Linkify Regex Function PHP Daring Fireball Method

心已入冬 提交于 2019-11-28 00:23:27

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", "<a href=\"\\0\" rel=\"nofollow\" target=\"_blank\">\\0</a>", $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 '<a href="' . $url . '" rel="nofollow" target="_blank">' . "$input</a>";
}, $str); 

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

I was looking to just get the urls from a string using the same regex from the answer above by d_inevitable and wasn't looking to turn them into links or care about the rest of the string, I only wanted the urls with in the string so this is what I did. Hope it helps.

/**
 * Returns the urls in an array from a string.
 * This dos NOT return the string, only the urls with-in.
 */
function get_urls($str){

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

}
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