How to know if a PHP variable exists, even if its value is NULL?

Question

$a = NULL;
$c = 1;
var_dump(isset($a)); // bool(false)
var_dump(isset($b)); // bool(false)
var_dump(isset($c)); // bool(true)

How can I distinguish $a, which exists but has a value of NULL, from the “really non-existent” $b?

Why? Just initialize your variables properly and you _know_ which one exists and which doesn't — KingCrunch, Oct 06 '11 at 12:52
I wouldn't even use isset. Initialize your variables so you can assume they exist. — GolezTrol, Oct 06 '11 at 12:57
http://stackoverflow.com/questions/418066/best-way-to-test-for-a-variables-existence-in-php-isset-is-clearly-broken — aziz punjani, Oct 06 '11 at 13:01
@KingCrunch: Theorical question or specific edge case, either ways the question was not about if it's appropriate or not. Please stay constructive. — FMaz008, Oct 06 '11 at 13:05
@Insterstellar_Coder: Thanks, I had not found that topic :) ( Even if I don't really like the _GLOBAL solution ) — FMaz008, Oct 06 '11 at 13:08

score 9 · Accepted Answer · answered Oct 06 '11 at 12:53

9

Use the following:

$a = NULL;
var_dump(true === array_key_exists('a', get_defined_vars()));

answered Oct 06 '11 at 12:53

Dmytro Shevchenko

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Paul Dixon · Answer 2 · 2011-10-06T13:18:33.513

It would be interesting to know why you want to do this, but in any event, it is possible:

Use get_defined_vars, which will contain an entry for defined variables in the current scope, including those with NULL values. Here's an example of its use

function test()
{
    $a=1;
    $b=null;

    //what is defined in the current scope?
    $defined= get_defined_vars();

    //take a look...
    var_dump($defined);

    //here's how you could test for $b
    $is_b_defined = array_key_exists('b', $defined);
}

test();

This displays

array(2) {
  ["a"] => int(1)
  ["b"] => NULL
}

How to know if a PHP variable exists, even if its value is NULL?

2 Answers2