v arg="$value" '

  match($2,/[0-9]+/){

    val=substr($2,RSTART,RLENGTH)

    val=val<arg?val+1:val

    $2=val substr($2,RSTART+RLENGTH)

}

1'

Here is an example when I run script.ksh it gives following output.

/script.ksh 7

/foo/bar/baz59_ 6stuff.thing

share|improve this answer

edited Jan 25 at 15:10

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

if it's greater than a script argument.

If I get it correctly(I am assuming you are passing an argument through a script and if its value is greater than string's 2nd field digit then increase 1 into that 2nd field's digit), could you please try following once.

cat script.ksh

value=$1

echo "/foo/bar/baz59_ 5stuff.thing" | 

awk -v arg="$value" '

  match($2,/[0-9]+/){

    val=substr($2,RSTART,RLENGTH)

    val=val<arg?val+1:val

    $2=val substr($2,RSTART+RLENGTH)

}

1'

Here is an example when I run script.ksh it gives following output.

/script.ksh 7

/foo/bar/baz59_ 6stuff.thing

share|improve this answer

edited Jan 25 at 15:10

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

share|improve this answer

share|improve this answer

edited Jan 25 at 15:10

edited Jan 25 at 15:10

edited Jan 25 at 15:10

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

answered Jan 25 at 14:56

RavinderSingh13RavinderSingh13

30.1k41639

30.1k41639

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, could you please check this one also once and let me know if this works for you?
  
  – RavinderSingh13
  Jan 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  works fine! thanks
  
  – UKMonkey
  Jan 25 at 15:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, cool glad that it helped you, why I have specifically asked is no other solution have taken argument from script and compared it, so I wanted to make sure I understood question correctly, if you tested by passing argument then cool :)
  
  – RavinderSingh13
  Jan 25 at 15:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  it was the principle behind the replacement that I cared about - the full implementation such as how to plug in the string etc is a worry I didn't want to infect the question with.
  
  – UKMonkey
  Jan 25 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, well(if I get it correctly) if that is complete requirement then you need not to worry now, you have complete solution now :)
  
  – RavinderSingh13
  Jan 25 at 15:29
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, could you please check this one also once and let me know if this works for you?
  
  – RavinderSingh13
  Jan 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  works fine! thanks
  
  – UKMonkey
  Jan 25 at 15:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, cool glad that it helped you, why I have specifically asked is no other solution have taken argument from script and compared it, so I wanted to make sure I understood question correctly, if you tested by passing argument then cool :)
  
  – RavinderSingh13
  Jan 25 at 15:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  it was the principle behind the replacement that I cared about - the full implementation such as how to plug in the string etc is a worry I didn't want to infect the question with.
  
  – UKMonkey
  Jan 25 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @UKMonkey, well(if I get it correctly) if that is complete requirement then you need not to worry now, you have complete solution now :)
  
  – RavinderSingh13
  Jan 25 at 15:29
  
  
  

  
  
  
  

@UKMonkey, could you please check this one also once and let me know if this works for you?

– RavinderSingh13
Jan 25 at 15:09

@UKMonkey, could you please check this one also once and let me know if this works for you?

– RavinderSingh13
Jan 25 at 15:09

1

1

works fine! thanks

– UKMonkey
Jan 25 at 15:12

works fine! thanks

– UKMonkey
Jan 25 at 15:12

@UKMonkey, cool glad that it helped you, why I have specifically asked is no other solution have taken argument from script and compared it, so I wanted to make sure I understood question correctly, if you tested by passing argument then cool :)

– RavinderSingh13
Jan 25 at 15:18

@UKMonkey, cool glad that it helped you, why I have specifically asked is no other solution have taken argument from script and compared it, so I wanted to make sure I understood question correctly, if you tested by passing argument then cool :)

– RavinderSingh13
Jan 25 at 15:18

it was the principle behind the replacement that I cared about - the full implementation such as how to plug in the string etc is a worry I didn't want to infect the question with.

– UKMonkey
Jan 25 at 15:26

it was the principle behind the replacement that I cared about - the full implementation such as how to plug in the string etc is a worry I didn't want to infect the question with.

– UKMonkey
Jan 25 at 15:26

@UKMonkey, well(if I get it correctly) if that is complete requirement then you need not to worry now, you have complete solution now :)

– RavinderSingh13
Jan 25 at 15:29

@UKMonkey, well(if I get it correctly) if that is complete requirement then you need not to worry now, you have complete solution now :)

– RavinderSingh13
Jan 25 at 15:29

add a comment | 

1

Here is a shorter gnu awk approach:

cat incr.awk

{

   n = split($0, a, /[0-9]+/, b)

   for(i=1; i<n; i++)

      s = s a[i] b[i] + (b[i] < max && i == n-1 ? 1 : 0)

   print s a[i]

}

Then use it as:

awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 5stuff.thing'

/foo/bar/baz59_ 6stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 79stuff.thing'

/foo/bar/baz59_ 80stuff.thing





awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 90stuff.thing'

/foo/bar/baz59_ 90stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 80stuff.thing'

/foo/bar/baz59_ 80stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/stuff.thing'

/foo/bar/stuff.thing

share|improve this answer

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

add a comment | 

1

Here is a shorter gnu awk approach:

cat incr.awk

{

   n = split($0, a, /[0-9]+/, b)

   for(i=1; i<n; i++)

      s = s a[i] b[i] + (b[i] < max && i == n-1 ? 1 : 0)

   print s a[i]

}

Then use it as:

awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 5stuff.thing'

/foo/bar/baz59_ 6stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 79stuff.thing'

/foo/bar/baz59_ 80stuff.thing





awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 90stuff.thing'

/foo/bar/baz59_ 90stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 80stuff.thing'

/foo/bar/baz59_ 80stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/stuff.thing'

/foo/bar/stuff.thing

share|improve this answer

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

add a comment | 

1

1

1

Here is a shorter gnu awk approach:

cat incr.awk

{

   n = split($0, a, /[0-9]+/, b)

   for(i=1; i<n; i++)

      s = s a[i] b[i] + (b[i] < max && i == n-1 ? 1 : 0)

   print s a[i]

}

Then use it as:

awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 5stuff.thing'

/foo/bar/baz59_ 6stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 79stuff.thing'

/foo/bar/baz59_ 80stuff.thing





awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 90stuff.thing'

/foo/bar/baz59_ 90stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 80stuff.thing'

/foo/bar/baz59_ 80stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/stuff.thing'

/foo/bar/stuff.thing

share|improve this answer

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

Here is a shorter gnu awk approach:

cat incr.awk

{

   n = split($0, a, /[0-9]+/, b)

   for(i=1; i<n; i++)

      s = s a[i] b[i] + (b[i] < max && i == n-1 ? 1 : 0)

   print s a[i]

}

Then use it as:

awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 5stuff.thing'

/foo/bar/baz59_ 6stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 79stuff.thing'

/foo/bar/baz59_ 80stuff.thing





awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 90stuff.thing'

/foo/bar/baz59_ 90stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/baz59_ 80stuff.thing'

/foo/bar/baz59_ 80stuff.thing



awk -v max=80 -f incr.awk <<< '/foo/bar/stuff.thing'

/foo/bar/stuff.thing

share|improve this answer

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

share|improve this answer

share|improve this answer

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

answered Jan 25 at 15:28

anubhavaanubhava

532k47330408

532k47330408

add a comment | 

add a comment | 

1

An awk:

$ echo /foo/bar/baz59_ 99stuff.thing |

awk '

/[0-9]/ {

    rstart=1                                    # keep track of the start

    while(match(substr($0,rstart),/[0-9]+/)) {  # while numbers last

        rstart+=RSTART+RLENGTH-1                # increase rstart

        rlength=RLENGTH                         # remember length too

    }

    v=substr($0,rstart-rlength,rlength)+1                    # increase last number

    print substr($0,1,rstart-rlength-1) v substr($0,rstart)  # print in parts

    next

}1'                                             # in case there was no number

/foo/bar/baz59_ 100stuff.thing

Edit:

Whoops, I missed the argument requirement (increase the last number - - by a one, if it's greater than a script argument):

$ echo /foo/bar/baz59_ 99stuff.thing |

awk -v arg=100 '

/[0-9]/ {

    rstart=1

    while(match(substr($0,rstart),/[0-9]+/)) {

        rstart+=RSTART+RLENGTH-1

        rlength=RLENGTH

    }

    v=substr($0,rstart-rlength,rlength)

    if(0+v>arg) {                           # test if v greater that argument

        print substr($0,1,rstart-rlength-1) v+1 substr($0,rstart)

        next

    }

}1'

Output now:

/foo/bar/baz59_ 99stuff.thing

share|improve this answer

edited Jan 25 at 15:56

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated with the script argument requirement. Missed it at first.
  
  – James Brown
  Jan 25 at 15:58
  

  
  
  
  

add a comment | 

1

An awk:

$ echo /foo/bar/baz59_ 99stuff.thing |

awk '

/[0-9]/ {

    rstart=1                                    # keep track of the start

    while(match(substr($0,rstart),/[0-9]+/)) {  # while numbers last

        rstart+=RSTART+RLENGTH-1                # increase rstart

        rlength=RLENGTH                         # remember length too

    }

    v=substr($0,rstart-rlength,rlength)+1                    # increase last number

    print substr($0,1,rstart-rlength-1) v substr($0,rstart)  # print in parts

    next

}1'                                             # in case there was no number

/foo/bar/baz59_ 100stuff.thing

Edit:

Whoops, I missed the argument requirement (increase the last number - - by a one, if it's greater than a script argument):

$ echo /foo/bar/baz59_ 99stuff.thing |

awk -v arg=100 '

/[0-9]/ {

    rstart=1

    while(match(substr($0,rstart),/[0-9]+/)) {

        rstart+=RSTART+RLENGTH-1

        rlength=RLENGTH

    }

    v=substr($0,rstart-rlength,rlength)

    if(0+v>arg) {                           # test if v greater that argument

        print substr($0,1,rstart-rlength-1) v+1 substr($0,rstart)

        next

    }

}1'

Output now:

/foo/bar/baz59_ 99stuff.thing

share|improve this answer

edited Jan 25 at 15:56

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated with the script argument requirement. Missed it at first.
  
  – James Brown
  Jan 25 at 15:58
  

  
  
  
  

add a comment | 

1

1

1

An awk:

$ echo /foo/bar/baz59_ 99stuff.thing |

awk '

/[0-9]/ {

    rstart=1                                    # keep track of the start

    while(match(substr($0,rstart),/[0-9]+/)) {  # while numbers last

        rstart+=RSTART+RLENGTH-1                # increase rstart

        rlength=RLENGTH                         # remember length too

    }

    v=substr($0,rstart-rlength,rlength)+1                    # increase last number

    print substr($0,1,rstart-rlength-1) v substr($0,rstart)  # print in parts

    next

}1'                                             # in case there was no number

/foo/bar/baz59_ 100stuff.thing

Edit:

Whoops, I missed the argument requirement (increase the last number - - by a one, if it's greater than a script argument):

$ echo /foo/bar/baz59_ 99stuff.thing |

awk -v arg=100 '

/[0-9]/ {

    rstart=1

    while(match(substr($0,rstart),/[0-9]+/)) {

        rstart+=RSTART+RLENGTH-1

        rlength=RLENGTH

    }

    v=substr($0,rstart-rlength,rlength)

    if(0+v>arg) {                           # test if v greater that argument

        print substr($0,1,rstart-rlength-1) v+1 substr($0,rstart)

        next

    }

}1'

Output now:

/foo/bar/baz59_ 99stuff.thing

share|improve this answer

edited Jan 25 at 15:56

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

An awk:

$ echo /foo/bar/baz59_ 99stuff.thing |

awk '

/[0-9]/ {

    rstart=1                                    # keep track of the start

    while(match(substr($0,rstart),/[0-9]+/)) {  # while numbers last

        rstart+=RSTART+RLENGTH-1                # increase rstart

        rlength=RLENGTH                         # remember length too

    }

    v=substr($0,rstart-rlength,rlength)+1                    # increase last number

    print substr($0,1,rstart-rlength-1) v substr($0,rstart)  # print in parts

    next

}1'                                             # in case there was no number

/foo/bar/baz59_ 100stuff.thing

Edit:

Whoops, I missed the argument requirement (increase the last number - - by a one, if it's greater than a script argument):

$ echo /foo/bar/baz59_ 99stuff.thing |

awk -v arg=100 '

/[0-9]/ {

    rstart=1

    while(match(substr($0,rstart),/[0-9]+/)) {

        rstart+=RSTART+RLENGTH-1

        rlength=RLENGTH

    }

    v=substr($0,rstart-rlength,rlength)

    if(0+v>arg) {                           # test if v greater that argument

        print substr($0,1,rstart-rlength-1) v+1 substr($0,rstart)

        next

    }

}1'

Output now:

/foo/bar/baz59_ 99stuff.thing

share|improve this answer

edited Jan 25 at 15:56

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

share|improve this answer

share|improve this answer

edited Jan 25 at 15:56

edited Jan 25 at 15:56

edited Jan 25 at 15:56

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

answered Jan 25 at 15:10

James BrownJames Brown

19.9k42037

19.9k42037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated with the script argument requirement. Missed it at first.
  
  – James Brown
  Jan 25 at 15:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated with the script argument requirement. Missed it at first.
  
  – James Brown
  Jan 25 at 15:58
  

  
  
  
  

Updated with the script argument requirement. Missed it at first.

– James Brown
Jan 25 at 15:58

Updated with the script argument requirement. Missed it at first.

– James Brown
Jan 25 at 15:58

add a comment | 

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