For some reason this function won't return the value ciao:
$a = "ciao";
function a() {
return $a;
}
I have no idea why.
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3Read php.net/manual/en/language.variables.scope.php. It is explained right at the top.Tomalak– Tomalak2011-07-02 13:24:52 +00:00Commented Jul 2, 2011 at 13:24
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3 Answers
Functions can only return variables they have in their local space, called scope:
$a = "ciao";
function a() {
$a = 'hello`;
return $a;
}
Will return hello, because within a(), $a is a variable of it's own. If you need a variable within the function, pass it as parameter:
$a = "ciao";
function a($a) {
return $a;
}
echo a($a); # "ciao"
BTW, if you enable NOTICES to be reported (error_reporting(-1);), PHP would have given you notice that return $a in your original code was using a undefined variable.
Comments
In PHP, functions don't have access to global variables. Use global $a in body of the function or pass the value of $a as parameter.
1 Comment
hakre
Don't suggest using
global, suggest using a function parameter instead.$a is not in scope within the function.
PHP does not work with a closure like block scope that JS works with for instance, if you wish to access an external variable in a function, you must pass it in which is sensible, or use global to make it available, which is frowned on.
$a = "ciao";
function a() {
global $a;
return $a;
}
or with a closure style in PHP5.3+
function a() use ($a) {
return $a;
}
