How do I fix "Undefined variable" error in PHP?

Question

Today, I have started to learn PHP. And, I have created my first PHP file to test different variables. You can see my file as follows.

<?php
    $x = 5; // Global scope

    function myTest()
    {
        $y = 10; // Local scope
        echo "<p>Test variables inside the function:<p>";
        echo "Variable x is: $x";
        echo "<br>";
        echo "Variable y is: $y";
    }

    myTest();

    echo "<p>Test variables outside the function:<p>";
    echo "Variable x is: $x";
    echo "<br>";
    echo "Variable y is: $y";
?>

I have found the following errors when I have run this file in the browser.

Notice: Undefined variable: x in /opt/lampp/htdocs/anand/php/index.php on line 19

Notice: Undefined variable: y in /opt/lampp/htdocs/anand/php/index.php on line 29

How can I fix the issue regarding it?

Answer 1

<?php
    $a = 1; /* Global scope */

    function test()
    {
        echo $a; /* Reference to local scope variable */
    }

    test();
?>

You are getting the first error because the variable $a can't access the global variable's value unless you explicitly declare global $a inside the function.

Example #1 Using a global

<?php
    $a = 1;
    $b = 2;

    function Sum()
    {
        global $a, $b; // If you want to access a global variable,
                       // you have to use the 'global' keyword

        $b = $a + $b;
    }

    Sum();
    echo $b;
?>

And the last error you are getting because $y is defined inside the function mytest() so its scope will be limited to that function only.

For a detailed explanation, read Variable scope.

Answer 2

The first error ($x is undefined) is because globals are not imported into functions by default (as opposed to "super globals", which are).

You need to tell your function you're referencing the global variable $x:

function myTest()
{
  global $x; // $x refers to the global variable

  $y=10; // local scope
  echo "<p>Test variables inside the function:<p>";
  echo "Variable x is: $x";
  echo "<br>";
  echo "Variable y is: $y";
}

Otherwise, PHP cannot tell whether you are shadowing the global variable with a local variable of the same name.

The second error ($y is undefined), is because local scope is just that, local. The whole point of it is that $y doesn't "leak" out of the function. Of course you cannot access $y later in your code, outside the function in which it is defined. If you could, it would be no different than a global.