Linkify Regex Function PHP Daring Fireball Method

Question

So, I know there are a ton of related questions on SO, but none of them are quite what I\'m looking for. I\'m trying to implement a PHP function that will convert text URLs

Answer 1

Try this:

$pattern = '(?xi)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`\!()\[\]{};:\'".,<>?«»“”‘’]))';     
return preg_replace("!$pattern!i", "\\0", $str); 

PHP's preg function do need delimiters. The i at the end makes it case-insensitive

Update

If you use # as the delimiter, you wan't need to escape the ! in the pattern as such use the original pattern string (the pattern does not have a #): "#$pattern#i"

Update 2

To ensure that the links are correct, do this:

$pattern = '(?xi)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:\'".,<>?«»“”‘’]))';
return preg_replace_callback("#$pattern#i", function($matches) {
    $input = $matches[0];
    $url = preg_match('!^https?://!i', $input) ? $input : "http://$input";
    return '' . "$input";
}, $str); 

This will now append http:// to the urls so that browser doesn't think it is a relative link.