Last updated: Fri, 27 Jun 2008

get_included_files

(PHP 4, PHP 5)

get_included_files — Returns an array with the names of included or required files

Description

array get_included_files ( void )

Gets the names of all files that have been included using include(), include_once(), require() or require_once().

Return Values

Returns an array of the names of all files.

The script originally called is considered an "included file," so it will be listed together with the files referenced by include() and family.

Files that are included or required multiple times only show up once in the returned array.

ChangeLog

Version	Description
4.0.1	In PHP 4.0.1 and previous versions this function assumed that the required files ended in the extension .php; other extensions would not be returned. The array returned by get_included_files() was an associative array and only listed files included by include() and include_once().

Examples

Example #1 get_included_files() example


<?php
// This file is abc.php

include 'test1.php';
include_once 'test2.php';
require 'test3.php';
require_once 'test4.php';

$included_files = get_included_files();

foreach ($included_files as $filename) {
    echo "$filename\n";
}

?>

The above example will output:

abc.php
test1.php
test2.php
test3.php
test4.php

Notes

Note: Files included using the auto_prepend_file configuration directive are not included in the returned array.


If you want to get the relative path of an included file, from within itself use this function. If you ever have a include file thats path may not be static, this can save some time.



 function get_current_files_path($file_name)

 {

   //find the current files directory

   $includes = get_included_files();

   $path = "";

   for ($i=0; $i < count($includes); $i++)

   {

     $path = strstr($includes[$i], $file_name);

     if ($path != false)

     { 

       $key = $i;

       break;

     }

   } 

   $path = str_replace(getcwd(), "", $includes[$key]);

   $path = str_replace("\\", "/", $path);

   $path = str_replace($file_name, "", $path);

   $path = ltrim($path, "/");

   

   return $path;

 }

sam at consideropen dot com
29-May-2008 01:57


If you want to avoid the filepaths, just wrap get_included_files() inside preg_replace() to get rid of path info:



<?php

$filenames = preg_replace("/\/.*\//", "", get_included_files());

?>

yarco dot w at gmail dot com
03-Jul-2007 08:27


If you have a MAIN php script which you don't want to be included by other scripts, you could use this function. For example:



main.php:

<?php

function blockit()

{

  $buf = get_included_files();

  return $buf[0] != __FILE__;

}



blockit() and exit("You can not include a MAIN file as a part of your script.");



print "OK";

?>



So other script couldn't include main.php to modify its internal global vars.

ahmetantmen at msn dot com
30-Apr-2007 09:09


Note that; you can't solve primary filename with get_included_files() to block directly accesses if you using a graphic file created with php or a stylesheet or a php script run under iframe and etc...

131 dot php at cloudyks dot org
28-Feb-2007 07:41


Actually, auto_prepend_files are listed with  get_included_files ( php 5.2 )

indigohaze at gmail dot com
28-Aug-2006 10:30


Something that's not noted in the docs, if a file is included remotely and you do a get_included_files() in the include itself it will *not* return the document that included it.



ie:

test2.php (server 192.168.1.14):

<?php



include("http://192.168.1.11/test/test3.php");



?>



test3.php (server 192.168.1.11):



<?php



$files = get_included_files();



print_r($files);

?>



returns:



Array ( [0] => /var/www/localhost/htdocs/test/test3.php ) 



Which means you can use get_included_files() to help intercept and prevent XSS-style attacks against your code.

warhog at warhog dot net
02-Aug-2006 03:42


The example is simply wrong as the behaviour of this function changed. It now in fact returns some absolut filenames (like you were using realpath() on them). In the past it returned the same string that was used to include/require the file.



example (file in /var/www ):

require('../www/somefile.php');

would be listed as ../www/somefile.php in the past but now as /var/www/somefile.php . The problem with the prior behaviour was that when you changed the working directory and used realpath() on the filenames than you got the wrong file or FALSE. I think the behviour was changed with PHP 5.0.0 (and therefor PHP 4.3.0) but I'm not sure. It is not mentioned here in the manual unfortunately.

quis -AT- maffiaworld -DOT- n e t
18-Mar-2006 09:08


If you wan`t to compare __FILE__ and $_SERVER['SCRIPT_NAME']

you could use realpath()

it strips out symlinks and things like that



realpath(__FILE__) == realpath($_SERVER['SCRIPT_NAME'])

gamblor at crazyhomer dot com
14-Mar-2006 05:08


In regards to

__FILE__ != $_SERVER['SCRIPT_FILENAME'] to check for a file as an include:



This only works if you are using PHP as an Apache module; when using PHP as a CGI binary on shared hosts, the filepaths may differ, even if they end up pointing to the exact same file.



For example, __FILE__ might be /home/SERVER/USER/SITE/test.php

and $_SERVER['SCRIPT_FILENAME'] might be /home/USER/SITE/test.php



Because of the SERVER included in the __FILE__ path, the comparison returns true, even though the file is not being included by any other file.

RPaseur at NationalPres dot org
08-Mar-2006 08:04


As is often the case, YMMV.  I tried the __FILE__ and SCRIPT_FILENAME comparison and found this:



SCRIPT_FILENAME: /var/www/cgi-bin/php441

__FILE__: /raid/home/natpresch/natpresch/RAY_included.php



As an alternative:



count(get_included_files()); 



Gives one when the script is standalone and always more than one when the script is included.

keystorm :at: gmail dotcom
07-Sep-2004 02:08


As of PHP5, this function seems to return an array with the first index being the script all subsequent scripts are included to.

If index.php includes b.php and c.php and calls get_included_files(), the returned array looks as follows:



index.php

a.php

b.php



while in PHP<5 the array would be:



a.php

b.php



If you want to know which is the script that is including current script you can use $_SERVER['SCRIPT_FILENAME'] or any other similar server global.



If you also want to ensure current script is being included and not run independently you should evaluate following expression:



__FILE__ != $_SERVER['SCRIPT_FILENAME']



If this expression returns TRUE, current script is being included or required.

php at bronosky dot com
02-Jun-2004 05:46


Just FYI, the given example will return this ONLY if executed from the filesystem root:

abc.php

test1.php

test2.php

test3.php

test4.php



What makes this function useful is that it actually returns the complete path of each file.  Like this:

/path/including/document_root/to/abc.php

/path/including/document_root/to/test1.php

/path/including/document_root/to/test2.php

/path/including/document_root/to/test3.php

/path/including/document_root/to/test4.php

add a note

get_loaded_extensions

get_include_path