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 — 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(), 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。
add a note
User Contributed Notesget_included_files
cleggypdc at gmail dot com
24-Jun-2008 01:47
24-Jun-2008 01:47
sam at consideropen dot com
30-May-2008 05:57
30-May-2008 05:57
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 12:27
04-Jul-2007 12:27
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 01:09
01-May-2007 01:09
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
01-Mar-2007 12:41
01-Mar-2007 12:41
Actually, auto_prepend_files are listed with get_included_files ( php 5.2 )
indigohaze at gmail dot com
29-Aug-2006 02:30
29-Aug-2006 02:30
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 07:42
02-Aug-2006 07:42
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
19-Mar-2006 02:08
19-Mar-2006 02:08
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 10:08
15-Mar-2006 10:08
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
09-Mar-2006 01:04
09-Mar-2006 01:04
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
08-Sep-2004 06:08
08-Sep-2004 06:08
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 09:46
03-Jun-2004 09:46
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