将函数存储在变量中

Question

I don't understand how this thing works but I need to create a new function and store it in a variable. Then, by using array_filter, remove all dogs that are under certain age. Here is what I created so far:

$animals = [
  [ 'name' => 'Waffles', 'type' => 'dog', 'age'  => 12],
  [ 'name' => 'Fluffy',  'type' => 'cat', 'age'  => 14],
  [ 'name' => 'Spelunky', 'type' => 'dog', 'age'  => 4],
  [ 'name' => 'Hank'      , 'type' => 'dog', 'age'  => 11],
];


$youngDogs = function ($animals, $filter){    
    array_filter(
        $animals, 
        function($animals, $age = 5){
            $arr = [];
            if($animals['type'] == 'dog' && $animals['age'] < $age)
            {
                $arr = [
                    'name' => $animals['name'],
                    'type' => $animals['type'],
                    'age' => $animals['age'],
                ];
            }
            return $arr;
        }
    );
};


var_dump($youngDogs($animals, 5));

My idea is to create a new array and store the dogs that match the criteria and return it, however everything is a mess and the function is returning NULL and I don't know what is happening anymore. Could anyone give me a hint?

Answer 1

• You need use($age) as array_filter passes only one argument to the callback
• You need to return the result of array_filter from your function
• array_filter expects true or false to be returned from the callback

---

$youngDogs = function ($animals, $age){    
                return array_filter($animals, 
                    function($animal) use($age){
                        if($animal['type'] == 'dog' && $animal['age'] < $age) {
                            return false;
                        }
                        return true;
                    }
                );
             };