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.
<?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;
}
?>
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。
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 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.
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.
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.