Currently I am using parse_url, however the host item of the array also includes the 'WWW' part which I do not want. How would I go about removing this?
$parse = parse_url($url);
print_r($parse);
$url = $parse['host'] . $parse['path'];
echo $url;
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3 Answers
$url = preg_replace('#^www\.(.+\.)#i', '$1', $parse['host']) . $parse['path'];
This won't remove the www in www.com, but www.www.com results in www.com.
1 Comment
Sergey Korzhov
your solution is good, but it miss the case like www.www.test.com. updated version - preg_replace('#^(?:www\.)+(.+\.)#i', '$1', $domain);
preg_replace('#^(http(s)?://)?w{3}\.#', '$1', $url);
if you don't need a protocol prefix, leave the second parameter empty
1 Comment
Empty
new version: preg_replace('#^(http(s)?://)?w{3}\.(\w+\.\w+)#', '$1$3', $url); Remove $1 for skipping protocol
$url = preg_replace('/^www\./i', '', $parse['host']) . $parse['path'];
