PHP assign to an array only if the value exists

Question

I am currently using this code to assign a variable into the array only if it does not carry a null reference. Is their any shortcut/alternative method?

if (!is_null($foo)) {
    $var['foo'] = $foo;
}

if (!is_null($bar)) {
    $var['bar'] = $bar;
}

What is wrong with it? It is perfectly clear what the code does; what would be the point of a shortcut/alternative? — Sverri M. Olsen
– Sverri M. Olsen, Commented Jul 25, 2015 at 2:18
do you want it to look shorter? you can use ternary operators then. — Burning Crystals
– Burning Crystals, Commented Jul 25, 2015 at 2:19
@SverriM.Olsen nothing wrong with it, I was just expecting short assignment logic. :) — Dipendra Gurung
– Dipendra Gurung, Commented Jul 25, 2015 at 2:44
@BurningCrystals Ternary operators still expect something to be assigned to the index/variable, for eg. in my case if $var2 is null then $combinations['var2'] must not exist! — Dipendra Gurung
– Dipendra Gurung, Commented Jul 25, 2015 at 2:45
Do you need to do this with a bunch of variables in a row? Is that why you want to shorten it, or is this more of a general question? Anyway, if you do need to do it with multiple variables, you could loop over the variable names and use variable variables — Matt Browne
– Matt Browne, Commented Jul 25, 2015 at 3:12

C3roe · Accepted Answer · 2015-07-25 03:27:08Z

3

Well you could use the ternary operator as a “shortcut”:

is_null($foo) ?: $var['foo'] = $foo;

– but I would not really recommend this.

It works because $var['foo'] = $foo in itself is a valid expression – but is it kinda “perverting” the whole concept of the operator a bit.

Edit: As other have asked in comments as well, a little more context would be helpful. If you are not asking this out of pure curiosity, but for example because you have to do this for multiple variables – then putting them all into the array in any case, and then using array_filter to “throw away” the null values afterwards might be more straight forward …

edited Jul 25, 2015 at 3:27

answered Jul 25, 2015 at 3:04

C3roe

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Comments

Nate May · Accepted Answer · 2015-07-25 02:19:31Z

0

$a = 'var';
For ($x = 1; $x < 3; $x++){
  if (!is_null(${$a.$i})) {
    $combinations[$a.$i] = ${$a.$i};
  }
}

answered Jul 25, 2015 at 2:19

Nate May

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