$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?
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4Why? Just initialize your variables properly and you know which one exists and which doesn'tKingCrunch– KingCrunch2011-10-06 12:52:23 +00:00Commented Oct 6, 2011 at 12:52
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I wouldn't even use isset. Initialize your variables so you can assume they exist.GolezTrol– GolezTrol2011-10-06 12:57:09 +00:00Commented Oct 6, 2011 at 12:57
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2stackoverflow.com/questions/418066/…aziz punjani– aziz punjani2011-10-06 13:01:08 +00:00Commented Oct 6, 2011 at 13:01
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8@KingCrunch: Theorical question or specific edge case, either ways the question was not about if it's appropriate or not. Please stay constructive.FMaz008– FMaz0082011-10-06 13:05:30 +00:00Commented Oct 6, 2011 at 13:05
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@Insterstellar_Coder: Thanks, I had not found that topic :) ( Even if I don't really like the _GLOBAL solution )FMaz008– FMaz0082011-10-06 13:08:15 +00:00Commented Oct 6, 2011 at 13:08
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2 Answers
Use the following:
$a = NULL;
var_dump(true === array_key_exists('a', get_defined_vars()));
Comments
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
}
