php - How can I check if an array element exists?

Question

Example: I'm checking for the existence of an array element like this:

if (!self::$instances[$instanceKey]) {
    $instances[$instanceKey] = $theInstance;
}

However, I keep getting this error:

Notice: Undefined index: test in /Applications/MAMP/htdocs/mysite/MyClass.php on line 16

Of course, the first time I want an instance, $instances will not know the key. I guess my check for available instance is wrong?

Answer 1

You can use either the language construct isset, or the function array_key_exists.

isset should be a bit faster (as it's not a function), but will return false if the element exists and has the value NULL.

For example, considering this array :

$a = array(
    123 => 'glop', 
    456 => null, 
);

And those three tests, relying on isset :

var_dump(isset($a[123]));
var_dump(isset($a[456]));
var_dump(isset($a[789]));

The first one will get you (the element exists, and is not null) :

boolean true

While the second one will get you (the element exists, but is null) :

boolean false

And the last one will get you (the element doesn't exist) :

boolean false

On the other hand, using array_key_exists like this :

var_dump(array_key_exists(123, $a));
var_dump(array_key_exists(456, $a));
var_dump(array_key_exists(789, $a));

You'd get those outputs :

boolean true
boolean true
boolean false

Because, in the two first cases, the element exists -- even if it's null in the second case. And, of course, in the third case, it doesn't exist.

For situations such as yours, I generally use isset, considering I'm never in the second case... But choosing which one to use is now up to you ;-)

For instance, your code could become something like this :

if (!isset(self::$instances[$instanceKey])) {
    $instances[$instanceKey] = $theInstance;
}