Array Undefined offset Notice in if condition [php]

Question

i know my first element in my array[zero index] is NULL and i want to check that in my {if} condition. but i have a notice here. what can i do to resolve the notice? my simple code:

$arr = [];
$arr[1] = '1';
if($arr[0] === NULL)
    echo 'if';
else
    echo 'else';

you first sentense "my first element in my array[zero index] is NULL" is not 100% correct: It's not defined, which is something different to being NULL! — Jeff
– Jeff, Commented Nov 23, 2018 at 21:09

Omari Celestine · Accepted Answer · 2018-11-23 21:24:30Z

1

What you can do is check if the variable relative to that index of the array has been initialized/set. For example:

$arr = [];
$arr[1] = '1';
if(!isset($arr[0]))
    echo 'if';
else
    echo 'else';

edited Nov 23, 2018 at 21:24

answered Nov 23, 2018 at 21:04

Omari Celestine

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1 Comment

rickdenhaan Over a year ago

Careful with isset(), because it will return false if $arr[0] is set to a falsey value (like null). It would probably be better to use array_key_exists().

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