Issue
Regarding global variable initialization,
function hello_testing() {
global $conditional_random;
if (isset($conditional_random)) {
echo "foo is inside";
}
}
The global variable (conditional_random) may not be initialized before the hello_testing()
function is called.
So, what happens to my validation via isset()
when $conditional_random
is not initialized? Will it fail or it will always be true?
Solution
Well, why don’t you just test ? 😉
Note: It is not as easy as you’d think — read the full answer 😉
Calling the hello_testing();
function, without setting the variable:
hello_testing();
I get no output — which indicates isset
returned false
.
Calling the function, after setting the variable:
$conditional_random = 'blah';
hello_testing();
I get an output:
foo is inside
Which indicates global
works as expected, when the variable is set — well, one should not have any doubt about that ^^
But note that isset
will return false
if a variable is set, and null
!
See the manual page of isset()
Which means that a better test would be:
function hello_testing() {
global $conditional_random;
var_dump($conditional_random);
}
hello_testing();
And this displays:
null
No Notice: the variable exists! Even if null
.
As I didn’t set the variable outside of the function, it shows that global
sets the variable — but it doesn’t put a value into it; which means it’s null
if not already set outside the function.
While:
function hello_testing() {
//global $conditional_random;
var_dump($conditional_random);
}
hello_testing();
Gives:
Notice: Undefined variable: conditional_random
It proves that notices are enabled 😉
And, if global didn’t “set” the variable, the previous example would have given the same notice.
And, finally:
function hello_testing() {
global $conditional_random;
var_dump($conditional_random);
}
$conditional_random = 'glop';
hello_testing();
Gives:
string 'glop' (length=4)
(This is to purely to demonstrate my example is not tricked ^^)
Answered By – Pascal MARTIN
Answer Checked By – Mary Flores (BugsFixing Volunteer)