return array element if it exists, otherwise null

Question

Given is an array $array and a key $key. If the array has the key, then I want to return the value, otherwise null. Is there a better way in PHP7 now, then doing it like that:

return (empty($array[$key])) ? null : $array[$key];

Answer 1

For the question title, there indeed a null coalescing operator introduced in PHP7:

return $array[$key] ?? null;

which is a shorthand for

return (isset($array[$key])) ? $array[$key] : null;

however, in your code you are using empty() which is different from isset() and for this code there is no shorthand in PHP.

with empty(), your code will also return null if $array[$key] exists but contains an empty-like value, like false, 0, empty array and such