Last updated: Fri, 05 Sep 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.





<?php


 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 09:57


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
04-Jul-2007 04:27


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.

indigohaze at gmail dot com
28-Aug-2006 06:30


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.

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


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 10:08


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.

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