array_push() method expects parameter 1 to be array but null is given

Question

What exactly wrong in this code?

      $user_friends_bday_set = array();
      function isFullDate($target, $array) {
        array_filter($array, function($friendo){
          if(isset($friendo['bdate']) && strlen($friendo['bdate'])>5 ){
            array_push($target, $friendo);
            return $target;
          }//IF
        });//ARRAYFILTER
      };

  $user_friends_bday_set = isFullDate($user_friends_bday_set, $user_friends_arr);

Why $user_friends_bday_set is null?

`$target` is unknow is what we're saying, not `$user_friends_bday_set` — brombeer, Mar 20 '20 at 09:28
What is the purpose of calling `array_filter()` in this code? You seem to use it to create another array. — Nigel Ren, Mar 20 '20 at 09:37
Edit your question and add some example input. As Nigel suggests array_filter doesn't feel like the right fit here. — Progrock, Mar 20 '20 at 11:32

brombeer · Answer 1 · 2020-03-20T09:32:02.100

3

$target is unknown to your function($friendo) .... Change your code to

$user_friends_bday_set = array();
function isFullDate($target, $array) {
  array_filter($array, function($friendo) use (&$target) {
    if(isset($friendo['bdate']) && strlen($friendo['bdate'])>5 ){
      array_push($target, $friendo);
        return $target;
    }//IF
  });//ARRAYFILTER
};

use(&$target) has been added so your closure knows about $target.

Read more: https://www.php.net/manual/en/functions.anonymous.php

edited Mar 20 '20 at 09:32

answered Mar 20 '20 at 09:30

brombeer

8,716
5
21
27

2

`use(&$target)` to make it a reference as OP changes its value. – u_mulder Mar 20 '20 at 09:31
Nice catch, thanks – brombeer Mar 20 '20 at 09:32

array_push() method expects parameter 1 to be array but null is given

1 Answers1