Conditionally showing values in a PHP array

Question

I have a PHP array which has some values that am storing to an API,, I want to show the field that has a value and omit the others but aint sure how to write the logic in PHP,,

~ Kindly assist?

array:5 [
  "child1Dob" => "2018-11-07",
  "child2Dob" => null,
  "child3Dob" => null,
  "child4Dob" => null,
  "child5Dob" => null,
]

You want to do so in a loop or simply remove null from the array and loop the clean array? — Diogo Santo, Nov 21 '18 at 14:48
@FedericoklezCulloca How can I use it,, kindly explain via an example — Patweb, Nov 21 '18 at 14:51
Is it just me, or does it seem like this question is just someone's course work? — Difster, Nov 21 '18 at 14:53

score 1 · Answer 1 · answered Nov 21 '18 at 14:52

1

You can use array_filter to only keep the values that are not Null. Try like this.

array_filter($array);

answered Nov 21 '18 at 14:52

Igor Carvalho

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score 0 · Accepted Answer · answered Nov 21 '18 at 14:53

0

Try this

$b = array_filter($a, function($k) { return $k!=null; });

Working example here.

answered Nov 21 '18 at 14:53

Kamil Kiełczewski

85,173
29
368
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Conditionally showing values in a PHP array

2 Answers2