In a maintenance script for a web server, I need to get the database settings for each website/application on that server. As most of the websites are based on Drupal, the main settings file is settings.php, within that file various variables (such as $database) are defined, but sometime it also contains ini_set() statements for that specific site.
The script could include the settings.php to obtain the database setting but then the script will throw errors such as
Warning: ini_set(): A session is active. You cannot change the session module's ini settings at this time in include() (line 123 of /path/to/website/sites/default/settings.php).
Is there a way to only load (specific) variables from an included php file?
Of course, I mean something smooth. I know, I could load the file into a string, use regex to extract the database variable and eval() that, but I'd rather think there is a better way.
Instead of just including the settings.php (due to errors with ini_set() commands inside the included file I had to read the file into a string, get the desired part and eval() that.
In the example below, I am searching for the database settings inside a standard Drupal settings.php
#include $settings_path; // critical due to ini_set() commands.
$settings = file_get_contents($settings_path); // load file into string
$settings = preg_replace("#/\*.*?\*/#si", '', $settings); // get rid of comments to avoid confusion with preg_match
preg_match('#\$databases.*?;#si', $settings, $dbraw); // get the desired part of the string
eval("\$databases = " . $dbraw[0]); // eval() to create the variable
Related
I have created a database connection instance in PHP like:
$mysqli = mysqli_connect("localhost", "root", "blabla", "blabla");
I need to pass this database connection to other PHP files, since I have an HTML form that first inserts data into a database and then it goes to a PHP page that retrieves data.
I know that I must not instantiate the database connection each time but I do not know how to do it, because I am not so familiar with OO PHP.
You can just name a file connection.php and store your line for connection to database.
Then you can include your file like this :
require_once('connection.php');
This will be the content of connection.php :
$mysqli = mysqli_connect("localhost", "root", "blabla", "blabla");
And then this way, your file will never load twice.
If I understand you correctly, I think you misunderstand what you've read.
In PHP how most systems (phpBB, WordPress, etc) do it is there is a base include file that is included at the start of each PHP page. In phpBB it is called "Common.php" if I remember correctly.
Common.php goes through and does a few things:
It reads the database configuration files and connects via the most acceptable database connection type (mysql_connect, or mysqli_connect, depending, or another connection if you aren't using MySQL).
It then unsets the database password variable (so that someone, on an off chance, couldn't figure it out via SQL Injection)
includes other files necessary for the system to function - certain user-made functions or class definitions.
This is done via one of 4 functions:
include('common.php') will include the contents of the file at the point it is written in the file. E.G.
<?php
do_function();
include('common.php');
do_other_function();
?>
Will run do_function(), then run any script in the common.php file, which could theoretically hold the definition for do_other_function(), and then run do_other_function();
include_once('common.php') makes sure that a file is run through the process only once. This is good for things like class definitions as it ensures a class is not accidentally re-defined. E.G.
<?php
include_once('database_class.php');
//Code here code here
...
//Oh crap, did I include that database class? I can't remember. I think so...maybe not
include_once('database_class.php');
?>
Will only include the database class def one time. Where as the same example above with include('database_class.php'); instead of include_once would throw an error akin to the class 'database' is already defined
Note, include and include_once will only throw warnings if a file does not exist, and if you have error reporting set to only show fatal errors, you won't notice when something isn't included except for perhaps a "constant/function/class 'something' not defined" and you'll scratch your head for perhaps a long time. So if you try to include the file 'cmmon.php' when you meant 'common.php', you may or may not see an error.
then there is the require and require_once functions, they do effectively the same thing as include and include_once, but unlike the latter two the script will stop running if it cannot include the file and throw an error, like "could not include file 'something.php' on line # [line] in file [file path]"
Use require for scripts that are absolutely necessary - function/class/constants that your scripts will need in order run properly.
Use include for things that are voluntary, the first thought that came to mind would be plugins - things that don't necessarily have to be there in order for your system to work as intended.
If $mysqli is global, then all php files can see it by peeking the $GLOBALS array, assuming the php is included.
For example:
$mysqli = mysqli_connect(..); // in global space
another php file:
$GLOBALS["mysqli"]; // use that
inside a php function you could also do:
global $mysqli; // use that
as we know, return keyword will RETURN some value and exit current function. Mean, that this one used only inside some functions.
BUT, I saw some php-dev's use return keyword outside functions, even in index.php file (in root of web server). What is that mean???? By the way, maybe it's logical to require some file inside function, but this style isnt mine.
There's not much more to say than what the docs do.
About the common usage of return:
If called from within a function, the return statement immediately
ends execution of the current function, and returns its argument as
the value of the function call. return will also end the execution of
an eval() statement or script file.
About the less common usage:
If called from the global scope, then execution of the current script
file is ended. If the current script file was included or required,
then control is passed back to the calling file. Furthermore, if the
current script file was included, then the value given to return will
be returned as the value of the include call. If return is called from
within the main script file, then script execution ends. If the
current script file was named by the auto_prepend_file or
auto_append_file configuration options in php.ini, then that script
file's execution is ended.
Its documented somewhere within the manual
// myFile.php
return array( 'foo' => 'bar');
// somewhere else
$config = include 'myFile.php';
echo $config['foo'];
If you use return in the main scope php will leave the file inclusion and use the value as "return value" of the inclusion (include[_once](), require[_once]()).
BUT, I saw some php-dev's use return keyword outside functions, even
in index.php file (in root of web server). What is that mean???
You know the common purpose. But what you are asking is used to prevent code injection in php include files. Take a look at this post which explains it:
Prevent Code Injection in PHP include files
While discussing Coding Standards it was not long ago I argued against
adding ?> at the end of php files. But miqrogroove pointed to me an
interesting aspect why it actually can make sense to have it and an
additional return statement at the end of each file: That one (merely
the return statement) can prevent an attacker to append payload code
to existing PHP files, for example known include files. The
countermeasurement is pretty easy, just add a return statement at the
end of the file. It will end the include “subroutine”:
Example:
/* all the include file's php code */
return;
?>
This question already has answers here:
Check if a file was included or loaded
(12 answers)
Closed 2 years ago.
My webapp has a buch of modules. Each module has a 'main' php script which loads submodules based on a query sent to the main module:
//file: clientes.php
//check for valid user...
//import CSS and JS...
switch( $_GET["action"] )
{
case "lista" : require_once("clientes.lista.php"); break;
case "listaDeudores" : require_once("clientes.listaDeudores.php"); break;
case "nuevo" : require_once("clientes.nuevo.php"); break;
case "detalles" : require_once("clientes.detalles.php"); break;
case "editar" : require_once("clientes.editar.php"); break;
default : echo "<h1>Error</h1><p>El sitio ha encontrado un error.</p>";
}
This main module deals with security and imports many resources all submodules need. The big problem shows up when a user asks for any of the submodules, bypassing all the security measures on the main module! My idea was to add a line on every submodule to test if it was being called directly and deny access or if its been called via another script, and continue. The least thing I would like to do is redo the security checking on every single file, since it does a bunch of query's to the database.
Does a php script know if its been called via a require_once() or a direct call ? I've been trying to implement some sort of $_SERVER['REQUEST_URI'] and $_SERVER['PHP_SELF'] pitfall but I was wondering if there was some sort of an elegant way of doing this.
I was looking for a way to determine if a file have been included or called directly, all from within the file. At some point in my quest I passed through this thread. Checking various other threads on this and other sites and pages from the PHP manual I got enlightened and came up with this piece of code:
if ( basename(__FILE__) == basename($_SERVER["SCRIPT_FILENAME"]) ) {
echo "called directly";
}
else {
echo "included/required"
}
In essence it compares if the name of the current file (the one that could be included) is the same as the file that is beeing executed.
EXPLANATION:
__FILE__ is a PHP magic constant that stores the full path and filename of the file, the beauty of it is that if the file has been included or required it still returns the full path and filename of such file (the included file).
(Magic Constants Manual: http://php.net/manual/en/language.constants.predefined.php)
$_SERVER["SCRIPT_FILENAME"] returns the absolute pathname of the currently executing script. As when a file is included/required it's not executed (just included) it returns the path name of the (let's say) "parent" file (the one that includs the other file and the one that gets executed).
basename(string $path) is a function that returns the trailing name component of path, that in this case is the file name. You could also just compare the full path and filename, that would be indeed better, it isn't really neceseary to use this function but it feels cleaner this way, jajaj.
(basename(): http://php.net/manual/en/function.basename.php)
I know it's a "bit" late to be answering the main question but I guessed that it could be useful to anyone who's on the same situation that I was and that also passes by.
One elegant way is putting all your files which should only be accessed via include outside the web directory.
Say your web directory is /foo/www/, make an include directory /foo/includes and set this in your include_path:
$root = '/foo';
$webroot = $root.'/www'; // in case you need it on day
$lib = $root.'/includes';
// this add your library at the end of the current include_path
set_include_path(get_include_path() . PATH_SEPARATOR . $lib);
Then nobody will be able to access your libraries directly.
There's a lot of other things you could do (test a global variable is set, use only classes in libraries, etc) but this one is the most secure one. Every file which is not in your DocumentRoot cannot be accessed via an url,. But that does not mean PHP cannot get access to this file (check as well your open_basedir configuration if you have it not empty, to allow your include dir in it).
The only file you really need in your web directory is what we call the bootstrap (index.php), with a nice rewrite rule or a nice url managment you can limit all your requests on the application to this file, this will be a good starting point for security.
One popular method to make sure modules are not called directly is defining a constant in the main script, and checking for that constant in the module.
// index.php
define("LEGIT_REQUEST", true);
// in each module
if (!defined("LEGIT_REQUEST"))
die ("This module cannot be called directly.");
For the sake of completeness, the other possibility is to move such files to a directory that's not publicly available. However, some control panels used by hosting providers make this impossible. In such case, if you are using Apache you can place an .htaccess file inside the directory:
#
# Private directory
#
Order allow,deny
Deny from all
A common technique is to add this to the main module (before the includes)
define('TEST', true);
and to add something like that at the first line of every submodule
if (!defined('TEST')) {
die('Do not cheat.');
}
An alternative to defining a constant and checking it is to simply put the files that index.php includes outside of the document root area. That way the user can't directly access them via your web server at all. This is also obviously the most secure way, in case your web server has a configuration error in future that eg. displays PHP files as plain text.
You can define('SOMETHING', null) in clientes.php and then check if (!defined('SOMETHING')) die; in the modules.
global.php
if(!defined("in_myscript"))
{
die("Direct access forbidden.");
}
module.php
define("in_myscript", 1);
include("global.php");
A generic way that works without having to define a constant or use htaccess or use a specific directory structure or depend on the $_SERVER array that could theoretically be modified is to start each include-only (no direct access) file with this code:
<?php $inc = get_included_files(); if(basename(__FILE__) == basename($inc[0])) exit();
As practice of habit I have a console class built to send messages, errors, etc. to console with FirePHP. Inside the Console class write() method I have a check to see if a $_REQUEST[debug] == 1, that way I'm not exposing errors to users if something pops up on production and they would have to know what the request variable is to access the debug information.
At the top of every file I add:
Console::debug('fileName.php is loaded.');
here is a snippit from it to give you the right idea:
class Console{
public static function write($msg,$msg_type='info',$msg_label=''){
if(isset($_REQUEST['debug']) && $_REQUEST['debug'] == 'PANCAKE!'){
ob_start();
switch($msg_type){
case 'info':
FB::info($msg, $msg_label);
break;
case 'debug':
FB::info($msg, 'DEBUG')
break;
...
}
}
}
public static function debug($msg){
Console::write($msg, '');
}
}
Short and simple (for CLI):
if (__FILE__ == realpath($argv[0]))
main();
I really have read the other articles that cover this subject. But I seem to be in a slightly different position. I'm not using modrewrite (other articles).
I would like to 'include' a webpage its a 'Joomla php' generated page inside a php script. I'd hoped to make additions on the 'fly' without altering the original script. So I was going to 'precomplete' elements of the page by parasing the page once it was included I hadent wanted to hack the original script. To the point I can't include the file and its not because the path is wrong -
so
include ("/home/public_html/index.php"); this would work
include ("/home/public_html/index.php?option=com_k2&view=item&task=add"); this would not!
I've tried a variety of alternates, in phrasing, I can't use the direct route "http:etc..." since its a current php version so must be a reference to the same server. I tried relative, these work without the ?option=com_k2&view=item&task=add
It may be the simple answer that 'options' or variables can be passed.
Or that the include can't be used to 'wait' for a page to be generated - i.e. it will only return the html.
I'm not the biggest of coders but I've done alot more than this and I thought this was so basic.
this would work include ("/home/public_html/index.php?option=com_k2&view=item&task=add"); this would not!
And it never will: You are mixing a filesystem path with GET parameters, which can be passed only through the web server (utilizing a http:// call... But that, in turn, won't run the PHP code the way you want.)
You could set the variables beforehand:
$option = "com_k2";
$view = "item";
$task = "add";
include the file the normal way:
include ("/home/public_html/index.php");
this is assuming that you have access to the file, and can change the script to expect variables instead of GET parameters.
I'm not a PHP developer, but I'm currently hacking on an internal tool so my team can take advantage of its goodness. There's an index file that looks like so:
require( ($loader_path = "../../loaderapi/") . "loader.php" );
Used like this, $loader_path will retain its value within the loader.php file.
However, we want to access this API from our team's server like so:
require( ($loader_path = "http://remoteservername/loaderapi/") . "loader.php" );
In this case the $loader_path variable doesn't retain its value. I'm guessing it has something to do with it being a full blown URL, but I might be wrong. Any idea on how I can make this work, or why I can't do it this way?
If your accessing a PHP script over HTTP, only the output of that script is returned. So your script will try to interpret the output of that remote PHP script and not its source.
If there is a connection over the file system, you may want to try file://remoteservername/loaderapi/loader.php instead.
NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO!
Remote file inclusion is a BAD idea, probably one of the biggest security flaws you can open up. Even for an internal tool this is not acceptable even if only purely for contributing bad habits.
PHP by default disables this behavior, and there is a broad push to have the ability to perform an include on a URL completely stripped from PHP (as there is no compelling reason to have this ability).
If you want to load shared resources, go through a shared file system drive (as in, don't use http, ftp, anything but file://) or better yet distribute copies of loader.php through a version control system. Loading from a single file resource opens you up to problems in the future of say a new dev overwriting loader.php and breaking everyone else's code.
There shouldn't be any real difference between the two; what you're doing is defining $loader_path, concatenating the loader.php, and passing that to require.
HOWEVER: you're defining the variable within the scope of a require, which will halt processing of the script of require fails.
Try replacing 'require' with 'include' and see if it retains the variable.
Also, note that if you are running your PHP server on a windows machine, and the php version is less than 4.3.0, neither 'require' nor 'include' can handle remote files : http://us.php.net/manual/en/function.include.php
Also, as noted before, if the .php lives on a remote server that parses php, you will not get code, but the result of the remote server processing the code. You'll either have to serve it up as a .txt file, or write php that, when processed, outputs valid php.
Have you tried splitting it into two lines:
$loader_path = "http://remoteservername/loaderapi/";
require( $loader_path . "loader.php" );
It's easier to read this way as well.
Simplify the code reading by simply putting everything on 3 lines:
$loader_path = "http://remoteservername/loaderapi/";
$page = "loader.php";
require($loader_path . $page );
Much clearer and it works.
why not just put it above the require statement? would make it easier to read too.
<?php
$loader_path = "../../folderName/"
require($loader_path . "filename")
?>