Laravel自动注入和在构造函数体中手动指定依赖项之间有什么区别？(What's the difference between Laravel automatic injection an

编程入门行业动态更新时间:2024-10-28 12:16:12

Laravel自动注入和在构造函数体中手动指定依赖项之间有什么区别？(What's the difference between Laravel automatic injection and manually specifying the dependencies in the constructor body?)

我在Laravel项目中使用了Repository模式。除了此代码段之外，官方文档中没有真正解释此模式：

您可以在控制器的构造函数中键入提示由应用程序定义的存储库。存储库将自动解析并注入到类中。

这是我的代码，根据文档：

class CategoriesController extends Controller { protected $repo; public function __construct(CategoriesRepository $repo) { $this->repo = $repo; }

我已经打字提示了CategoriesRepository，因此服务容器会自动加载它。

但是，如果我直接创建CategoriesController类的新实例（不使用服务容器），我必须指定我还需要CategoriesRepository的新实例，如下所示：

$example = new CategoriesController(new CategoriesRepository());

现在，让我们假设我编写以下代码。

class CategoriesController extends Controller { protected $repo; public function __construct() { $this->repo = new CategoriesRepository(); }

这样，我不必通过服务容器加载类，也不必通过传递CategoriesRepository的新实例作为参数来调用它，因为它是在构造函数内自动创建的。

所以，我的问题是：这是不好的做法吗？ 类型提示作为参数和在构造函数内部创建新实例之间有什么区别？

I'm using a Repository pattern in my Laravel project. This pattern is not really explained in the official documentation, except for this snippet:

You may type-hint a repository defined by your application in a controller's constructor. The repository will automatically be resolved and injected into the class.

This is my code, in accordance with the documentation:

class CategoriesController extends Controller { protected $repo; public function __construct(CategoriesRepository $repo) { $this->repo = $repo; }

I've type-hinted the CategoriesRepository so it gets automatically loaded by the Service Container.

However, if I directly create a new instance of the CategoriesController class (without using the Service Container), I have to specify that I need a new instance of the CategoriesRepository too, like this:

$example = new CategoriesController(new CategoriesRepository());

Now, let's suppose I write the following code.

class CategoriesController extends Controller { protected $repo; public function __construct() { $this->repo = new CategoriesRepository(); }

This way, I don't have to load the class through the Service Container, nor call it by passing a new instance of CategoriesRepository as the argument, because it's automatically created inside of the constructor.

So, my question is: would this be bad practice? What's the difference between type-hinting as a parameter and creating a new instance inside of the constructor?

最满意答案

这是依赖注入的美妙之处：

复杂初始化

class MyController { public function __construct(A $a) { } } class A { public function __construct(B $b) { } } class B { public function __construct(C $c) { } } class C { public function __construct(D $d) { } } class D { public function __construct() { } }

现在您可以请求laravel为您创建该类，例如：

$controller = make(MyController::class);

或者你可以这样做：

$controller = new MyController(new A(new B(new C(new D())))));

此外，您可以指定有关如何创建变量的更复杂的规则：

app()->bind(D::class, function ($app) { $d = new D(); $d->setValueOfSomething($app->make(AnotherClass::class)); return $d; });

测试

这是依赖注入优于手动创建事物的一个优点。另一个是单元测试：

public function testSomeFunctionOfC() { $this->app->bind(D::class, function () { $dMock = $this->createMock(D::class); }); $c = make(C::class); }

现在，当您创建C时，类D将是模拟类，您可以根据您的规范确保工作。

Here's the beauty of dependency injection:

Complex initialization