RE: [PHP] Parse domain from URL

> From: Brad Fuller [mailto:bfuller@cpacampaigns.com] > Sent: Wednesday, June 06, 2007 5:44 PM > Subject: [PHP] Parse domain from URL > > Hey guys, > > I'm faced with an interesting problem, and wondering if there's an easy > solution. > > I need to strip out a domain name from a URL, and ignore subdomains (like > www) > > I can use parse_url to get the hostname. And my first thought was to take > the last 2 segments of the hostname to get the domain. So if the URL is > http://www.example.com/ > Then the domain is "example.com." If the URL is ÐÂÌúN”F,ߟ=”# > http://example.org/ then > the domain is "example.org." > > This seemed to work perfectly until I come across a URL like > http://www.example.co.uk/ > My script thinks the domain is "co.uk." > > So I added a bit of code to account for this, basically if the 2nd to last > segment of the hostname is "co" then take the last 3 segments. > > Then I stumbled across a URL like http://www.example.com.au/ > > So it occurred to me that this is not the best solution, unless I have a > definitive list of all exceptions to go off of. > > Does anyone have any suggestions? > > Any advice is much appreciated. > > Thanks, > Brad > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > Maybe use some regexp like this: $url = http://www.a-domain.co.uk; $domain = preg_match( "!^(http://)([-a-z0-9]+(?:\.[-a-z0-9]+)*)$!i" ); // you will get www.a-domain.co.uk // So u can search for "www." In front of the string to replace it or // you could edit the search for ignoring the 'www.' Echo $domain[2]; Greetings, Stefan Note: this is untested.