Neil Nand Neil Nand - 10 days ago 5
PHP Question

Undefined offset despite array_key_exists() check

When using the code below in a function (

$floor
&
$lift
are passed in but I've added them here for demonstration) I'm getting an
Notice: Undefined offset: 20
on the last
return
statement.

The last
return
is the mean that's mean to be used in this example but why am I getting the
Notice: Undefined offset: 20
? If I'm not mistake (which I clearly am) the
array_key_exists()
function should be preventing this? I've also tried it with
isset()
with no success.

$floor = 20;
$lift = false;

$moving = array(
"no_lift" => array(
1 => 0,
2 => 13,
3 => 17,
4 => 20
),
"lift" => array(
1 => 0,
2 => 10,
3 => 10,
4 => 20
)
);

switch ( $lift ) {
case true:
return ( isset( $moving["lift"][ $floor ] ) ? $moving["lift"][ $floor ] : $moving["lift"][ end( $moving["lift"] ) ] );
break;
case false:
return ( array_key_exists( $floor, $moving["no_lift"] ) ? $moving["no_lift"][ $floor ] : $moving["no_lift"][ end( $moving["no_lift"] ) ] );
break;
}

Answer

end returns the last value from an array, so

$moving["lift"][ end( $moving["lift"] ) ]

and

$moving["no_lift"][ end( $moving["no_lift"] ) ]

will both be, in effect,

$moving["no_lift"][ 20 ]

If you are intending to look for something in the array and return the last item if it isn't found, you can use in_array if you're looking for a value

if ($lift) {
    return in_array($floor, $moving['lift']) ? $floor : end($moving['lift']);
} else {
    return in_array($floor, $moving['no_lift']) ? $floor : end($moving['no_lift']);
}

or isset if you're looking for a key.

if ($lift) {
    return isset($floor, $moving['lift']) ? $moving['lift'][$floor] : end($moving['lift']);
} else {
    return isset($floor, $moving['no_lift']) ? $moving['no_lift'][$floor] : end($moving['no_lift']);
}

You can use array_key_exists rather than isset if some of the values in the array may be null. (You can check the answers here for a bit more of a description about that.)

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