Laravel 带参数自动解析

Question

我有这样的课：

class PostValidator 
{
     public function __construct(Validator $validator, $data) 
     {
          $this->validator = $validator;
          $this->data = $data;
     }
}

我阅读了有关 IoC 自动解析的 Laravel 文档，它给出了一个示例：

class FooBar {

    public function __construct(Baz $baz)
    {
        $this->baz = $baz;
    }

}

$fooBar = App::make('FooBar');

是否可以仅在没有App::bind（带闭包）的情况下使用App::make 来实例化我的类，上面还传递参数$data？

Answer 1

不，你不能那样做。

这个想法是你只将依赖项传递给构造函数，显然data 不是一个。 Validator 处理数据，但不依赖于数据。

改为对数据使用 setter。

class PostValidator 
{
     public function __construct(Validator $validator)
     {
          $this->validator = $validator;
     }

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

并简单地显式调用它：

$validator = App::make('PostValidator');
$validator->setData($data);

// or in the controller, which is what you're doing most likely
public function __construct(PostValidator $validator)
{
  $this->validaotr = $validator;
}

public function update($id)
{
  $data = Input::only([ input that you need ]);

  $this->validator->setData($data);

  // run the validation
  ...
}

---

编辑：根据评论，这是第二个参数 $parameters 所做的：

// Foo class with Eloquent Models as dependencies
public function __construct(User $user, Category $category, Post $post)
{
 $this->user = $user;
 $this->category = $category;
 $this->post = $post;
}

然后 IoC 容器会将依赖关系解析为新实例化的模型：

$foo = App::make('Foo');
$foo->user; // exists = false
$foo->category; // exists = false
$foo->post; // exists = false

但如果你愿意，你可以这样做：

$user = User::first();
$cat = Category::find($someId);

$foo = App::make('Foo', ['category' => $cat, 'user' => $user]);
$foo->user; // exists = true, instance you provided
$foo->category; // exists = true, instance you provided
$foo->post; // exists = false, newly instantiated like before