get_included_files

(PHP 4, PHP 5)

get_included_filesinclude または require で読み込まれたファイルの名前を配列として返す

説明

array get_included_files ( void )

この関数は、 includeinclude_oncerequire あるいは require_once によりスクリプトにロードされたすべてのファイルの名前を取得します。

返り値

すべてのファイル名を含む配列を返します。

最初にコールされたスクリプトは "include されたファイル" という扱いに なります。そのため、 include やその仲間たちにより 読み込まれたファイルの一覧に含めて表示されます。

複数回読み込まれているファイルも、 返される配列には一度しかあらわれません。

変更履歴

バージョン 説明
4.0.1 PHP 4.0.1および以前のバージョンにおいて、 get_included_files() は読み込まれるファイルが拡張子 .php となることを仮定しており、他の拡張子のファイルは返されませんでした。 get_included_files() により返される配列は 連想配列であり、 include および include_once で読み込まれたファイルのみが一覧に含まれていました。

例1 get_included_files() の例

<?php
// このファイルは abc.php です

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";
}

?>

上の例の出力は以下となります。

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

注意

注意:

設定ディレクティブ auto_prepend_file で読み込まれたファイルは、返される配列には含まれません。

参考

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User Contributed Notes 5 notes

up
1
RPaseur at NationalPres dot org
8 years ago
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.
up
1
donikuntoro at integraasp dot com
2 years ago
This function aims to perform filtering of files that have been included :

<?php
function setIncludeFiles($arrayInc = array()){
   
$incFiles = get_included_files();
    if((
count($arrayInc)>0)&&(count($incFiles)>0)){
       
$aInt = array_intersect($arrayInc,$incFiles);
        if(
count($aInt)>0){
            return
false;
       }elseif(
count($aInt)<1) {
        foreach(
$arrayInc as $inc){
            if(
is_file($inc))
                include(
$inc);
            else{
                return
false;
            }
        }
       }   
    }else{
        return
false;
    }
}
?>

Usage :

<?php
$toBeInclude
= array('/data/your_include_files_1.php',
'/data/your_include_files_2.php',
'/data/your_include_files_3.php',
);
setIncludeFiles($toBeInclude);
?>

Return false if something goes wrong.
up
1
yarco dot w at gmail dot com
6 years ago
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.
up
1
indigohaze at gmail dot com
7 years ago
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.
up
0
keystorm :at: gmail dotcom
9 years ago
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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