Regex for dropping http:// and www. from URLs

Question

I have a bunch of urls like these.

  $urls = array(
    \'https://site1.com\',
    \'https://www.site2.com\',
    \'http://www.site3.com\',
    \'https://sit         


        

Answer 1

preg_replace can also take an array, so you don't even need the loop. You can do this with a one liner:

$urls = preg_replace('/(?:https?:\/\/)?(?:www\.)?(.*)\/?$/i', '$1', $urls);

Answer 2

Short and sweet:

$urls = preg_replace('~^(?:https?://)?(?:www[.])?~i', '', $urls);

Answer 3

Depending on what exactly it is you want to do, it might be better to stick with PHP's own URL parsing facilities, namely parse_url:

foreach ($urls as &$url) {
    $url = preg_replace('~^www.~', '', parse_url($url, PHP_URL_HOST));
}
unset($url);

parse_url will give you the host of the URL, even if it will contain a port number or HTTP authentication data. (Whether this is what you need, depends on your exact use case though.)

Answer 4

/^(https?:\/\/)?(www\.)?(.*)\/$/i

And use what's on $3. Or, even better, change the first two parentheses to the non-capturing version (?:) and use what's on 1.