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
– Barmar

2013-07-20 22:01:42 +00:00
Commented Jul 20, 2013 at 22:01 — Barmar
– Barmar, Commented Jul 20, 2013 at 22:01

nickb · Accepted Answer · 2016-06-30 12:09:48Z

61

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, 2016 at 12:09

answered Jul 20, 2013 at 21:59

nickb

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4 Comments

Ivanka Todorova Over a year ago

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

pulzarraider Over a year ago

Because it's faster.

nickb Over a year ago

@pulzarraider - 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.

pulzarraider Over a year ago

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.

RiaD · Accepted Answer · 2013-07-20 21:58:27Z

-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

answered Jul 20, 2013 at 21:58

RiaD

47.8k12 gold badges85 silver badges128 bronze badges

6 Comments

Barmar Over a year ago

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

RiaD Over a year ago

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

Bartek Kosa Over a year ago

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.

dev-null-dweller Over a year ago

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

Bartek Kosa Over a year ago

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".

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