identify version from a string using php regex

Question

I want to get the version number for some string in php. Here is the sample code

<?php

    function getVersion($str) {
        preg_match("/.*((?:[0-9]+\.?)+)/i", $str, $matches);
        return $matches[1];
    }

    print_r(getVersion("ansitl1-isam-6.0 1.0 9.7.03 418614 +"));
    print_r(getVersion("ams-ef 9.6.06ef4 - 394867"));

?>

for input string ansitl1-isam-6.0 1.0 9.7.03 418614 + output should be 9.7.03 for input string ams-ef 9.6.06ef4 - 394867 output should be 9.6.06

How to achieve this?

Can you please let us know what you tried and what didn't worked? — R. Gurung
– R. Gurung, Commented Dec 4, 2019 at 10:06
I tried the below function getVersion($str) { preg_match("/.*((?:[0-9]+\.?)+)/i", $str, $matches); return $matches[1]; } — user12478765
– user12478765, Commented Dec 4, 2019 at 10:08
For pattern num.num.num preceded by a space try (?<= )\d+\.\d+\.\d+ — bobble bubble
– bobble bubble, Commented Dec 4, 2019 at 10:18
@bobble bubble Do we need to provide escape characters? I am getting error Unknown modifier '\\' — user12478765
– user12478765, Commented Dec 4, 2019 at 10:25
Oh you're using double quotes. Either use single quotes for pattern or double the backslash :) See this PHP demo — bobble bubble
– bobble bubble, Commented Dec 4, 2019 at 10:27

bobble bubble · Accepted Answer · 2019-12-04 10:44:23Z

1

If the pattern is always num.num.num preceded by a space.

(?<= )\d+\.\d+\.\d+

See this demo at Regex101 or a PHP demo at tio.run

There is not much Regex magic used here, just a lookbehind to check, there is a space before.
Instead it can also be done by a caturing group and getting $out[1] like in this demo.

answered Dec 4, 2019 at 10:44

bobble bubble

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