PHP switch语句变量范围

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

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 .

Answer 2

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 3

它将打印1或2. PHP中的变量具有整个功能的范围。

Answer 4

如果在switch语句中更改$someVariable的值，它将打印1或2，如果不改变，则打印0。