PHP preg_replace weirdness with custom urls

Question

LS,

I'm using the following code to add <span> tags behind <a> tags.

$html = preg_replace("~<a.*?href=\"$url\".*?>.*?</a>~i", "$0<span>test</span>", $html);

The code is working fine for regular links (ie. http://www.google.com/), but it will not perform a replace when the contents of $url are $link$/3/.

This is example code to show the (mis)behaviour:

<?php
    $urls = array();
    $urls[] = '$link$/3/';
    $urls[] = 'http://www.google.com/';

    $html = '<a href="$link/3/">Test Link</a>' . "\n" . '<a href="http://www.google.com/"&gt;Google&lt;/a&gt;';

    foreach($urls as $url) {
        $html = preg_replace("~<a.*?href=\"$url\".*?>.*?</a>~i", "$0<span>test</span>", $html);
    }

    echo $html;
?>

And this is the output it produces:

<a href="$link$/3/">Test Link</a>
<a href="http://www.google.com/"&gt;Google&lt;/a&gt;&lt;span&gt;test&lt;/span&gt;

Kind regards,

Matthias Vance

Answer 1

$url = preg_quote($url, '~'); the dollar signs are interpreted as usual: end-of-input.

Answer 2

$ has special meaning in regex. End of line. Your expression is being expanded like this:

$html = preg_replace("~<a.*?href=\"$link$/3/\".*?>.*?</a>~i", "$0<span>test</span>", $html);

Which fails because it can't find "link" between two end of lines. Try escaping the $ in the $urls array:

$urls[] = '\$link\$/3/';

Answer 3

$ is considered a special regex character and needs to be escaped. Use preg_quote() to escape $url before passing it to preg_replace().

$url = preg_quote($url, '~');

Answer 4

just somebody is correct; you must escape your special regex characters if you mean for them to be interpreted as literal.

It also looks to me like it can't perform the replace because it never makes a match.

Try replacing this line:

$urls[] = '$link$/3/';

With this:

$urls[] = '$link/3/';