get_include_path
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.
get_included_files
(PHP 4, PHP 5)
get_included_files — include 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、 include、 include_once、 require 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
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
- include - include
- include_once - include_once
- require - require
- require_once - require_once
- get_required_files() - get_included_files のエイリアス
add a note
User Contributed Notes
get_included_files - [8 notes]
donikuntoro at integraasp dot com ¶
1 year ago
keystorm :at: gmail dotcom ¶
8 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.
RPaseur at NationalPres dot org ¶
7 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.
yarco dot w at gmail dot com ¶
5 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.
indigohaze at gmail dot com ¶
6 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.
peter at olds dot co ¶
1 year ago
In case anyone is wondering this function can NOT read into other files like get_defined_functions does.
Meaning I have ScriptA.php inside ScriptA.php I require ScriptB.php.
ScriptB.php requires ScriptC.php, ScriptD.php, etc. When I run the results of get_included_files() it will only show me ScriptB.php. Just an FYI.
cleggypdc at gmail dot com ¶
4 years ago
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 ¶
5 years ago
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());
?>