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 from a user-generated post into links. I'm using the 'improved' Regex from Daring Fireball towards the bottom of the page: http://daringfireball.net/2010/07/improved_regex_for_matching_urls The function does not return anything, and I'm not sure why.
<?php
if ( false === function_exists('linkify') ):
function linkify($str) {
$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, "<a href=\"\\0\" rel=\"nofollow\" target=\"_blank\">\\0</a>", $str);
}
endif;
?>
Can someone please help me get this to work? Thanks!
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>", $str);
PHP's preg
function do need delimiters. The i
at the end makes it case-insensitive
If you use #
as the delimiter, you wan't need to escape the !
in the pattern as such use the original pattern string (the pattern does not have a #
): "#$pattern#i"
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.
http://
prefix in the href
attribute. I don't think you can do this with single function call. Try preg_replace_callback
and an if-statement that will prepend http://
when necessary - d_inevitable 2012-04-03 22:40
www.google.com
and https://www.google.com
but http://stackoverflow.com
ends up as http://http//stackoverflow.com/
Jeff 2012-04-03 23:01
$url = preg_match('!^http?s://!i', $input) ? $input : "http://$input";
should be changed to $url = preg_match('!^https?://!i', $input) ? $input : "http://$input";
The question mark just had to be moved one spot over - Jeff 2012-04-03 23:36
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;
}
error_reporting
would have told you instantly - mario 2012-04-03 22:29