Question

I have a bunch of urls like these.

  $urls = array(
    'https://site1.com',
    'https://www.site2.com',
    'http://www.site3.com',
    'https://site4.com',
    'site5.com',
    'www.site6.com',
    'www.site7.co.uk',
    'site8.tk'
  );

I wanted to remove the http, https, :// and www. from these strings so that the output will look like these.

  $urls = array(
    'site1.com',
    'site2.com',
    'site3.com',
    'site4.com',
    'site5.com',
    'site6.com',
    'site7.co.uk',
    'site8.tk'
  );

I came up with this solution.

foreach ($urls as $url) {
   $pattern = '/(http[s]?:\/\/)?(www\.)?/i';
   $replace = "";
   echo "before: $url after: ".preg_replace('/\/$/', '', preg_replace($pattern, $replace, $url))."\n";
}

I was wondering how I could avoid the second preg_replace. Any ideas?

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);