在 PHP 中调用父类构造函数

Question

我有一个控制器

use API\Transformer\DataTransformer;
use API\Data\DataRepositoryInterface;

class DataController extends APIController implements APIInterface {

protected $data;

public function __construct(DataRepositoryInterface $data)
{
    $this->data = $data;        
}

而在APIController

use League\Fractal\Resource\Collection;
use League\Fractal\Resource\Item;
use League\Fractal\Manager;

class APIController extends Controller
{
protected $statusCode = 200;

public function __construct(Manager $fractal)
{       
    $this->fractal = $fractal;

    // Are we going to try and include embedded data?
    $this->fractal->setRequestedScopes(explode(',', Input::get('embed')));

    $this->fireDebugFilters();
}

APIController __construct() 中的任何内容都没有被调用，我尝试过 parent::__construct();，但是当我尝试从 APIController 调用类时出现此错误（请参阅下面的错误）

Argument 1 passed to APIController::__construct() must be an instance of League\Fractal\Manager, none given, called in /srv/app.dev/laravel/app/controllers/DataController.php on line 12 and defined

换句话说，它试图在 DataController 中实例化 APIController 构造函数。如何让它在DataController 之前调用APIController 构造函数？

Answer 1

您的构造函数需要将所有需要的对象传递给父构造函数。父构造函数需要一个Manager对象，所以要调用就必须传入。如果 DataRepositoryInterface 不是管理器，则需要将管理器传递给子构造函数或实例化一个对象，该对象是传递给父级的必要类。

 class DataController extends APIController implements APIInterface {

       protected $data;

       public function __construct(Manager $fractal,  DataRepositoryInterface $data) {
            parent::__construct($fractal);
            $this->data = $data;        
        }
  }

或者你可以在你的构造函数中实例化一个 Manager

     class DataController extends APIController implements APIInterface {

       protected $data;

       public function __construct(DataRepositoryInterface $data) {
            $fractal = new Manager(); //or whatever gets an instance of a manager
            parent::__construct($fractal);
            $this->data = $data;        
        }
  }