正则表达式-匹配字符，但不包含在结果中

Question

I have got the following Regex, which ALMOST works...

(?:^https?:\/\/)(?:www|[a-z]+)\.([^.]+)

I need the result to be the only result, or within the same position in the Array.

So for example this http://m.facebook.com/ matches perfect, there is only 1 group.

However, if I change it to http://facebook.com/ then I get com/ in place of where Facebook should be. So I need to have (?:www|[az]+) as an optional check really.

Edit:

What I expect is just to match facebook , if ANY of the strings are as follows:

http://www.facebook.com

http://facebook.com

http://m.facebook.com

And obviously the https counterparts.

This is my Regex now

(?:^https?:\/\/)(?:www)?\.?([^.]+)

This is close, however it matches the m on when I try ` http://m.facebook.com

https://regex101.com/r/GDapY5/1

Answer 1

So I need to have (?:www|[az]+) as an optional check really.

A ? at the end of a pattern is generally used for "optional" bits -- it means "match zero or one" of that thing, so your subpattern would be something like this:

(?:www|[a-z]+)?

If you're simply trying to get the second level domain, I wouldn't bother with regex, because you'll be constantly adjusting it to handle special cases you come across. Just split on dots and take the penultimate value:

$domain = array_reverse(explode('.', parse_url($str)['host']))[1];

Or:

$domain = array_reverse(explode('.', parse_url($str, PHP_URL_HOST)))[1];

Answer 2

Perhaps you could make the first m. part optional with (?:\\w+\\.)? . Instead of a capturing group you could use \\K to reset the starting point of the reported match.

Then match one or more word characters \\w+ and use a positive lookahead to assert that what follows is a dot (?=\\.)

For example:

^https?://(?:www)?(?:\\w+\\.)?\\K\\w+(?=\\.)

Edit: Or you could match for m. or www. using an alternation:

^https?://(?:m\\.|www\\.)?\\K\\w+(?=\\.)

Demo Php