Question

I have been looking around for a regex expression that will spit out just the stackoverflow part and no www. or .com etc. All I could find was to check if the url s were valid... I have used php s url filter to determine that much I now am looking to determine which site it is.

I have never written an expression before so I am hoping someone can check it/recommend a better solution.

preg_match( @^(?:http://)(?:www.)?([^.]+)@i , $url, $matches)

edit: All the url s I am dealing with are .com if that helps

Answer 1

one-liner without using regular expressions!!!

$url =  http://stackoverflow.com ;
$d = array_shift( explode(  . , str_replace( www. ,   , parse_url( $url, PHP_URL_HOST )) ) );
echo $d;

Answer 2

Have you considered PHP s parse_url() function?

Answer 3

If you know that the thing you re having is a URL and the domain end in .com a simple thing like

preg_match( #([^.]*).com# , $url, $matches);

should do the trick.

While this will fail on a domain like www.com.foo.com but might be enough, depending on your situation.

Answer 4

preg_match( #https?://(?:www.)?(.*?).com#i , $str, $match);

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