Access a global variable in a PHP function

PhpScope

Php Problem Overview


According to the most programming languages scope rules, I can access variables that are defined outside of functions inside them, but why doesn't this code work?

<?php
    $data = 'My data';

    function menugen() {
        echo "[" . $data . "]";
    }

    menugen();
?>

The output is [].

Php Solutions


Solution 1 - Php

It is not working because you have to declare which global variables you'll be accessing:

$data = 'My data';

function menugen() {
    global $data; // <-- Add this line

    echo "[" . $data . "]";
}

menugen();

Otherwise you can access it as $GLOBALS['data']. See Variable scope.

Even if a little off-topic, I'd suggest you avoid using globals at all and prefer passing as parameters.

Solution 2 - Php

You can do one of the following:

<?php
    $data = 'My data';

    function menugen() {
        global $data;
        echo "[" . $data . "]";
    }

    menugen();

Or

<?php
    $data = 'My data';

    function menugen() {
        echo "[" . $GLOBALS['data'] . "]";
    }

    menugen();

That being said, overuse of globals can lead to some poor code. It is usually better to pass in what you need. For example, instead of referencing a global database object you should pass in a handle to the database and act upon that. This is called dependency injection. It makes your life a lot easier when you implement automated testing (which you should).

Solution 3 - Php

Another way to do it:

<?php

$data = 'My data';

$menugen = function() use ($data) {

    echo "[".$data."]";
};

$menugen();

UPDATE 2020-01-13: requested by Peter Mortensen

As of PHP 5.3.0 we have anonymous functions support that can create closures. A closure can access the variable which is created outside of its scope.

In the example, the closure is able to access $data because it was declared in the use clause.

Solution 4 - Php

It's a matter of scope. In short, global variables should be avoided so:

You either need to pass it as a parameter:

$data = 'My data';

function menugen($data)
{
    echo $data;
}

Or have it in a class and access it

class MyClass
{
    private $data = "";

    function menugen()
    {
        echo this->data;
    }

}

See @MatteoTassinari answer as well, as you can mark it as global to access it, but global variables are generally not required, so it would be wise to re-think your coding.

Solution 5 - Php

For many years I have always used this format:

<?php
    $data = "Hello";

    function sayHello(){
        echo $GLOBALS["data"];
    }

    sayHello();
?>

I find it straightforward and easy to follow. The $GLOBALS is how PHP lets you reference a global variable. If you have used things like $_SERVER, $_POST, etc. then you have reference a global variable without knowing it.

Solution 6 - Php

<?php

    $data = 'My data';

    $menugen = function() use ($data) {

        echo "[ $data ]";
    };

    $menugen();
?>

You can also simplify

echo "[" . $data . "]"

to

echo "[$data]"

Solution 7 - Php

PHP can be frustrating for this reason. The answers above using global did not work for me, and it took me awhile to figure out the proper use of use.

This is correct:

$functionName = function($stuff) use ($globalVar) {
 //do stuff
}
$output = $functionName($stuff);
$otherOutput = $functionName($otherStuff);

This is incorrect:

function functionName($stuff) use ($globalVar) {
 //do stuff
}
$output = functionName($stuff);
$otherOutput = functionName($otherStuff);

Using your specific example:

    $data = 'My data';

    $menugen = function() use ($data) {
        echo "[" . $data . "]";
    }

    $menugen();

Solution 8 - Php

You need to pass the variable into the function:

$data = 'My data';

function menugen($data)
{
    echo $data;
}

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAmin GholibeigianView Question on Stackoverflow
Solution 1 - PhpMatteo TassinariView Answer on Stackoverflow
Solution 2 - PhpjcbwlkrView Answer on Stackoverflow
Solution 3 - PhpdonvercetyView Answer on Stackoverflow
Solution 4 - PhpwebnoobView Answer on Stackoverflow
Solution 5 - PhpPsalms KaluView Answer on Stackoverflow
Solution 6 - PhpSanjeev BudhaView Answer on Stackoverflow
Solution 7 - PhpMichael Aaron WilsonView Answer on Stackoverflow
Solution 8 - PhpMajor ProductionsView Answer on Stackoverflow