php get value of array by unknown index in one statement

Question

if one is uncertain whether an array index exists, he normally does something like

if (isset($array[$key]))
    $val = $array[$key];

for large arrays, is it faster to not do this look up two times?

If yes, how would one go about doing this in one look up?

$val = $array[$key] ?? null;?

Jeto
– Jeto

2020-01-03 09:35:26 +00:00
Commented Jan 3, 2020 at 9:35 — Jeto
– Jeto, Commented Jan 3, 2020 at 9:35
@Jeto doesnt that throw an error if key does not exist?

user813801
– user813801

2020-01-03 09:36:42 +00:00
Commented Jan 3, 2020 at 9:36 — user813801
– user813801, Commented Jan 3, 2020 at 9:36
It does not: 3v4l.org/572Zk. See null coalescing operator.

Jeto
– Jeto

2020-01-03 09:38:17 +00:00
Commented Jan 3, 2020 at 9:38 — Jeto
– Jeto, Commented Jan 3, 2020 at 9:38
@Jeto looks like you have an answer

user813801
– user813801

2020-01-03 09:42:47 +00:00
Commented Jan 3, 2020 at 9:42 — user813801
– user813801, Commented Jan 3, 2020 at 9:42

Jeto · Accepted Answer · 2020-01-03 09:45:46Z

1

You may use the null-coalescing operator:

$val = $array[$key] ?? null;

which is equivalent to:

$val = isset($array[$key]) ? $array['key'] : null;

The only minor difference is that $val gets defined no matter what (to null), where in your original code it would stay undefined if you don't have an else.

I can't say for sure that it executes faster, but it's definitely handier/cleaner to write/maintain (since you don't write the same $array[$key] part twice within the statement).

Demo: https://3v4l.org/mtG7S

answered Jan 3, 2020 at 9:45

Jeto

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