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?

I don't get any notice warnings with your code. – Barmar Jul 20 '13 at 22:01 — Barmar, Jul 20 '13 at 22:01

nickb · Accepted Answer · 2016-06-30T12:09:48.133

53

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';
        }
    }
}

edited Jun 30 '16 at 12:09

answered Jul 20 '13 at 21:59

nickb

56,839
11
91
130

@pulzarraider why did you replace `is_null()` with `===`? – Ivanka Todorova Jun 29 '16 at 08:38
Because it's faster. – pulzarraider Jun 29 '16 at 08:42
@pulzarraider - Any evidence to back up that claim? Because [this post](http://stackoverflow.com/questions/8228837/is-nullx-vs-x-null-in-php) and [this post](http://php.net/manual/en/function.is-null.php#84161) seems to heavily disagree with you, saving nanoseconds at best and making your edit a superfluous micro-optimization. – nickb Jun 29 '16 at 13:20
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. – pulzarraider Jun 30 '16 at 10:52

score -1 · Answer 2 · answered Jul 20 '13 at 21:58

-1

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

answered Jul 20 '13 at 21:58

RiaD

42,649
10
67
110

1

The original code also doesn't give a notice. Passing by reference is unnecessary. – Barmar Jul 20 '13 at 22:01
@nickb, you post answer while I was checking if things I remember were right:) – RiaD Jul 20 '13 at 22:03
@Barmar, yeah, my bad – RiaD Jul 20 '13 at 22:04
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. – Bartek Kosa Jul 20 '13 at 22:10
@BartekKosa your implementation is flawed. It will also return true and notice for keys that don't exists. – dev-null-dweller Jul 20 '13 at 22:15
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". – Bartek Kosa Jul 20 '13 at 22:18
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. – thomasrutter Jan 09 '14 at 23:35

Check if array value isset and is null

2 Answers2

Linked