返回URL preg_match的特定部分

Question

Imagine you have this cases

$d='http://www.example.com/';
$d1='http://example.com/';
$d2='http://www.example.com';
$d3='www.example.com/';
$d4='http://www.example.com/';
$d5='http://www.example.com/blabla/blabla.php';

I need to get only example.com and nothing else. I've tried using parse_url to no avail.

Using parse_url($d1, PHP_URL_HOST); returns nothing in $d3, for example.

Can any of you provide a ereg to match this?

Thank you very much in advance!

Answer 1

There is no path_url function, but you can use the parse_url function to get the host (domain name) out of a URL string:

if(!preg_match('#^https?://#', $str))
{
    $domain = 'http://' . $domain;
}

$domain = parse_url($str, PHP_URL_HOST);