PHP switch statement variable scope

Question

In PHP, how is variable scope handled in switch statements?

For instance, take this hypothetical example:

$someVariable = 0;

switch($something) {

    case 1:
        $someVariable = 1;
        break;

    case 2:
        $someVariable = 2;
        break;
}

echo $someVariable;

Would this print 0 or 1/2?

Answer 1

It will print 1 or 2 if you change the value of $someVariable in the switch statement, and 0 if you don't.

Answer 2

It will print 1 or 2. Variables in PHP have the scope of the whole function.

Answer 3

PHP does only have a global and function/method scope. So $someVariable inside the switch block refers to the same variable as outside.

But since $something is not defined (at least not in the code you provided), accessing it raises a Undefined variable notice, none of the cases match (undefined variables equal null), $someVariable will stay unchanged and 0 will be printed out.

Answer 4

The variable will be the same in your whole portion of code : there is not variable scope "per block" in PHP.

So, if $something is 1 or 2, so you enter in one of the case of the switch, your code would output 1 or 2.

On the other hand, if $something is not 1 nor 2 (for instance, if it's considered as 0, which is the case with the code you posted, as it's not initialized to anything), you will not enter in any of the case block ; and the code will output 0.