PHP return the value but false - php

Question

I have this simple function:
function isMember($uID, $pdo) {
$status = getUserStatus($uID, $pdo);
if(isAllowed($status['status']))
return $status['status'];
return false;
}
Now I am looking for a way to return false yes, but to return also the value of the variable.
I tried the following, but it makes it empty anyway:
return $status['status'] == false;
So the logi is return false anyway but give me back also the value of the variable, even if it's false, because false should not mean empty :)

Answer 1

Return an array, and use the list() method to get your results:
function isMember($uID, $pdo) {
$status = getUserStatus($uID, $pdo);
$statusString = $status['status'];
$statusFlag = false;
if(isAllowed($status['status']))
$statusFlag = true;
return array($statusFlag,statusString);
}
//usage
list($flag,$msg) = isMember(5,"whatever");
echo "Access: $flag, with message $msg";

Answer 2

A function can not return multiple values, but similar results can be obtained by (1) returning an array or by (2) passing a variable by reference and storing the value you want returned in that variable.
You will need to write your function in a way that it returns an array containing the following:
The value you wan't returned
A flag that signifies true/false
Pass a variable by reference into your function and store the value of the status in that variable.
function isMember($uID, $pdo, &statByRef) {
$status = getUserStatus($uID, $pdo);
if(isAllowed($status['status'])) {
return $status['status'];
}
$statByRef = $status['status'];
return false;
}

Answer 3

False returns empty in PHP, see http://php.net/manual/en/function.empty.php
From documentation:
Determine whether a variable is considered to be empty. A variable is considered empty if it does not exist or if its value equals FALSE. empty() does not generate a warning if the variable does not exist.

Answer 4

Try using something like this:
function isMember($uID, $pdo) {
$status = getUserStatus($uID, $pdo);
if(isAllowed($status['status'])){
return $status['status'];
}
return false;
} // If isAllowed returns true, then PHP will return $Status['Status'];, if not then PHP will by default return false.
I have noticed you haven't used braces which makes the code a little awkward to debug. Then validate like:
if (isMember($Var,$AnotherVar) !== false){
//isMember has not returned false, so PHP is operating within these braces
}
Such a simple thing, which should be most effective.
If your wanting to assign true/false to $status['status']; then you are performing the right method, but wrong operator.
== is a comparision operator. Not assignment
= is an assignment operator, so your assignment should be:
$status['status'] = false;