get_included_files

(PHP 4, PHP 5, PHP 7)

get_included_filesinclude나 require한 파일명의 배열을 반환

설명

array get_included_files ( void )

include, include_once, require, require_once를 사용하여 포함한 모든 파일명을 얻습니다.

반환값

모든 파일명의 배열을 반환합니다.

호출한 스크립트 파일은 "포함한 파일"로 처리하기에, include 등으로 참조한 파일과 함께 나옵니다.

여러번 include나 require한 파일은 반환하는 배열에 한번만 나옵니다.

변경점

버전 설명
4.0.1 PHP 4.0.1과 이전 버전에서 이 함수는 .php 확장자로 끝나는 파일만 반환하고, 다른 확장자를 가지는 파일은 반환하지 않았습니다. get_included_files()includeinclude_once로 포함한 파일만 가지는 연관 배열을 반환했습니다.

예제

Example #1 get_included_files() 예제

<?php
// 이 파일은 abc.php 입니다

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";
}

?>

위 예제의 출력:

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

주의

Note:

auto_prepend_file 설정 지시어를 사용하여 포함한 파일은 반환 배열에 나오지 않습니다.

참고

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User Contributed Notes 5 notes

up
12
keystorm :at: gmail dotcom
15 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.
up
2
yarco dot w at gmail dot com
12 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
RPaseur at NationalPres dot org
13 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
8 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
-9
indigohaze at gmail dot com
13 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.
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