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 — Betiğe dahil edilen dosyaların isimlerini içeren bir dizi döndürür
Açıklama
array get_included_files
( void
)
include(), include_once(), require() veya require_once() kullanarak betiğe dahil edilen tüm dosyaların isimlerini döndürür.
Dönen Değerler
Betiğe dahil edilen dosyaların isimlerini içeren bir dizi döndürür.
Betiğin kendisi de içerilmiş bir dosya olarak ele alınır ve ismi dönen diziye dahil edilir.
Betiğe birden fazla dahil edilen dosyaların isimleri dönen dizi içinde bir kere zikredilir.
Sürüm Bilgisi
| Sürüm: | Açıklama |
|---|---|
| 4.0.1 | PHP 4.0.1 ve öncesinde, bu işlev betiğe dahil edilen dosyaların .php uzantısına sahip olduklarını varsayar, diğer dosyaların isimlerini döndürmezdi. get_included_files() tarafından döndürülen ilişkisel dizi sadece include() ve include_once() kullanılarak betiğe dahil edilen dosyaların isimlerini döndürürdü. |
Örnekler
Örnek 1 - get_included_files() örneği
<?php
// Dosyanın kendi ismi: abc.php
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";
}
?>
Yukarıdaki örneğin çıktısı:
abc.php test1.php test2.php test3.php test4.php
Notlar
Bilginize: auto_prepend_file yönergesinde belirtilerek dahil edilen dosyaların isimleri dönen dizide bulunmaz.
Ayrıca Bakınız
- include() - include
- include_once() - include_once
- require() - require
- require_once() - require_once
- get_required_files() - get_included_files işlevinin takma adıdır
add a note
User Contributed Notesget_included_files
cleggypdc at gmail dot com
23-Jun-2008 04:47
23-Jun-2008 04:47
sam at consideropen dot com
29-May-2008 08:57
29-May-2008 08:57
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 03:27
04-Jul-2007 03:27
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
28-Aug-2006 05:30
28-Aug-2006 05:30
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
08-Mar-2006 04:04
08-Mar-2006 04:04
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
07-Sep-2004 09:08
07-Sep-2004 09:08
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.