PHP Get the correct path from URL

Question

Please consider the following URLs:

http://www.mydomain.com/a/test.php
https://www.mydomain.org/a/b/test.php
http://www.mydomain.co.nr/a/b/c/test.php
https://www.mydomain.com/a/b/c/d/test.php
http://www.mydomain.co.uk/a/b/c/d/e/test.php
https://www.mydomain.co.au.nm/a/b/c/d/e/f/test.php?var1=test1&var2=test2

Now I want to declare a constant called ACTUAL_URL in test.php file so that it contains the following outputs respectively (Lets assume that these URLs represent the main directory of the website, respectively in the order mentioned above):

http://www.mydomain.com/
https://www.mydomain.org/a/
http://www.mydomain.co.nr/a/b/
https://www.mydomain.com/a/b/c/
http://www.mydomain.co.uk/a/b/c/d/
https://www.mydomain.co.au.nm/a/b/c/d/e/

Now consider a file, 1.php located in the following locations:

http://www.mydomain.com/1.php
https://www.mydomain.org/a/1.php
http://www.mydomain.co.nr/a/b/1.php
https://www.mydomain.com/a/b/c/1.php
http://www.mydomain.co.uk/a/b/c/d/1.php
https://www.mydomain.co.au.nm/a/b/c/d/e/1.php

The PHP code for 1.php would be:

//1.php
//The folder name may remain as "a" or may change. 
//But for most cases, it will be "a"
require_once('a/test.php');

//Rest of the code for the page 

Few things to note:

1. Ideally test.php will never be executed on its own. Rather, it is an include file that would be included in other files within the main directory and/or within any other files in the sub-directories.
2. Note that the code is smart enough to detect http and https from the received url and use the same.
3. It could be a .com domain or a .co.au.nm domain and it would still fetch it correctly.
4. Strips out any parameters received via $_GET.

I tried the following, but failed to get a satisfactory answer:

define("ACTUAL_URL", basename($_SERVER['PHP_SELF']) );
define("ACTUAL_URL", dirname($_SERVER['PHP_SELF']) );
define("ACTUAL_URL", $_SERVER["SERVER_NAME"].$_SERVER["REQUEST_URI"] );

The first 2 dont provide the website URL, rather they seem to provide the physical path of the folder, which is not required. I need the accessible URL. The third one gives the domain name + folders (which is good) but contains the file name of the currently executing page and any $_GET params associated with it. For example, consider the following structure of my files in a website:

https://www.mydomain.co.au.nm/a/b/c/d/e/1.php?var1=test1&var2=test2
https://www.mydomain.co.au.nm/a/b/c/d/e/f/test.php

In the above case, I want the constant ACTUAL_URL to give me:

https://www.mydomain.co.au.nm/a/b/c/d/e/

Relative path to test.php will always be included in 1.php file.

So how can I achieve the above indicated functionality?

Answer 1

You can use the dirname() function twice. Once to get rid of the filename (and any querystring variables) and another time to traverse up one directory.

dirname(dirname($_SERVER["REQUEST_URI"]));

Please note, this won't work on files in the root directory of the website. All your paths will need to be at least 1 folder deep.

Really though, this whole approach is questionable from a security standpoint. Allowing the client to pass in paths which ultimately get chopped up and passed into require_once opens your application up to path traversal attacks. If you are going to do this, make sure to validate/sanitize the input.

Answer 2

Have a look into the parse_url function. It should give you a good starting point.