匹配没有无效字符的URL的PHP​​正则表达式

Question

I'm looking for a Regex that will find URLs in a string but ignore pre/following characters which are not part of the URL.

for example, from the string:

example.co.uk (main site: example.com),

The Regex will find:

example.co.uk and exaple.com.

In order to find URLs within a given string, I use the Regex '#(www\.|https?://)?[a-z0-9]+\.[a-z0-9]{2,4}\S*#i'. The problem is that if I use this regex with the given string above, it finds example.co.uk and example.com) with the closing bracket at the end.

Is there any Regex that can find URLs in a string, not matter what characters it has from both sides?

Thanks!

Answer 1

You may have to use a word boundary (\b) ...

(?:www\.|https?:\/\/)?[a-z0-9]+\.[a-z0-9]{2,4}\S*\b
                                                  ^

regex demo