Putting If() inside an array()

Question

Is it possible to put if statement inside an array like this

array(1 => 1, 2 => 2 if(!empty(3)){echo ", 3 => 3";});

instead of going

if(!empty(3)){ array (1 => 1, 2 => 2, 3 => 3); } else { array (1 =>1, 2 => 2); }

Did you try it and see? Also, there are other options besides the one you showed for doing this. In PHP there's usually more than one way to do it. — John Conde
– John Conde, Commented Apr 20, 2017 at 12:15
As an (abhorrent) one-liner: [1 => 1, 2 => 2] + (!empty(3) ? [3 => 3] : [])… — deceze
– deceze ♦, Commented Apr 20, 2017 at 12:32
Bad readability. Explicit and easily readable should always trump terseness. — deceze
– deceze ♦, Commented Apr 20, 2017 at 13:05

Philipp · Accepted Answer · 2017-04-20 12:16:31Z

You could add the value after you created the array

$array = [1 => 1, 2 => 2]; if (...) { $array[3] = 3; }

if you don't need the key, you could also just write

$array[] = 3;

Александр Резниченко · Accepted Answer · 2017-04-20 12:18:16Z

1

$array = [ 1 => 1, 2 => 2 ]; if ( !empty(3) ) { $array[3] = 3; }

answered Apr 20, 2017 at 12:18

644 bronze badges

Comments

Bhupat Bheda · Accepted Answer · 2017-04-20 12:36:44Z

You can use the conditional operator for value like below

$array = array( '1' => '1', '2' => '2', '3' => $cond ? '3' : '' );

if you need to add condition in key you can do below way

if ($cond) { $array[3] = 3; }

you can also use array union operator or array_merge:

array('1' => '1') + ($cond ? array('3' => '3') : array()) array_merge(array('1' => '1'), $cond ? array('3' => '3') : array())

now you have to decide what is better for you.