Using require/require_once() in a functions file - php

my functions.php file is just a placeholder for many different functions. I have another file settings.php that I use to store different settings. Some of the functions in functions.php will need to access some of the variables in settings.php, so I tried:
require("settings.php")
on top of the functions.php file, but that does not seem to give my functions access to the variable in settings.php. I need to put require("settings.php") in every function separately.
Is there a way I can declare the variables in settings.php to be accessible globally to all functions defined in functions.php?
Thank you,

You need to modify your functions in functions.php to use global variables defined in settings.php.
Inside settings.php;
$var1 = "something";
$var2 = "something else";
By using the global keyword:
function funcName() {
global $var1;
global $var2;
// Rest of the function...
}
Or more preferably, using the $GLOBALS[] superglobal array. Any variables declared at global scope, like I assume your settings.php vars are, will be included as array keys in $GLOBALS[]. It's a little better for readability than using the global keyword inside each function.
function funcName() {
$y = $GLOBALS['var1'];
$x = $GLOBALS['var2'];
// Etc...
}

Although you've set these variables in your settings.php file, when you try to reference them in your functions it won't work. This is because variables in functions a locally scoped to the function.
To access these global variables you need to use the global keyword to bring your global variables into scope.
For example:
// Declared in settings.php
$x = 1234;
// One of your functions
function MyFunc()
{
global $x;
DoSomething($x);
}
You need to do this for every function where $x is accessed.
For more information see:
Variable scope - php.net docs

I do not know the details of your issue, but you may wish to use require_once() within each function. This seems to be better idea than using some global variables.

You're thinking global variables.
However that's not the best way to go.
Can you perhaps create a class?
class SomeClass
{
private $settings;
function __construct()
{
require_once($settings);
$this->settings = $variable(s) from settings file
}
function some_function()
{
print($this->settings['something']);
}
}

Related

How to use variables from php file where require/include is declared

So basically, I have a config.php file for connecting to the mysql database, and a functions.php file that includes some functions. In all my files (eg. index, login, register), I use the line require('config.php'); and inside config.php I have this line require('functions.php'); for including the config + functions together using the line require('config.php'); in my index, login, and register.php files.
So basically my problem is, the variables that I've declared in config.php are not recognized inside functions.php. How do I make it work?
Thanks in advance.
Use the global statement to declare variables inside functions as global variables.
function myfunction() {
global $myvar; // $myvar set elsewhere can be read within this function
// and if its value changes in this function, it changes globally
}
You can use global variavlename or $GLOBAL['variavlename without $']
<?php
$a = 1;
$b = 2;
function Sum()
{
global $a;
$a = $a + $GLOBALS['b'];
}
Sum();
echo $a;
?>
It's very likely your functions don't work because their scope does not include the variables you are trying to use.
First, make sure functions.php is being included after the variables are set
Also, make your functions Public Functions, OR, declare global variables inside functions by doing this:
$testVariable = "test";
function testFunction() {
global $testVariable;
}

require_once to global scope within a function

It seems that if require_once is called within function, the included file doesn't extend the global variable scope. How to require_once a file to global scope from within a function?
What I'm trying to do is some dynamic module loader:
function projects_init()
{
...
foreach ($projects as $p) {
require_once($p['PHPFile']);
$init_func = $p['init'];
if ($init_func)
$init_func();
}
}
If it is not possible to use require_once that way, what is the simplest solution for this? (Please no heavy frameworks.)
EDIT: it should also work for PHP 5.2.
To summarize all the information:
functions are not an issue, they will be global anyway this way
for global variables, there are 2 options:
declare them as global in the included file
declare them as global in that function (projects_init() in my case)
The above answer is right, you can use global to get what you need.
In the included file just declare the variables global at the beginning of the file, this way the code will run in the function scope but it will change the global variables(yes, you have to be careful and declare everything you need to change as global but it should work), example:
function a() {
require_once("a.php");
}
a();
echo $globalVariable;
and in the a.php file:
global $globalVariable;
$globalVariable="text";
Functions are not an issue (ref):
All functions and classes in PHP have the global scope - they can be called outside a function even if they were defined inside and vice versa.
About global variables: As in an existing question regarding the scope of require and the like, the scope is defined where the use is. If you need something else, there are numerous answers (my take) that show how to deal with global variables, most making use of get_defined_vars.
You can use global to put a variable in the global scope.
http://php.net/manual/en/language.variables.scope.php

Why can't I pass variables into an included file in PHP?

I've had this problem before before with no real resolution. It's happening again so I'm hoping you experts can help.
I set a variable on my index.php based on the $_GET array. So I use this code:
$admin = isset($_GET['admin']) ? $_GET['admin'] : "dashboard";
Below that, I use a function to include my layout:
include_layout("admin_header.php");
which calls this function:
function include_layout($template="") {
include(SITE_ROOT.DS.'layouts'.DS.$template);
}
So far, so good, everything works. But if I try to echo the $admin variable from within the admin_header.php file, I get nothing. It's as if it's not set. I even test it by using echo $admin; right before I include the header file, and it outputs there, but it's not set from the admin_header.php file's perspective.
Why would this happen? Is it something to do with using a function to include, rather than including it directly? And if so, WHY would that matter?
Its because that variable is out of scope because you are including the file using a function. You either need to pass the variable to the function in another parameter or declare the variable as global within the function
either:
function include_layout($template="") {
global $admin;
include(SITE_ROOT.DS.'layouts'.DS.$template);
}
or:
function include_layout($template="",$admin="") {
include(SITE_ROOT.DS.'layouts'.DS.$template);
}
include_layout("admin_header.php",$admin);
It's not possible to access global variables unless you say so explicitly. You have to put global $admin in the function to be able to access the variable.
The reason is you include the file inside of a function, so the $admin variable lives in the global scope, while the included file lives in the functions scope.
function include_layout($template="") {
global $admin;
include(SITE_ROOT.DS.'layouts'.DS.$template);
}
Another possibility is to use the extract method.
function include_layout($template="", $variables) {
extract($variables, EXTR_SKIP);
include(SITE_ROOT.DS.'layouts'.DS.$template);
}
include_layout("test.php", array('admin' => $admin));
It's a variable scoping issue. $admin is not defined inside the function, include_layout(). You can adjust your code a few ways. One is to say "global $admin;" inside include_layout(). Another is to set the variable, $admin, inside of "admin_header.php" if, for instance, the variable is only needed in that one layout context.
It might matter because PHP doesn't give functions access to global variables unless you explicitly ask for it. Sanitizing your input would also be a good idea.

PHP Preserve scope when calling a function

I have a function that includes a file based on the string that gets passed to it i.e. the action variable from the query string. I use this for filtering purposes etc so people can't include files they shouldn't be able to and if the file doesn't exist a default file is loaded instead.
The problem is that when the function runs and includes the file scope, is lost because the include ran inside a function. This becomes a problem because I use a global configuration file, then I use specific configuration files for each module on the site.
The way I'm doing it at the moment is defining the variables I want to be able to use as global and then adding them into the top of the filtering function.
Is there any easier way to do this, i.e. by preserving scope when a function call is made or is there such a thing as PHP macros?
Edit: Would it be better to use extract($_GLOBALS); inside my function call instead?
Edit 2:
For anyone that cared. I realised I was over thinking the problem altogether and that instead of using a function I should just use an include, duh! That way I can keep my scope and have my cake too.
Edit: Okay, I've re-read your question and I think I get what you're talking about now:
you want something like this to work:
// myInclude.php
$x = "abc";
// -----------------------
// myRegularFile.php
function doInclude() {
include 'myInclude.php';
}
$x = "A default value";
doInclude();
echo $x; // should be "abc", but actually prints "A default value"
If you are only changing a couple of variables, and you know ahead of time which variables are going to be defined in the include, declare them as global in the doInclude() function.
Alternatively, if each of your includes could define any number of variables, you could put them all into one array:
// myInclude.php
$includedVars['x'] = "abc";
$includedVars['y'] = "def";
// ------------------
// myRegularFile.php
function doInclude() {
global $includedVars;
include 'myInclude.php';
// perhaps filter out any "unexpected" variables here if you want
}
doInclude();
extract($includedVars);
echo $x; // "abc"
echo $y; // "def"
original answer:
this sort of thing is known as "closures" and are being introduced in PHP 5.3
http://steike.com/code/php-closures/
Would it be better to use extract($_GLOBALS); inside my function call instead?
dear lord, no. if you want to access a global variable from inside a function, just use the global keyword. eg:
$x = "foo";
function wrong() {
echo $x;
}
function right() {
global $x;
echo $x;
}
wrong(); // undefined variable $x
right(); // "foo"
When it comes to configuration options (especially file paths and such) I generally just define them with absolute paths using a define(). Something like:
define('MY_CONFIG_PATH', '/home/jschmoe/myfiles/config.inc.php');
That way they're always globally accessible regardless of scope changes and unless I migrate to a different file structure it's always able to find everything.
If I understand correctly, you have a code along the lines of:
function do_include($foo) {
if (is_valid($foo))
include $foo;
}
do_include(#$_GET['foo']);
One solution (which may or may not be simple, depending on the codebase) is to move the include out in the global scope:
if (is_valid(#$_GET['foo']))
include $_GET['foo'];
Other workarounds exists (like you mentioned: declaring globals, working with the $_GLOBALS array directly, etc), but the advantage of this solution is that you don't have to remember such conventions in all the included files.
Why not return a value from your include and then set the value of the include call to a variable:
config.php
return array(
'foo'=>'bar',
'x'=>23,
'y'=>12
);
script.php
$config = require('config.php');
var_dump($config);
No need to mess up the place with global variables
Is there any easier way to do this, i.e. by preserving scope when a function call is made
You could use:
function doInclude($file, $args = array()) {
extract($args);
include($file);
}
If you don't want to explicitly pass the variables, you could call doInclude with get_defined_vars as argument, eg.:
doInclude('test.template.php', get_defined_vars());
Personally I would prefer to pass an explicit array, rather than use this, but it would work.
You can declare variables within the included file as global, ensuring they have global scope:
//inc.php
global $cfg;
$cfg['foo'] = bar;
//index.php
function get_cfg($cfgFile) {
if (valid_cfg_file($cfgFile)) {
include_once($cfgFile);
}
}
...
get_cfg('inc.php');
echo "cfg[foo]: $cfg[foo]\n";

global variables in php not working as expected

I'm having trouble with global variables in php. I have a $screen var set in one file, which requires another file that calls an initSession() defined in yet another file. The initSession() declares global $screen and then processes $screen further down using the value set in the very first script.
How is this possible?
To make things more confusing, if you try to set $screen again then call the initSession(), it uses the value first used once again. The following code will describe the process. Could someone have a go at explaining this?
$screen = "list1.inc"; // From model.php
require "controller.php"; // From model.php
initSession(); // From controller.php
global $screen; // From Include.Session.inc
echo $screen; // prints "list1.inc" // From anywhere
$screen = "delete1.inc"; // From model2.php
require "controller2.php"
initSession();
global $screen;
echo $screen; // prints "list1.inc"
Update:
If I declare $screen global again just before requiring the second model, $screen is updated properly for the initSession() method. Strange.
Global DOES NOT make the variable global. I know it's tricky :-)
Global says that a local variable will be used as if it was a variable with a higher scope.
E.G :
<?php
$var = "test"; // this is accessible in all the rest of the code, even an included one
function foo2()
{
global $var;
echo $var; // this print "test"
$var = 'test2';
}
global $var; // this is totally useless, unless this file is included inside a class or function
function foo()
{
echo $var; // this print nothing, you are using a local var
$var = 'test3';
}
foo();
foo2();
echo $var; // this will print 'test2'
?>
Note that global vars are rarely a good idea. You can code 99.99999% of the time without them and your code is much easier to maintain if you don't have fuzzy scopes. Avoid global if you can.
global $foo doesn't mean "make this variable global, so that everyone can use it". global $foo means "within the scope of this function, use the global variable $foo".
I am assuming from your example that each time, you are referring to $screen from within a function. If so you will need to use global $screen in each function.
You need to put "global $screen" in every function that references it, not just at the top of each file.
If you have a lot of variables you want to access during a task which uses many functions, consider making a 'context' object to hold the stuff:
//We're doing "foo", and we need importantString and relevantObject to do it
$fooContext = new StdClass(); //StdClass is an empty class
$fooContext->importantString = "a very important string";
$fooContext->relevantObject = new RelevantObject();
doFoo($fooContext);
Now just pass this object as a parameter to all the functions. You won't need global variables, and your function signatures stay clean. It's also easy to later replace the empty StdClass with a class that actually has relevant methods in it.
You must declare a variable as global before define values for it.
The global scope spans included and required files, you don't need to use the global keyword unless using the variable from within a function. You could try using the $GLOBALS array instead.
It is useless till it is in the function or a class. Global means that you can use a variable in any part of program. So if the global is not contained in the function or a class there is no use of using Global

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