[php] Access a global variable in a PHP function

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 [].

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).

If you want, you can use the "define" function, but this function creates a constant which can't be changed once defined.

<?php
    define("GREETING", "Welcome to W3Schools.com!");

    function myTest() {
        echo GREETING;
    }

    myTest();
?>

PHP Constants

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.

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();

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.

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.

<?php

    $data = 'My data';

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

        echo "[ $data ]";
    };

    $menugen();
?>

You can also simplify

echo "[" . $data . "]"

to

echo "[$data]"

You need to pass the variable into the function:

$data = 'My data';

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