array_key_exists is not working well

Question

I have this array:

$variableNames = [
        'x1',
        'x2',
        'x3',
        'x4',
        'x5',
        'x6',
        'x7'
    ];

But, when i use array_key_exists function like this:

array_key_exists('x3', $this->variableNames)

It return false. But, if i have this array:

$variableNames = [
        'x1' => null,
        'x2' => null,
        'x3' => null,
        'x4' => null,
        'x5' => null,
        'x6' => null,
        'x7' => null
    ];

It return true. How can i use the first array, and get true? In the first array, the value is null also, like the second array. So, why the first array return false and the second array return true?

in above array use in_array() cause key is 0,1,2... not 'x1'..... — Maninderpreet Singh, Apr 20 '16 at 09:54
I believe that the code thinks in your first case that the "keys" is values and makes the keys 0,1,2 etc. Try to var_dump the frist case — Andreas, Apr 20 '16 at 09:55
you are confusing key with values,in your array the keys are implcit 0=>x1 and so on — Mihai, Apr 20 '16 at 09:57

score 6 · Accepted Answer · answered Apr 20 '16 at 09:55

6

array_key_exists() search for keys not values.

In your first case, you x3 is in value.

So, its not searching.

In this case you can use in_array(), this function searches for values.

In second case, x3 is key, therefore, searching properly.

answered Apr 20 '16 at 09:55

Pupil

23,834
6
44
66

score 2 · Answer 2 · answered Apr 20 '16 at 09:55

The keys are not null, never.

$variableNames = [
        'x1',
        'x2',
        'x3',
        'x4',
        'x5',
        'x6',
        'x7'
    ];

means

$variableNames = [
        0 => 'x1',
        1 => 'x2',
        2 => 'x3',
        3 => 'x4',
        4 => 'x5',
        5 => 'x6',
        6 => 'x7'
    ];

use

in_array('x3', $this->variableNames)

instead.

score 0 · Answer 3 · answered Apr 20 '16 at 10:03

0

Use in_array() instead of array_key_exists()

In your case,

$variableNames = ['x1',
        'x2',
        'x3',
        'x4',
        'x5',
        'x6',
        'x7'];

if (in_array("x3", $this->variableNames)) {
    echo "Found x3";
}

answered Apr 20 '16 at 10:03

6339

475
3
16

score 0 · Answer 4 · answered Apr 20 '16 at 10:11

No your incorrect. The function works well, you're just using it incorrectly. array_key_exists looks for the key, not value.

The first array you provided, is actually treated as a array of value. They have index keys, which are automatically added by PHP. I you print_r($variableNames), you'll see it will return the following.

$variableNames = [
        0 => 'x1',
        1 => 'x2',
        2 => 'x3',
        3 => 'x4',
        4 => 'x5',
        5 => 'x6',
        6 => 'x7'
    ];

You would need to search for the value instead. Use in_array() or isset(), both ways are right, one is just more convenient than the other.

array_key_exists is not working well

4 Answers4