Revision 05c4c718-4832-4569-a107-e3a5794ba9bf - Stack Overflow

Well, the first step for a robust solution would be to extract the domain from the string (using [`parse_url`](http://us3.php.net/manual/en/function.parse-url.php) ):

 $string = 'http://bla.ttt.something.com/blabla';
 $domain = parse_url($string, PHP_URL_HOST);

Then, assuming you want to match on the top two levels of the domain only (`something.com`), you need to extract that from the domain...

 $parts = explode('.', $domain);
 $top2parts = array_slice($parts, -2);
 $rootDomain = implode('.', $top2parts);

Then, we build the query:
 
 $sql = "SELECT * 
 FROM TestTable 
 WHERE domain LIKE '%".mysql_real_escape_string($rootDomain)."%'";

That should do it for you. For the input given, it'll generate:

 SELECT * 
 FROM TestTable 
 WHERE domain LIKE '%something.com%'
 
Of course you could do it the other way, and search by concating wildcards to the domain:

 SELECT * 
 FROM TestTable 
 WHERE CONCAT('%', domain, '%') LIKE 'http://bla.ttt.something.com/blabla'

But the problem with that is shown by the sample URL: `http://foo.bar/blah/something.com.html`... Which will return a false positive for `something.com` since it's in the URL, but it's not in the domain... So, one method of avoiding that (while still searching for the full domain) would be to do a combination:

 $string = 'http://bla.ttt.something.com/blabla';
 $domain = parse_url($string, PHP_URL_HOST);
 $sql = "SELECT * 
 FROM TestTable 
 WHERE CONCAT('%', domain, '%') LIKE '".mysql_real_escape_string($domain)."'";

Which for this case will generate:

 SELECT * 
 FROM TestTable 
 WHERE CONCAT('%', domain, '%') LIKE 'bla.ttt.something.com'