Check if array value isset and is null

Question

How to check if array variable

$a = array('a'=>1, 'c'=>null);

is set and is null.

function check($array, $key)
{
    if (isset($array[$key])) {
        if (is_null($array[$key])) {
            echo $key . ' is null';
        }
        echo $key . ' is set';
    }
}

check($a, 'a');
check($a, 'b');
check($a, 'c');

Is it possible in PHP to have function which will check if $a['c'] is null and if $a['b'] exist without "PHP Notice: ..." errors?

Illinium

I don't get any notice warnings with your code. – Locomotion

Answer 1

Use array_key_exists() instead of isset(), because isset() will return false if the variable is null, whereas array_key_exists() just checks if the key exists in the array:

function check($array, $key)
{
    if(array_key_exists($key, $array)) {
        if (is_null($array[$key])) {
            echo $key . ' is null';
        } else {
            echo $key . ' is set';
        }
    }
}

Coonskin

@pulzarraider why did you replace is_null() with ===? – Indenture

Because it's faster. – Towill

@Towill - Any evidence to back up that claim? Because this post and this post seems to heavily disagree with you, saving nanoseconds at best and making your edit a superfluous micro-optimization. – Coonskin

Yes, it's microoptimization, but why not make your function the fastest as it can be? Isset() is a bit slower because it's function call. – Towill

Answer 2

You may pass it by reference:

function check(&$array, $key)
{
    if (isset($array[$key])) {
        if (is_null($array[$key])) {
            echo $key . ' is null';
        }
        echo $key . ' is set';
    }
}

check($a, 'a');
check($a, 'b');
check($a, 'c');

SHould give no notice

But isset will return false on null values. You may try array_key_exists instead

Extensible

The original code also doesn't give a notice. Passing by reference is unnecessary. – Locomotion

@nickb, you post answer while I was checking if things I remember were right:) – Extensible

It is what I wanted. Because of isset returning false on null I will do it like this: if(!isset($array[$key]) && is_null($array[$key])) which equals true if $array[$key] is null. – Illinium

@BartekKosa your implementation is flawed. It will also return true and notice for keys that don't exists. – Hendren

My final version would be if(is_null(...){ catch all nulls}elseif(isset(...)){ catch all set variables}. But the most important trick is passing an array by reference to get rid of "Notice". – Illinium

Have you tested this? I believe isset() always returns false for null values even if they "exist" thus the is_null() will never be reached. The problem from the OP question remains. – Stanleigh