What is the logic behind the isset() function in PHP?

Question

I accidentally experience this issue when I was writing code in one of my application.

$ar = [
   'first' => 1,
   'second' => 2,
   ......
];

when I tried to check that index in $array which doesn't exist

if(isset($ar['third']) && !empty($ar['third'])){
    echo "Found";
}else{
    echo "Not Found";
}

It worked without error as expected, but when I put this conditions in the common function and then checked

function sanitize($value){
   if(isset($value) && !empty($value)){
       return true;
   }else{
       return false;
   }
}

if(sanitize($ar['third'])){
   echo "Found";
}else{
   echo "Not Found";
}

Above sample throws an exception undefined index error, can someone explain why this is causing the error.

1. `sanitize` is a bad name for this function, since it doesn't "sanitise" anything. Just saying, hope it's just an example. 2. `isset && !empty` is redundant. 3. If `$ar['third']` doesn't exist here: `sanitize($ar['third'])`, you can't have any expectation that code *inside* the function will suppress that error. — deceze, Oct 09 '18 at 12:29
[The Definitive Guide To PHP's isset And empty](http://kunststube.net/isset/) — deceze, Oct 09 '18 at 12:30
The key thing to remember is that `isset` is not a function, but a special language construct. Since its purpose is to detect variables, array elements, or object properties which are uninitialized, it bypasses the normal check that would raise a Notice. But in your code, the Notice *was already raised* before the `isset` line is reached, so it's too late for it to suppress the check. — IMSoP, Oct 09 '18 at 12:31

score 1 · Answer 1 · answered Oct 09 '18 at 12:21

$value is always set because it is declared as a function parameter so it always exists. So in this case using isset() is pointless.

function sanitize($value){ // <-- Variable is declared so it exists 
                           // within the scope of this function

score 1 · Answer 2 · answered Oct 09 '18 at 12:22

1

You are trying to refer to the $ar array at that the 'third' index before the isset/empty check is actually executed (as that is within the sanitise function).

PHP is therefore showing the error for the if(sanitize($ar['third'])){ line.

answered Oct 09 '18 at 12:22

psx

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What is the logic behind the isset() function in PHP?

2 Answers2