使用当前范围的值定义匿名函数？

Question

I am trying to create an anonymous function but need to access variables from the current scope in it's definition:

class test {
    private $types = array('css' => array('folder' => 'css'));

    public function __construct(){

        //define our asset types
        foreach($this->types as $name => $attrs){
            $this->{$name} = function($file = ''){
                //this line is where is falls over!
                //undefined variable $attrs!
                return '<link href="'.$attrs['folder'].'/'.$file.'" />';                
            }
        }
    }
}

$assets = new test();

Obviously this example is very very minimalistic but it gets across what I am trying to do. So, my question is, How can I access the parent scope only for the definition of the function? (once defined I obviously don't need that context when the function is called).

---

Edit #1

Ok so after using Matthew's answer I have added use as below; but now my issue is that when I call the function I get no output.

If i add a die('called') in the function then that is produced, but not if I echo or return something.

class test {
    private $types = array('css' => array('folder' => 'css'));

    public function __construct(){

        //define our asset types
        foreach($this->types as $name => $attrs){
            $this->{$name} = function($file = '') use ($attrs){
                //this line is where is falls over!
                //undefined variable $attrs!
                return '<link href="'.$attrs['folder'].'/'.$file.'" />';                
            }
        }
    }

    public function __call($method, $args)
{
    if (isset($this->$method) === true) {
        $func = $this->$method;
        //tried with and without "return"
        return $func($args);
    }
}
}

$assets = new test();
echo 'output: '.$assets->css('lol.css');

Answer 1

function($file = '') use ($attrs)