Use external variable in array_filter
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;});
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
Because your closure function can't see $id. You need the use keyword:
$foo = array_filter($bar, function($obj) use ($id) {