PHP: replace relative top URL "../" with absolute domain URL - php

Question

I want to convert relative URLs that starts with ../stuff/more.php to http://www.example.com/stuff/more.php in my RSS feed.
I used this PHP code to do so is the following:
$content = preg_replace("#(<\s*a\s+[^>]*href\s*=\s*[\"'])(?!http)([^\"'>]+)([\"'>]+)#", '$1http://www.example.com/$2$3', $content);
The result is wrong thought, it returns the URL like this
http://www.example.com/../stuff/more.php
Notice the ../ part hasn't been removed, please help!
So Basically..
This what I have: ../stuff/more.php
This is what I get (after running the code above): http://www.example.com/../stuff/more.php
This what I WANT: http://www.example.com/stuff/more.php

Answer 1

Adding (\.|\.\.|\/)* should work.
$content = preg_replace("#(<\s*a\s+[^>]href\s=\s*[\"'])(?!http)(../|../|/)*([^\"'>]+)([\"'>]+)#", '$1http://www.example.com/$3$4', $content);
Also, note $2$3 has been changed to $3$4
Edit:
Reduced to one alternative:
$content = preg_replace("#(<\s*a\s+[^>]*href\s*=\s*[\"'])(?!http)(\.\.\/)*([^\"'>]+)([\"'>]+)#", '$1http://www.example.com/$3$4', $content);

Answer 2

Why don't you just replace the first 2 dots with the domain?
$result = str_replace('..', 'http://www.example.com', $contet, 1);

Answer 3

Use $_SERVER[HTTP_HOST] $_SERVER[REQUEST_URI] is the global variable in PHP to get the absolute url.

Answer 4

Well, I'll start looking at the regex. Most of it looks good (in fact, you've got a good enough regex here I'm a little surprised you're having trouble otherwise!) but the end is a bit weird -- better like this:
#(<\s*a\s+[^>]*href\s*=\s*[\"'])(?!http)([^\"'>]+)([\"']>)#
(Technically it would be better to capture the starting quote and make sure it's a matching ending quote, but chances are you won't have any problems there.
To remove the ../ I would do it apart from regex entirely:
foreach (array("<a href=\"http://../foo/bar\">",
"<a href=\"../foo/bar\">") as $content) {
echo "A content=$content<br />\n";
########## copy from here down to...
if (preg_match("#(<\s*a\s+[^>]*?href\s*=\s*[\"'])(?!http)([^\"'>]+)([\"']>)#", $content, $m)) {
echo "m=<pre>".print_r($m,true)."</pre><br />\n";
if (substr($m[2], 0, 3) == '../')
$m[2] = substr($m[2], 3);
$content = $m[1].'http://www.example.com/'.$m[2].$m[3];
}
######### copy from above down to HERE
echo "B content=$content<br />\n";
}
(I included a mini-test suite around what you're looking for - you will need to take just the marked lines inside for your code.)

Answer 5

I found the solution thanks to everyone who helped me on this.
Here's the code I used:
$content = preg_replace("#(<a href=\"\.\.\/)#", '<a href="http://www.example.com/', $content);
it searches for <a href="../ and replace it with http://www.example.com/ it's not general but this works for me.