I have a variable, such as this:
$domain = "http://test.com"
I need to use preg_replace or str_place to get the variable like this:
$domain = "test.com"
I have tried using the following, but they do not work.
1) $domain = preg_replace('; ((ftp|https?)://|www3?\.).+? ;', ' ', $domain);
2) $domain = preg_replace(';\b((ftp|https?)://|www3?\.).+?\b;', ' ', $domain);
Any suggestions?
$domain = ltrim($domain, "http://");
Did you try the str_replace?
$domain = "http://test.com"
$domain = str_replace('http://','',$domain);
You regular expressions probably don't find a match for the pattern.
preg_replace('~(https://|http://|ftp://)~',, '', $domain);
preg_match('/^[a-z]+:[/][/](.+)$/', $domain, $matches);
echo($matches[1]);
Should be what you are looking for, should give you everything after the protocol... http://domain.com/test becomes "domain.com/test". However, it doesn't care about the protocol, if you only want to support specific protocols such as HTTP and FTP, then use this instead:
preg_match('/^(http|ftp):[/][/](.+)$/', $domain, $matches);
If you only want the domain though, or similar parts of the URI, I'd recommend PHP's parse_url() instead. It does all the hard work for you and does it the proper way. Depending on your needs, I would probably recommend you use it anyway and just put it all back together instead.
simple regex:
preg_replace('~^(?:f|ht)tps?://~i','', 'https://www.site.com.br');
