A regex for version number parsing

Question

I have a version number of the following form   version release modification  where version  release and modification are either a set of digits or the     wildcard character  Additionally  any of these numbers  and any preceding    may be missing   So the following are valid and parse as   1 23 456   version 1  release 23  modification 456 1 23       version 1  release 23  any modification 1 23       version 1  release 23  any modification 1          version 1  any release  any modification 1          version 1  any release  any modification            any version  any release  any modification   But these are not valid     12  123 1 12  12   34   Can anyone provide me a not-too-complex regex to validate and retrieve the release  version and modification numbers

User · Answer

Specifying XSD elements    lt xs simpleType gt       lt xs restriction base  xs string  gt           lt xs pattern value   0-9  1 3    0-9  1 3    0-9  1 3           gt       lt  xs restriction gt   lt  xs simpleType gt

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

My 2 cents  I had this scenario  I had to parse version numbers out of a string literal   I know this is very different from the original question  but googling to find a regex for parsing version number showed this thread at the top  so adding this answer here   So the string literal would be something like   Service version 1 2 35 564 is running    I had to parse the 1 2 35 564 out of this literal  Taking a cue from  ajborley  my regex is as follows        d           d           d     d     A small C  snippet to test this looks like below   void Main         Regex regEx   new Regex        d           d           d     d     RegexOptions Compiled        Match version   regEx Match  The Service SuperService 2 1 309 0  is Running         version Value Dump  Version using RegEx         Prints 2 1 309 0

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

Sometimes version numbers might contain alphanumeric minor information  e g  1 2 0b or 1 2 0-beta   In this case I am using this regex    0-9  1 4     0-9a-z  1 6   1 5

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

I ve seen a lot of answers  but    i have a new one  It works for me at least  I ve added a new restriction  Version numbers can t start  major  minor or patch  with any zeros followed by others      01 0 0   is not valid   1 0 0    is valid   10 0 10  is valid    1 0 0000 is not valid        0      1-9    d            0      1-9    d          0   1-9    d        It s based in a previous one  But i see this solution better    for me     Enjoy

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

One more solution     1-9   d     1-9   d

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

My take on this  as a good exercise - vparse  which has a tiny source  with a simple function   function parseVersion v        var m   v match   d     d  g             v             major   m 0     0          minor   m 1     0          patch   m 2     0          build   m 3     0            v isEmpty    v major  amp  amp   v minor  amp  amp   v patch  amp  amp   v build      v parsed    v major  v minor  v patch  v build       v text   v parsed join           return v

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

This should work for what you stipulated   It hinges on the wild card position and is a nested regex            0-9             0-9             0-9

User · Answer

One more solution     1-9   d     1-9   d

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

For parsing version numbers that follow these rules   - Are only digits and dots  - Cannot start or end with a dot  - Cannot be two dots together  This one did the trick to me      d       1  d        0      Valid cases are   1  0 1  1 2 1

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

My take on this  as a good exercise - vparse  which has a tiny source  with a simple function   function parseVersion v        var m   v match   d     d  g             v             major   m 0     0          minor   m 1     0          patch   m 2     0          build   m 3     0            v isEmpty    v major  amp  amp   v minor  amp  amp   v patch  amp  amp   v build      v parsed    v major  v minor  v patch  v build       v text   v parsed join           return v

User · Answer

This should work for what you stipulated   It hinges on the wild card position and is a nested regex            0-9             0-9             0-9

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

For parsing version numbers that follow these rules   - Are only digits and dots  - Cannot start or end with a dot  - Cannot be two dots together  This one did the trick to me      d       1  d        0      Valid cases are   1  0 1  1 2 1

User · Answer

I ve seen a lot of answers  but    i have a new one  It works for me at least  I ve added a new restriction  Version numbers can t start  major  minor or patch  with any zeros followed by others      01 0 0   is not valid   1 0 0    is valid   10 0 10  is valid    1 0 0000 is not valid        0      1-9    d            0      1-9    d          0   1-9    d        It s based in a previous one  But i see this solution better    for me     Enjoy

User · Answer

One more solution     1-9   d     1-9   d

User · Answer

Specifying XSD elements    lt xs simpleType gt       lt xs restriction base  xs string  gt           lt xs pattern value   0-9  1 3    0-9  1 3    0-9  1 3           gt       lt  xs restriction gt   lt  xs simpleType gt

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

I had a requirement to search match for version numbers  that follows maven convention or even just single digit  But no qualifier in any case  It was peculiar  it took me time then I came up with this      0-9  0-9        This makes sure the version    Starts with a digit Can have any number of digit Only digits and     are allowed   One drawback is that version can even end with     But it can handle indefinite length of version  crazy versioning if you want to call it that   Matches    1 2 3 1 09 5 3 4 4 5 7 8 8  23 6 209 234 3   If you are not unhappy with     ending  may be you can combine with endswith logic

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

For parsing version numbers that follow these rules   - Are only digits and dots  - Cannot start or end with a dot  - Cannot be two dots together  This one did the trick to me      d       1  d        0      Valid cases are   1  0 1  1 2 1

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

Sometimes version numbers might contain alphanumeric minor information  e g  1 2 0b or 1 2 0-beta   In this case I am using this regex    0-9  1 4     0-9a-z  1 6   1 5

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

I found this  and it works for me             d x         d x         d x

User · Answer

My 2 cents  I had this scenario  I had to parse version numbers out of a string literal   I know this is very different from the original question  but googling to find a regex for parsing version number showed this thread at the top  so adding this answer here   So the string literal would be something like   Service version 1 2 35 564 is running    I had to parse the 1 2 35 564 out of this literal  Taking a cue from  ajborley  my regex is as follows        d           d           d     d     A small C  snippet to test this looks like below   void Main         Regex regEx   new Regex        d           d           d     d     RegexOptions Compiled        Match version   regEx Match  The Service SuperService 2 1 309 0  is Running         version Value Dump  Version using RegEx         Prints 2 1 309 0

User · Answer

For parsing version numbers that follow these rules   - Are only digits and dots  - Cannot start or end with a dot  - Cannot be two dots together  This one did the trick to me      d       1  d        0      Valid cases are   1  0 1  1 2 1

User · Answer

Sometimes version numbers might contain alphanumeric minor information  e g  1 2 0b or 1 2 0-beta   In this case I am using this regex    0-9  1 4     0-9a-z  1 6   1 5

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

I ve seen a lot of answers  but    i have a new one  It works for me at least  I ve added a new restriction  Version numbers can t start  major  minor or patch  with any zeros followed by others      01 0 0   is not valid   1 0 0    is valid   10 0 10  is valid    1 0 0000 is not valid        0      1-9    d            0      1-9    d          0   1-9    d        It s based in a previous one  But i see this solution better    for me     Enjoy

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

My 2 cents  I had this scenario  I had to parse version numbers out of a string literal   I know this is very different from the original question  but googling to find a regex for parsing version number showed this thread at the top  so adding this answer here   So the string literal would be something like   Service version 1 2 35 564 is running    I had to parse the 1 2 35 564 out of this literal  Taking a cue from  ajborley  my regex is as follows        d           d           d     d     A small C  snippet to test this looks like below   void Main         Regex regEx   new Regex        d           d           d     d     RegexOptions Compiled        Match version   regEx Match  The Service SuperService 2 1 309 0  is Running         version Value Dump  Version using RegEx         Prints 2 1 309 0

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

I had a requirement to search match for version numbers  that follows maven convention or even just single digit  But no qualifier in any case  It was peculiar  it took me time then I came up with this      0-9  0-9        This makes sure the version    Starts with a digit Can have any number of digit Only digits and     are allowed   One drawback is that version can even end with     But it can handle indefinite length of version  crazy versioning if you want to call it that   Matches    1 2 3 1 09 5 3 4 4 5 7 8 8  23 6 209 234 3   If you are not unhappy with     ending  may be you can combine with endswith logic

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

I found this  and it works for me             d x         d x         d x

User · Answer

I found this  and it works for me             d x         d x         d x

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

Thanks for all the responses  This is ace     Based on OneByOne s answer  which looked the simplest to me   I added some non-capturing groups  the       parts - thanks to VonC for introducing me to non-capturing groups    so the groups that do capture only contain the digits or   character          d           d           d      Many thanks everyone

User · Answer

My take on this  as a good exercise - vparse  which has a tiny source  with a simple function   function parseVersion v        var m   v match   d     d  g             v             major   m 0     0          minor   m 1     0          patch   m 2     0          build   m 3     0            v isEmpty    v major  amp  amp   v minor  amp  amp   v patch  amp  amp   v build      v parsed    v major  v minor  v patch  v build       v text   v parsed join           return v

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

This might work         d     d   0 2             At the top level      is a special case of a valid version number  Otherwise  it starts with a number  Then there are zero  one  or two   nn  sequences  followed by an optional       This regex would accept 1 2 3   which may or may not be permitted in your application   The code for retrieving the matched sequences  especially the     d   0 2  part  will depend on your particular regex library

User · Answer

I had a requirement to search match for version numbers  that follows maven convention or even just single digit  But no qualifier in any case  It was peculiar  it took me time then I came up with this      0-9  0-9        This makes sure the version    Starts with a digit Can have any number of digit Only digits and     are allowed   One drawback is that version can even end with     But it can handle indefinite length of version  crazy versioning if you want to call it that   Matches    1 2 3 1 09 5 3 4 4 5 7 8 8  23 6 209 234 3   If you are not unhappy with     ending  may be you can combine with endswith logic

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

I had a requirement to search match for version numbers  that follows maven convention or even just single digit  But no qualifier in any case  It was peculiar  it took me time then I came up with this      0-9  0-9        This makes sure the version    Starts with a digit Can have any number of digit Only digits and     are allowed   One drawback is that version can even end with     But it can handle indefinite length of version  crazy versioning if you want to call it that   Matches    1 2 3 1 09 5 3 4 4 5 7 8 8  23 6 209 234 3   If you are not unhappy with     ending  may be you can combine with endswith logic

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

Specifying XSD elements    lt xs simpleType gt       lt xs restriction base  xs string  gt           lt xs pattern value   0-9  1 3    0-9  1 3    0-9  1 3           gt       lt  xs restriction gt   lt  xs simpleType gt

User · Answer

My 2 cents  I had this scenario  I had to parse version numbers out of a string literal   I know this is very different from the original question  but googling to find a regex for parsing version number showed this thread at the top  so adding this answer here   So the string literal would be something like   Service version 1 2 35 564 is running    I had to parse the 1 2 35 564 out of this literal  Taking a cue from  ajborley  my regex is as follows        d           d           d     d     A small C  snippet to test this looks like below   void Main         Regex regEx   new Regex        d           d           d     d     RegexOptions Compiled        Match version   regEx Match  The Service SuperService 2 1 309 0  is Running         version Value Dump  Version using RegEx         Prints 2 1 309 0

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

Specifying XSD elements    lt xs simpleType gt       lt xs restriction base  xs string  gt           lt xs pattern value   0-9  1 3    0-9  1 3    0-9  1 3           gt       lt  xs restriction gt   lt  xs simpleType gt

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

This matches 1 2 3   too          d    d   0 2           I would propose the less elegant       d    d           d   d   d

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

Sometimes version numbers might contain alphanumeric minor information  e g  1 2 0b or 1 2 0-beta   In this case I am using this regex    0-9  1 4     0-9a-z  1 6   1 5

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

Don t know what platform you re on but in  NET there s the System Version class that will parse  n n n n  version numbers for you

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

It seems pretty hard to have a regex that does exactly what you want  i e  accept only the cases that you need and reject all others and return some groups for the three components   I ve give it a try and come up with this          d      d      d                   IMO  I ve not tested extensively  this should work fine as a validator for the input  but the problem is that this regex doesn t offer a way of retrieving the components  For that you still have to do a split on period   This solution is not all-in-one  but most times in programming it doesn t need to  Of course this depends on other restrictions that you might have in your code

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

User · Answer

I d express the format as       1-3 dot-separated components  each numeric except that the last one may be      As a regexp  that s       d       d          d       Edit to add  this solution is a concise way to validate  but it has been pointed out that extracting the values requires extra work  It s a matter of taste whether to deal with this by complicating the regexp  or by processing the matched groups    In my solution  the groups capture the     characters  This can be dealt with using non-capturing groups as in ajborley s answer   Also  the rightmost group will capture the last component  even if there are fewer than three components  and so for example a two-component input results in the first and last groups capturing and the middle one undefined  I think this can be dealt with by non-greedy groups where supported   Perl code to deal with both issues after the regexp could be something like this    version        groups     1   2   3   foreach   groups        next if  defined      s           push  version          major   minor   mod      version               Which isn t really any shorter than splitting on

User · Answer

This should work for what you stipulated   It hinges on the wild card position and is a nested regex            0-9             0-9             0-9

User · Answer

Use regex and now you have two problems  I would split the thing on dots        then make sure that each part is either a wildcard or set of digits  regex is perfect now   If the thing is valid  you just return correct chunk of the split

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

User · Answer

This should work for what you stipulated   It hinges on the wild card position and is a nested regex            0-9             0-9             0-9

User · Answer

I found this  and it works for me             d x         d x         d x

User · Answer

ms       d                d               d              d      Does exactly match your 6 first examples  and rejects the 4 others   group 1  major or major minor or     group 2 if exists  minor or   group 3 if exists      You can remove    ms   I used it to indicate to this regexp to be applied on multi-lines through QuickRex

User · Answer

I ve seen a lot of answers  but    i have a new one  It works for me at least  I ve added a new restriction  Version numbers can t start  major  minor or patch  with any zeros followed by others      01 0 0   is not valid   1 0 0    is valid   10 0 10  is valid    1 0 0000 is not valid        0      1-9    d            0      1-9    d          0   1-9    d        It s based in a previous one  But i see this solution better    for me     Enjoy

User · Answer

Another try        d        d        d         This gives the three parts in groups 4 5 6 BUT  They are aligned to the right  So the first non-null one of 4 5 or 6 gives the version field    1 2 3 gives 1 2 3 1 2   gives 1 2   1 2 gives null 1 2     gives null null   1   gives null 1

User · Answer

One more solution     1-9   d     1-9   d

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

Keep in mind regexp are greedy  so if you are just searching within the version number string and not within a bigger text  use   and   to mark start and end of your string  The regexp from Greg seems to work fine  just gave it a quick try in my editor   but depending on your library language the first part can still match the     within the wrong version numbers  Maybe I am missing something  as I haven t used Regexp for a year or so   This should make sure you can only find correct version numbers         d     d              edit  actually greg added them already and even improved his solution  I am too slow

User · Answer

d           d           d      Perhaps a more concise one could be          d      0 2      d      This can then be enhanced to 1 2 3 4 5   or restricted exactly to X Y Z using   or  2  instead of  0 2

User · Answer

My take on this  as a good exercise - vparse  which has a tiny source  with a simple function   function parseVersion v        var m   v match   d     d  g             v             major   m 0     0          minor   m 1     0          patch   m 2     0          build   m 3     0            v isEmpty    v major  amp  amp   v minor  amp  amp   v patch  amp  amp   v build      v parsed    v major  v minor  v patch  v build       v text   v parsed join           return v

User · Answer

I tend to agree with split suggestion   Ive created a  tester  for your problem in perl     usr bin perl -w    strings      1 2 3    1 2      1             regexp     svrist   gt  qr      d      d      d     d      d     d                            onebyone   gt  qr    d       d          d                  greg   gt  qr       d     d   0 2                         vonc   gt  qr       d                d               d              d                  ajb   gt  qr       d           d           d                  jrudolph   gt  qr      d        d        d                        foreach my  r  keys  regexp       my  reg    regexp  r       print  Using  r regexp n   foreach my  s   strings     print   s           if   s   m  reg        my   main   maj   min  rev  ex1  ex2  ex3      any   any   any   any   any   any   any         main    1 if   1  amp  amp   1 ne             maj    2 if   2  amp  amp   2 ne             min    3 if   3  amp  amp   3 ne             rev    4 if   4  amp  amp   4 ne             ex1    5 if   5  amp  amp   5 ne             ex2    6 if   6  amp  amp   6 ne             ex3    7 if   7  amp  amp   7 ne            print   main  maj  min  rev  ex1  ex2  ex3 n       else    print   nomatch n           print  ------------------------ n       Current output    gt  perl regex pl Using onebyone regexp 1 2 3   1  2  3 any any any any 1 2     1  2  any any any any any 1     1  any any any any any any     any any any any any any any ------------------------ Using svrist regexp 1 2 3   1 2 3 any any any any 1 2     any any any 1 2 any any 1     any any any any any 1 any     any any any any any any any ------------------------ Using vonc regexp 1 2 3   1 2  3 any any any any any 1 2     1  2    any any any any 1     any any any 1 any any any     any any any any any any any ------------------------ Using ajb regexp 1 2 3   1 2 3 any any any any 1 2     1 2 any any any any any 1     1 any any any any any any     any any any any any any any ------------------------ Using jrudolph regexp 1 2 3   1 2  1  1 2 3 any any 1 2     1 2  1  1 2 any any any 1     1  any any 1 any any any     any any any any any any any ------------------------ Using greg regexp 1 2 3   1 2 3  3 any any any any any 1 2     1 2    2    any any any any 1     1   any    any any any any     any any any any any any any ------------------------

[regex] A regex for version number parsing

Examples related to regex

Examples related to versioning