On 15/02/2022 12:54, Andreas Leathley wrote:
The problem with your way of writing code is that it is ambiguous in
meaning, which is why this is a warning.
I think that's a good way of looking at it. There's actually quite a lot
of code hiding a check like "if ($array['key'])"; roughly speaking, it's
short-hand for:
if ( array_key_exists('key', $array) && $array['key'] !== null &&
$array['key'] !== false && $array['key'] !== 0 && $array['key'] !== 0.0
&& $array['key'] !== '' && $array['key'] !== [] )
Now, if that's what you intended, there's a syntax that's slightly more
explicit while still being reasonably short: the empty() pseudo-function:
if ( ! empty($array['key']) )
On the other hand, if what you actually meant was this:
if ( array_key_exists('key', $array) && $array['key'] !== null )
Then you might have some bugs lurking, and actually want the isset()
pseudo-function instead:
if ( isset($array['key']) )
For other cases, you do have to spend a few more characters to be
explicit, often using the "??" operator; for instance, if you expect
$array['key'] to be either unset or a boolean, and want this:
if ( array_key_exists('key', $array) && $array['key'] === true )
Then the shortest is probably something like this:
if ( $array['key'] ?? false === true )
Regards,
--
Rowan Tommins
[IMSoP]
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