PHP: is there a good syntax to initialize an array element depending on whether it is undefined or not, without using if?

Question

The double question mark syntax in PHP

$foo = $bar ?? null;

will return null if $bar is undefined. This syntax is extremely useful to simplify code and avoid massive if statements.

Now I have the following situation

if (!isset($bar2)) {
    $fooArr = [$bar1];
} else {
    $fooArr = [$bar1, $bar2];
}

It seems so likely that there exists a one-line simplification for the statement but I just couldn't come up with it.

The closest I come up with is the following, but it will create an array with size 2 where the second element is null, which is not what I want.

$fooArr = [$bar1, $bar2 ?? null];

Is there a way to simplify the aforementioned nested if without using if statement?

Edit : the ? : syntax works for the above case but I will still have to write down $bar1 twice in that situation, which is not much neater than the if statement and can grow really big when the array consists of 5 elements.

Answer 1

The array_filter() method only returns the non-empty values from an array by default.

This code shows the various outcomes:

$bar1=1;
$fooArr = [$bar1, $bar2 ?? null];
print_r($fooArr);

$bar2=2;
$fooArr = [$bar1, $bar2 ?? null];
print_r($fooArr);

unset($bar1,$bar2);

$bar1=1;
$fooArr = array_filter([$bar1, $bar2 ?? null]);
print_r($fooArr);

$bar2=2;
$fooArr = array_filter([$bar1, $bar2 ?? null]);
print_r($fooArr);

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