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 i开发者_运维知识库sset()
when $conditional_random
is not initialized? Will it fail or it will always be true?
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 ^^)
You can check to see if the global has been created by checking if the key exists in $GLOBALS:
echo array_key_exists('fooBar', $GLOBALS) ? "true\n" : "false\n";
// Outputs false
global $fooBar;
echo array_key_exists('fooBar', $GLOBALS) ? "true\n" : "false\n";
// Outputs true
echo isset($fooBar) ? "true\n" : "false\n";
// Outputs false
This is the only way I know of to check the existence of a global without triggering a warning.
As Manos Dilaverakis mentioned, you should avoid using globals whenever possible.
SOLUTION 1
No need to use any function. If a global variable is not set, it will result as null
, and null
is equal to false
in PHP statements. So you can use something like this and it will work:
global $GLOBAL_VARIABLE;
if ($GLOBAL_VARIABLE) {
// Global variable exists
} else {
// Global variable does not exists
}
SOLUTION 2
if (isset($GLOBALS["name"])) {
// Global variable "name" exists
} else {
// Global variable does not exists
}
Global
sets the variable. Therefore isset($some_globald_variable)
will always return true.
The better option is empty()
:
if(empty($globald_variable))
{
// Variable is not set
}
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