[RESOLVED] Using preg_replace?

john1704

Hi,

Can somebody show me how to write a code that replaces "[WILDCARD].css" with "http://www.domain.com/somefile.ccs"?

I'm don't understand how patterns work in preg_replace. That's why I'm asking.

john1704

ONE MORE THING I'm using this code:

$url = $_GET['url'];
preg_replace("/$url/", "http://www.johninteractive.net/proxy/proxy.php?url=$url", $sourcecode);

But when I post the URL to the page (for example, [url]http://www.jmarshall.com)[/url], it only returns "http://www.johninteractive.net/tools/proxy" instead of the result I wanted, "http://www.johninteractive.net/proxy/proxy.php?url=http://www.jmarshall.com/tools/proxy".

What's wrong.

Thanks for all of your help! This community does a great job.

NogDog

You need to assign the value of preg_replace() to something. Often it is reassigned to the string variable being searched. Also, if using a variable as part of the regex, you may need to escape it:

$sourcecode = preg_replace('/'.preg_quote($url, '/').'/',
   "http://www.johninteractive.net/proxy/proxy.php?url=$url", $sourcecode);

However, [man]str_replace/man would be a better option for this, as you are not dealing with anything that requires a regular expression.

$sourcecode = str_replace($url, "http://www.johninteractive.net/proxy/proxy.php?url=$url", 
   $sourcecode);

john1704

New problem:

I have a source code and I want to print it out, right? It's seems good but whoever coded that web page uses partial URLs. For example, they have:

And it thinks the CSS file is on my site, which it's not:

http://www.johninteractive.net/css/global.css

When it should look something like this:

http://www.nintendo.com/ccs/global.css

How can I get it so that it replaces all partial URLs with the full URL? Remember, the variable for the posted url is $url.

laserlight

A simple [man]str_replace/man of 'css/global.css' with 'http://www.nintendo.com/ccs/global.css' will do. Pretty much all text editors provide such functionality, so why not use that instead of a PHP script?

john1704

I don't think you understand what I'm trying to get at here.

I don't know what users are going to enter as a URL in my circumventor. Therefore, I CANNOT use a simple str_replace() in my PHP script! I need to use preg_replace to search for a pattern for a partial URL, and replace that with the fulll URL of the domain.

How can I get it so that it replaces partial URLs with the full version, using preg_replace()?

Thanks much!

laserlight

I don't know what users are going to enter as a URL in my circumventor. Therefore, I CANNOT use a simple str_replace() in my PHP script! I need to use preg_replace to search for a pattern for a partial URL, and replace that with the fulll URL of the domain.

A "partial" URL is not a pattern, so you simply do not need regular expressions for it. I would check if the URL provided is an absolute URL. If it is not, I would modify it to become an absolute URL. This certainly does not need regular expressions, unless you want to use preg_match() to check if it is an absolute URL (but even that may be overkill).

I don't think you understand what I'm trying to get at here.

Ah, the reason is that you did not provide a context for your problem. Earlier, you ask "how to replace X with Y?", and then you asked the same question, phrased differently, so without reading NogDog's reply, I gave pretty much the same reply. Nowhere was it mentioned anything about a "circumventor", whatever that is.

Instead of asking "How can I get it so that it replaces partial URLs with the full version, using preg_replace()?", for which I would just give you an answer that would be better implemented with str_replace(), present what you are really trying to do, and ask for suggestions on how to solve it appropriately.

john1704

I see. So, how would you use a wildcard in str_replace? Because there are multiple .css files in the document, and I have to make sure that those display correctly.

NogDog

<?php
function trueUrl($text, $url)
{
   $search = array(
      '#((?:href|src)=(["\']))(/.*)(?:\2)#isU',
      '#((?:href|src)=(["\']))([^/][^:]*)(?:\2)#isU'
   );
   $replace = array(
      '\1'.array_shift(preg_split('#(?<!/)/(?!/)#', $url)).'\3\2',
      '\1'.dirname($url).'/\3\2'
   );
   return(preg_replace($search, $replace, $text));
}
// TEST:
header("Content-Type: text/plain");
$url = "http://www.example.com/directory/page.html";
$testData = <<<EOD
This is <a href="http://do.not.change/this/link.html">a test</a>.
It is only <a href="/from/web/root/link.html">a test</a>.
<img src='relative/to/current_page.jpg' alt='test image'>
EOD;

echo "URL:
$url

BEFORE:
$testData

AFTER:
";
echo trueUrl($testData, $url);

Output:

URL:
http://www.example.com/directory/page.html

BEFORE:
This is <a href="http://do.not.change/this/link.html">a test</a>.
It is only <a href="/from/web/root/link.html">a test</a>.
<img src='relative/to/current_page.jpg' alt='test image'>

AFTER:
This is <a href="http://do.not.change/this/link.html">a test</a>.
It is only <a href="http://www.example.com/from/web/root/link.html">a test</a>.
<img src='http://www.example.com/directory/relative/to/current_page.jpg' alt='test image'>

john1704

Thank you so much! It worked!