get_include_path
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;
}
get_included_files
(PHP 4, PHP 5)
get_included_files — Devuelve una matriz con los nombres de los archivos incluidos o requeridos
Descripción
array get_included_files
( void
)
Obtiene los nombres de todos los archivos que han sido incluidos usando include(), include_once(), require() o require_once().
Valores retornados
Devuelve una matriz con los nombres de los archivos.
El script llamado originalmente es considerado un "archivo incluido", así que será listado junto con los archivos referenciados por la familia de funciones include().
Los archivos que son incluidos o requeridos múltiples veces solo aparecen una vez en la matriz devuelta.
Registro de cambios
| Versión | Descripción |
|---|---|
| 4.0.1 | En PHP 4.0.1 y versiones anteriores esta función asumía que los archivos requeridos finalizaban en la extensión .php; otras extensiones no serán devueltas. La matriz devuelta por get_included_files() era una matriz asociativa y solo listaba los archivos incluidos por include() y include_once(). |
Ejemplos
Example #1 Ejemplo de get_included_files()
<?php
// Este archivo es abc.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";
}
?>
El resultado del ejemplo seria:
abc.php test1.php test2.php test3.php test4.php
Notes
Note: Los archivos incluidos usando la directiva de configuración auto_prepend_file no son incluidos en la matriz devuelta.
add a note
User Contributed Notesget_included_files
cleggypdc at gmail dot com
23-Jun-2008 06:47
23-Jun-2008 06:47
sam at consideropen dot com
29-May-2008 10:57
29-May-2008 10:57
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 05:27
04-Jul-2007 05:27
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
01-May-2007 06:09
01-May-2007 06:09
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 04:41
28-Feb-2007 04:41
Actually, auto_prepend_files are listed with get_included_files ( php 5.2 )
indigohaze at gmail dot com
28-Aug-2006 07:30
28-Aug-2006 07:30
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 12:42
02-Aug-2006 12:42
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 06:08
18-Mar-2006 06:08
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
15-Mar-2006 02:08
15-Mar-2006 02:08
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 05:04
08-Mar-2006 05:04
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 11:08
07-Sep-2004 11:08
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
03-Jun-2004 02:46
03-Jun-2004 02:46
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