PHP Get version from url

Question

How can I download the versions from the link? I tried to do it with regEx but it didn't work as I wanted.

I wish it would work like this (Where I gave NULL, there is nothing to choose from)

https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js > 1.11.2
https://ajax.googleapis.com/ajax/libs/jquery/v1.11.2/jquery.min.js > 1.11.2
https://ajax.googleapis.com/ajax/libs/jquery/version-1.11.2/jquery.min.js > 1.11.2
https://ajax.googleapis.com/ajax/libs/jquery/1-11-2/jquery.min.js > NULL
https://ajax.googleapis.com/ajax/libs/jquery/1112/jquery.min.js > NULL
https://www.google.com/recaptcha/api.js?render=6LdBsjgVaoTJ8rC-Npzz16bnAE > NULL

My regEx: /([0-9]+\.?)+/

Thank you in advance for your help :)

What do you exactly want to do? please clarify your problem. — Adison Masih
– Adison Masih, Commented Jun 8, 2021 at 17:46
@AdisonMasih I would like to get versions from links as shown on the example of several links. I gave the arrow to illustrate what to download from the link — Kondzio
– Kondzio, Commented Jun 8, 2021 at 17:51
My current regEx does not work as it needs because it selects every number and not so many which should be because, for example, when I give a link to recaptcha, it selects values from the key and should not return anything — Kondzio
– Kondzio, Commented Jun 8, 2021 at 17:53

Wiktor Stribiżew · Accepted Answer · 2021-06-08 17:50:03Z

2

You can use

\/(?:v(?:ersion)?-?)?\K\d+(?:\.\d+)+

See the regex demo. NOTE: if you want to make sure after the version number there is a / or end of string, add a (?![^\/]) lookahead at the end of the pattern, \/(?:v(?:ersion)?-?)?\K\d+(?:\.\d+)+(?![^\/]) (see this regex demo).

Details:

\/ - a / char
(?:v(?:ersion)?-?)? - an optional sequence of a v and then an optional ersion and then an optional - char - \K - omit the matched text
\d+(?:\.\d+)+ - match and consumer one or more digits and then one or more sequences of a dot and one or more digits.

answered Jun 8, 2021 at 17:50

Wiktor Stribiżew

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