>PHP: get_included_files

DescripciÃ³n

array get_included_files ( void )

Devuelve una matriz de nombres de todos los archivos que han sido incluidos usando include(), include_once(), require() o require_once().

El script llamado originalmente es considerado un "archivo incluido", asÃ que serÃ¡ listado junto con los archivos referenciados por la familia de funciones include().

Los archivos que son incluidos o requeridos mÃºltiples veces solo aparecen una vez en la matriz devuelta.

Nota: Los archivos incluidos usando la directiva de configuraciÃ³n auto_prepend_file no se incluyen en la matriz devuelta.

Ejemplo 1. Ejemplo de get_included_files() (abc.php)

<?php include 'test1.php'; include_once 'test2.php'; require 'test3.php'; require_once 'test4.php'; $archivos_incluidos = get_included_files(); foreach ($archivos_incluidos as $nombre_archivo) { echo "$nombre_archivo\n"; } ?>
generarÃ¡ la siguiente salida:
abc.php test1.php test2.php test3.php test4.php

Nota: En PHP 4.0.1pl2 y versiones anteriores, get_included_files() asumÃa que los archivos requeridos finalizaban en la extensiÃ³n .php; otras extensiones no serÃ¡n devueltas. La matriz devuelta por get_included_files() era una matriz asociativa y solo listaba los archivos incluidos por include() y include_once().

Vea tambiÃ©n include(), include_once(), require(), require_once(), y get_required_files().

add a note User Contributed Notes
get_included_files

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 http://indices.com.es/index.html includes b.php and c.php and calls get_included_files(), the returned array looks as follows:



http://indices.com.es/index.html

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

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MANUAL DE PHP

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get_included_files

DescripciÃ³n