php perform isset in if statement without grouped statement

Question

I have this if statement

if ($something !== $array[$key]) {
    // do something
}

But I can't be 100% sure that $array[$key] exists. So I'd like to do something like this:

if ($something !== $array[$key] ?? null) {
    // do something
}

The if statement should run when the array key does not exist or or different from the variable. Of course, I could do something like this:

if ($something !== (isset($array[$key]) ? $array[$key] : null)) {
    // do something
}

But I want to avoid this since it makes the code less readable. Are there any other options except this one:

$compare = $array[$key] ?? null;
if ($something !== $compare) {
    // do something
}

so why exactly `$something !== $array[$key] ?? null` doesn't work for you? — Alex Blex, Mar 23 '17 at 13:42
@AlexBlex, probably because of "Notice: Undefined offset:" if $array[$key] isn't set. — xlordt, Mar 23 '17 at 13:51
The `$something !== $array[$key] ?? null` doesn't work for me because I get a notice: `Undefined index:`. — , Mar 23 '17 at 14:34
ah, fair enough. A pair of parenthesis should fix it: `$something !== ($array[$key] ?? null)` — Alex Blex, Mar 23 '17 at 15:09

score 2 · Accepted Answer · answered Mar 23 '17 at 13:41

2

Following code might help:

if (!array_key_exists($key, $array) || $array[$key] !== $something) {
   // Do it
}

answered Mar 23 '17 at 13:41

KiwiJuicer

1 Answers1