PHP Regex to Remove http:// from string - php

Question

I have full URLs as strings, but I want to remove the http:// at the beginning of the string to display the URL nicely (ex: www.google.com instead of http://www.google.com)
Can someone help?

Answer 1

$str = 'http://www.google.com';
$str = preg_replace('#^https?://#', '', $str);
echo $str; // www.google.com
That will work for both http:// and https://

Answer 2

You don't need regular expression at all. Use str_replace instead.
str_replace('http://', '', $subject);
str_replace('https://', '', $subject);
Combined into a single operation as follows:
str_replace(array('http://','https://'), '', $urlString);

Answer 3

Better use this:
$url = parse_url($url);
$url = $url['host'];
echo $url;
Simpler and works for http:// https:// ftp:// and almost all prefixes.

Answer 4

Why not use parse_url instead?

Answer 5

To remove http://domain ( or https ) and to get the path:
$str = preg_replace('#^https?\:\/\/([\w*\.]*)#', '', $str);
echo $str;

Answer 6

If you insist on using RegEx:
preg_match( "/^(https?:\/\/)?(.+)$/", $input, $matches );
$url = $matches[0][2];

Answer 7

Yeah, I think that str_replace() and substr() are faster and cleaner than regex. Here is a safe fast function for it. It's easy to see exactly what it does. Note: return substr($url, 7) and substr($url, 8), if you also want to remove the //.
// slash-slash protocol remove https:// or http:// and leave // - if it's not a string starting with https:// or http:// return whatever was passed in
function universal_http_https_protocol($url) {
// Breakout - give back bad passed in value
if (empty($url) || !is_string($url)) {
return $url;
}
// starts with http://
if (strlen($url) >= 7 && "http://" === substr($url, 0, 7)) {
// slash-slash protocol - remove https: leaving //
return substr($url, 5);
}
// starts with https://
elseif (strlen($url) >= 8 && "https://" === substr($url, 0, 8)) {
// slash-slash protocol - remove https: leaving //
return substr($url, 6);
}
// no match, return unchanged string
return $url;
}

Answer 8

<?php
// (PHP 4, PHP 5, PHP 7)
// preg_replace — Perform a regular expression search and replace
$array = [
'https://lemon-kiwi.co',
'http://lemon-kiwi.co',
'lemon-kiwi.co',
'www.lemon-kiwi.co',
];
foreach( $array as $value ){
$url = preg_replace("(^https?://)", "", $value );
}
This code output :
lemon-kiwi.co
lemon-kiwi.co
lemon-kiwi.co
www.lemon-kiwi.co
See documentation PHP preg_replace