views:

529

answers:

2

Hello,

I want to find a method to include some files based on the current file path.. for example:

I have "website.com/templates/name1/index.php", this "index.php should be a unique file that I will use in many different directories on different depths, so I want to make the code universal so I don't need to change anything inside this file..

So if this "index.php" is located in

"website.com/templates/name1/index.php"

than it should include the file located here:

"website.com/content/templates/name1/content.php"

another example:

"website.com/templates/name2/index.php"

than it should include the file located here:

"website.com/content/templates/name2/content.php"

Also I want to overrun "*Warning: include_once() [function.include-once]: http:// wrapper is disabled in the server configuration by allow_url_include=0*" kind of error.. because is disabled and unsafe..

Is there a way to achieve this? Thank you!

+2  A: 

I think you need to use __FILE__ (it has two underscores at the start and at the end of the name) and DIRECTORY_SEPARATOR constants for working with files based on the current file path.

For example:

<?php
  // in this var you will get the absolute file path of the current file
  $current_file_path = dirname(__FILE__);
  // with the next line we will include the 'somefile.php'
  // which based in the upper directory to the current path
  include(dirname(__FILE__) . DIRECTORY_SEPARATOR . '..' . DIRECTORY_SEPARATOR . 'somefile.php');

Using DIRECTORY_SEPARATOR constant is more safe than using "/" (or "\") symbols, because Windows and *nix directory separators are different and your interpretator will use proper value on the different platforms.

Sergey Kuznetsov
A: 

I'd just do something as simple as:

$dir = str_replace('website.com/','website.com/content/',__DIR__);
include "$dir/content.php";

And your HTTP wrapper issue doesn't seem to have anything to do with this. What do you mean by overrun it? You generally never want to do a remote include.

Rasmus