PHP Preserve scope when calling a function - php

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";

Related

PHP use one central variables file inside functions

In PHP, what is the right way forward for the following situation:
I have a 'central' vars.php which contains some parameters like MySQL hostname, user, pass, a $Test variable which indicates if the code is running in a test environment or not, and some other variables which I use throughout my site.
Then I have a functions.php which contains all the functions I could use throughout my site.
In some of these functions, I might need some the variables out of the vars.php file.
What is the right way to make the vars.php available to 'everywhere' in PHP?
Now I am doing it like this:
vars.php:
<?php
if(strstr($_SERVER['HTTP_HOST'],"localhost"))
{
$Test = true;
}
?>
functions.php:
<?php
function DoSomething()
{
include "vars.php";
if($Test)
{
$String = "Test is true!";
}
else
{
$String = "Test is false!";
}
return $String;
}
?>
index.php:
<?php
include "vars.php";
include "functions.php;
$DoSomething = DoSomething();
echo $DoSomething;
?>
While this does work, I have to include the vars.php in each function I define, this doesn't seem like a 'nice' solution to me.
Isn't there a way to define the variables in vars.php directly in the global scope, and be able to use them inside functions directly without having to worry about including them?
Thanks!
The right way would be using constants instead of variables.
You're looking for global variables. If you include the file at the top of your page and then declare each one as a global within your function like this:
include('vars.php');
function whatHost()
{
global $host;
echo $host;
}
it will work. This is very bad practice though, as global state is very bad 95% of the time. PHP has powerful object-oriented features which allow you to encapsulate functionality and state within objects. I suggest you read up on that and learn about more proper ways of storing application configuration. Note that for this particular situation, you could just use constants (:
Use constants. Define all together and always on top of classes/files.
Constants are uniques and we use them as config info or static/const data.
If u are sue DEFINE remember use const as a string not a var, example:
define("BASE_PATH", "C://base/");
$dir = BASE_PATH . "images/dropbox.txt";
http://php.net/manual/en/language.constants.php

PHP Variable Scope Issue

In my code, I use a public load_snippet function of a class when I need to include HTML or PHP snippets. (I do this instead of a direct include_once because the directory structure varies depending on certain variables).
I had some issues with variable scopes, so I've narrowed down the problem to this: let's say I define a variable within my page:
$variable = 'Hello World!";
Then, I need load in a snippet:
$APP->load_snippet("slider");
The snippet renders perfectly, except that PHP gives an undefined variable error if I try to reference $variable in the slider code. If I directly include the php file, it works as expected, so I don't understand why I'm having this problem, since this is the load_snippet function:
public function load_snippet($snippet){
if(file_exists("app/".$this->APP_TYPE."/snippets/".$snippet.".php")){
include "app/".$this->APP_TYPE."/snippets/".$snippet.".php";
}
else{
include 'common/txt/404.txt';
}
}
Any help you can give me is much appreciated.
The file is being included within the context of the load_snippet() function, and therefore has only those variables which exist within that function. One way to modify this is to make your function accept two variables: the filename and an array of values.
public function load_snippet($snippet, $content) {
if (is_array($content)) extract($content);
if (file_exists("app/".$this->APP_TYPE."/snippets/".$snippet.".php")) {
include "app/".$this->APP_TYPE."/snippets/".$snippet.".php";
} else {
include 'common/txt/404.txt';
}
}
Then
$arr = array('variable' => 'Hello world!');
load_snippet('slider', $arr);
I think include inside a function makes no sense to me... I think that you should put in function
global $variable;
Note that include will put the code inside the function(include will be replaced by code) as i know..
The way you are doing it is an ugly one, but you can use global $variable inside the snipped to refer to the variable. However if you include the snipped inside a function or a method, you'll have to make the variables in that function/method global as well
If you need $variable inside of the App::load_snippet() method, it would probably be best to pass it in:
public function load_snippet($snippet, $var='Hello world'){
if(file_exists("app/".$this->APP_TYPE."/snippets/".$snippet.".php")){
include "app/".$this->APP_TYPE."/snippets/".$snippet.".php";
}else{
include 'common/txt/404.txt';
}
}
//do something with $var
}
You can set a default for when $variable hasn't been set. No globals, no out of scope variables.
Instead you can use the constants like define('VARIALABLE','value'). which will be available to you anywhere in your file
You are including inside a class. Which means that the included file has the same variable scope as the line of code which includes it has. TO fix this all you need to do is put
global $variable;
Above the include.

Using require/require_once() in a functions file

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']);
}
}

PHP include_once inside a function to have global effect

I have a function in php:
function importSomething(){
include_once('something.php');
}
How do i make it sot that the include_once has a global effect? That everything imported will be included in the global scope?
You can return all the variables in the file like so...
function importSomething(){
return include_once 'something.php';
}
So long as something.php looks like...
<?php
return array(
'abc',
'def'
);
Which you could assign to a global variable...
$global = importSomething();
echo $global[0];
If you wanted to get really crazy, you could extract() all those array members into the scope (global in your case).
include() and friends are scope-restricted. You can't change the scope that the included content applies to unless you move the calls out of the function's scope.
I guess a workaround would be to return the filename from your function instead, and call it passing its result to include_once()...
function importSomething() {
return 'something.php';
}
include_once(importSomething());
It doesn't look as nice, and you can only return one at a time (unless you return an array of filenames, loop through it and call include_once() each time), but scoping is an issue with that language construct.
If you want ordinary variable definitions to be teleported into the global scope automatically, you could also try:
function importSomething(){
include_once('something.php');
$GLOBALS += get_defined_vars();
}
However, it if it's really just a single configuration array, I would also opt for the more explicit and reusable return method.
I know this answer is really late for this user, but this is my solution:
inside your function, simply declare any of the vars you need to ovewrite as global.
Example:
Need to have the $GLOBALS['text'] set to "yes":
Contents of index.php:
function setText()
{
global $text;
include("setup.php");
}
setText();
echo 'the value of $text is "'.$text.'"'; // output: the value of $text is "yes"
Contents of setup.php:
$text = "yes";
The solution is similar to mario's, however, only explicitely globally-declared vars are overwritten.
All variables brought in from an included file inherit current variable scope of the including line. Classes and functions take on global scope though so it depends what your importing.
http://uk.php.net/manual/en/function.include.php
(Final Paragraph before example)

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.

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