In PHP, how 开发者_运维知识库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?
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
.
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.
It will print 1 or 2. Variables in PHP have the scope of the whole function.
It will print 1 or 2 if you change the value of $someVariable
in the switch statement, and 0 if you don't.
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