Checking if an array key exists

Question

I have a multidimensional array produced by json_decode(). The json is dynamically generated, that means some keys will be present randomly.

I would like to avoid Undefined index: notice, so i encapsulated the calls to the array in a function like this:

function exists($value) {
    if (isset($value)) {
        return $value;
    }
}

I then call data:

$something = exists($json_array['foo']['bar']['baz']);

But i still get the Undefined index: baz notice. Any suggestions?

I choose if(strlen($value) > 0) because sometimes array isn't set so it just show black but in isset it returns true ) — Ankit Pise
– Ankit Pise, Commented Feb 25, 2014 at 2:59
possible duplicate of Check if a "run-time" multidimensional array key exists — David d C e Freitas
– David d C e Freitas, Commented Feb 25, 2014 at 3:00

Levi Morrison · Accepted Answer · 2014-02-25 03:04:23Z

2

It seems you are new to PHP, so I'll give a bit lengthier answer than normal.

$something = exists($json_array['foo']['bar']['baz']);

This is equivalent to what you wrote:

$baz = $json_array['foo']['bar']['baz'];
$something = exists($baz);

As you may have noticed, this means that $json_array['foo']['bar']['baz'] is evaluated before it's passed to exists(). This is where the undefined index is coming from.

The correct idiom would be more like this:

$something = NULL;
if (isset($json_array['foo']['bar']['baz'])) {
    $something = $json_array['foo']['bar']['baz'];
}

The following is also identical to the above lines:

$something = isset($json_array['foo']['bar']['baz'])
    ? $json_array['foo']['bar']['baz']
    : NULL;

answered Feb 25, 2014 at 3:04

Levi Morrison

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Community · Accepted Answer · 2017-05-23 10:26:11Z

1

You would have to chain the exists calls one by one, because you are trying to dereference the array before you send it to the exists function.

See this question for more info: Check if a "run-time" multidimensional array key exists

edited May 23, 2017 at 10:26

CommunityBot

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answered Feb 25, 2014 at 2:57

David d C e Freitas

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StackSlave · Accepted Answer · 2014-02-25 02:59:46Z

0

$json_array['foo']['bar']['baz'] fails when you pass it as an argument, before it's passed to isset(). That is your problem.

answered Feb 25, 2014 at 2:59

StackSlave

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