Use external variable in array_filter

Question

I've got an array, which I want to filter by an external variable. The situation is as follows:

$id = '1'; var_dump($id); $foo = array_filter($bar, function($obj){     if (isset($obj->foo)) {         var_dump($id);         if ($obj->foo == $id) return true;     }     return false; }); 

The first var_dump returns the ID (which is dynamically set ofcourse), however, the second var_dump returns NULL.

Can anyone tell me why, and how to solve it?

Answer 1

The variable $id isn't in the scope of the function. You need to use the use clause to make external variables accessible:

$foo = array_filter($bar, function($obj) use ($id) {     if (isset($obj->foo)) {         var_dump($id);         if ($obj->foo == $id) return true;     }     return false; }); 

Answer 2

Variable scope issue!

Simple fix would be :

$id = '1'; var_dump($id); $foo = array_filter($bar, function($obj){     global $id;     if (isset($obj->foo)) {         var_dump($id);         if ($obj->foo == $id) return true;     }     return false; });  

or, since PHP 5.3

$id = '1'; var_dump($id); $foo = array_filter($bar, function($obj) use ($id) {     if (isset($obj->foo)) {         var_dump($id);         if ($obj->foo == $id) return true;     }     return false; }); 

Hope it helps