A regex for version number parsing

RegexVersioning

Regex Problem Overview


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?

Regex Solutions


Solution 1 - Regex

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 "." ]

Solution 2 - Regex

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.

Solution 3 - Regex

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!

Solution 4 - Regex

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.

Solution 5 - Regex

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    	
}

Solution 6 - Regex

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.

Solution 7 - Regex

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,

  1. Starts with a digit
  2. Can have any number of digit
  3. 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

Solution 8 - Regex

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 => qr/(?:(\d+)\.(\d+)\.(\d+)|(\d+)\.(\d+)|(\d+))?(?:\.\*)?/,
            onebyone => qr/^(\d+\.)?(\d+\.)?(\*|\d+)$/,
            greg => qr/^(\*|\d+(\.\d+){0,2}(\.\*)?)$/,
            vonc => qr/^((?:\d+(?!\.\*)\.)+)(\d+)?(\.\*)?$|^(\d+)\.\*$|^(\*|\d+)$/,
            ajb => qr/^(?:(\d+)\.)?(?:(\d+)\.)?(\*|\d+)$/,
            jrudolph => 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 && $1 ne "*") ;
    $maj = $2 if ($2 && $2 ne "*") ;
    $min = $3 if ($3 && $3 ne "*") ;
    $rev = $4 if ($4 && $4 ne "*") ;
    $ex1 = $5 if ($5 && $5 ne "*") ;
    $ex2 = $6 if ($6 && $6 ne "*") ;
    $ex3 = $7 if ($7 && $7 ne "*") ;
    print "$main $maj $min $rev $ex1 $ex2 $ex3\n";

  }else{
  print " nomatch\n";
  }
  }
print "------------------------\n";
}

Current output:

> 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
------------------------

Solution 9 - Regex

^(?:(\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}

Solution 10 - Regex

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!!!

Solution 11 - Regex

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]+)))?)))?))$

http://imgur.com/3E492.png

Solution 12 - Regex

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

Solution 13 - Regex

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,*

Solution 14 - Regex

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 && !v.minor && !v.patch && !v.build;
    v.parsed = [v.major, v.minor, v.patch, v.build];
    v.text = v.parsed.join('.');
    return v;
}

Solution 15 - Regex

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})

Solution 16 - Regex

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 :)

Solution 17 - Regex

(?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

Solution 18 - Regex

This matches 1.2.3.* too

>^(*|\d+(.\d+){0,2}(.*)?)$

I would propose the less elegant:

(*|\d+(.\d+)?(.*)?)|\d+.\d+.\d+)

Solution 19 - Regex

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.

Solution 20 - Regex

Specifying XSD elements:

<xs:simpleType>
	<xs:restriction base="xs:string">
		<xs:pattern value="[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}(\..*)?"/>
	</xs:restriction>
</xs:simpleType>

Solution 21 - Regex

One more solution:

^[1-9][\d]*(.[1-9][\d]*)*(.\*)?|\*$

Solution 22 - Regex

I found this, and it works for me:

/(\^|\~?)(\d|x|\*)+\.(\d|x|\*)+\.(\d|x|\*)+

Solution 23 - Regex

/^([1-9]{1}\d{0,3})(\.)([0-9]|[1-9]\d{1,3})(\.)([0-9]|[1-9]\d{1,3})(\-(alpha|beta|rc|HP|CP|SP|hp|cp|sp)[1-9]\d*)?(\.C[0-9a-zA-Z]+(-U[1-9]\d*)?)?(\.[0-9a-zA-Z]+)?$/
  • A normal version: ([1-9]{1}\d{0,3})(\.)([0-9]|[1-9]\d{1,3})(\.)([0-9]|[1-9]\d{1,3})
  • A Pre-release or patched version: (\-(alpha|beta|rc|EP|HP|CP|SP|ep|hp|cp|sp)[1-9]\d*)? (Extension Pack, Hotfix Pack, Coolfix Pack, Service Pack)
  • Customized version: (\.C[0-9a-zA-Z]+(-U[1-9]\d*)?)?
  • Internal version: (\.[0-9a-zA-Z]+)?

enter image description here

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAndrew BorleyView Question on Stackoverflow
Solution 1 - RegexSteve JessopView Answer on Stackoverflow
Solution 2 - RegexPaweł HajdanView Answer on Stackoverflow
Solution 3 - RegexAndrew BorleyView Answer on Stackoverflow
Solution 4 - RegexGreg HewgillView Answer on Stackoverflow
Solution 5 - RegexSudhanshu MishraView Answer on Stackoverflow
Solution 6 - RegexDuncan SmartView Answer on Stackoverflow
Solution 7 - RegexShivaView Answer on Stackoverflow
Solution 8 - RegexsvristView Answer on Stackoverflow
Solution 9 - RegexofauraxView Answer on Stackoverflow
Solution 10 - RegexIsrael RomeroView Answer on Stackoverflow
Solution 11 - RegexnomuusView Answer on Stackoverflow
Solution 12 - RegexPau BalladaView Answer on Stackoverflow
Solution 13 - RegexjrudolphView Answer on Stackoverflow
Solution 14 - Regexvitaly-tView Answer on Stackoverflow
Solution 15 - RegexMarc RuefView Answer on Stackoverflow
Solution 16 - RegexFrankSView Answer on Stackoverflow
Solution 17 - RegexVonCView Answer on Stackoverflow
Solution 18 - RegexVictorView Answer on Stackoverflow
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Solution 20 - RegexEmmersonView Answer on Stackoverflow
Solution 21 - RegexOleksandr YarushevskyiView Answer on Stackoverflow
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Solution 23 - Regex山茶树和葡萄树View Answer on Stackoverflow