How to make an exception in regular expression?

Question

PHP

There is this code:

$foo = preg_replace('/\[url=([^\s]*)\]\[img(.*?)\]([^\s]*)\[\/img\]\[\/url\]/i', '[img$2]$3[/img]', $foo);

It removes links from images.
How to make it so that for a specific site it does not remove links from images? And removed for the rest.

Thanks in advance for the replies!

Answer 1

With lookahead (?!...)

$text = preg_replace(
    '#\[url=(?!https?://site\.ru)\S*\](\[img.*?\]\S*\[/img\])\[/url\]#i',
    '$1',
    $text
);

Or with a callback function

$white_list = ['site3.ru', 'site4.ru'];
$text = '
[url=https://site1.ru/page][img]https://site1.ru/file.jpg[/img][/url]
[url=https://site2.ru/page][img]https://site2.ru/file.jpg[/img][/url]
[url=https://site3.ru/page][img]https://site3.ru/file.jpg[/img][/url]
[url=https://site4.ru/page][img]https://site4.ru/file.jpg[/img][/url]
';

$text = preg_replace_callback(
    '#\[url=(\S*)\](\[img.*?\]\S*\[/img\])\[/url\]#i',
    function ($m) use($white_list) {
        $host = parse_url($m[1], PHP_URL_HOST);
        if (in_array($host, $white_list))
            return $m[0];
        
        return $m[2];
    },
    $text
);

sandbox.onlinephpfunctions.com/code/48cddfef32affd...

Answer 2

EM
IF SITE !== "Desired site"
CHECK