如何正确地将数据从控制器传递到视图? [英] How to correctly pass data from controller to view?

问题描述

我当前的实现方式:

class SomeController extends AppController
{
    function someaction()
    {   
        $d['text'] = "ahoy!";
        $this->render("someactionView", $d);
    }
}

在AppController中:

function render($file, $data = "")
{
    require "views/" . $file . ".php";
}

$data将在视图文件中可用.这是正确的实现吗?此实现有任何谬论吗?

And the $data will be available in the views file. Is this a correct implementation? Are there any fallacies with this implementation?

推荐答案

$data将在视图文件中可用.这是正确的吗 执行?此实现有任何谬论吗?

And the $data will be available in the views file. Is this a correct implementation? Are there any fallacies with this implementation?

基本上，您可以像大多数框架一样实现它.有两个问题:

Basically you do implement it like the most frameworks do. There's a couple of problems with that:

• 控制器获取输入并发送输出(这违反了单一职责原则)
• 视图与HTML紧密耦合.因此，您不能将同一视图用于其他内容，例如XML，JSON.
• 如果在render()方法中执行require "views/" . $file . ".php";，则再次将其紧密耦合.如果您更改视图的位置怎么办?然后，您将不得不稍微重写您的方法.这种方法只会破坏重用能力.

• A controller takes an input and sends an output (which breaks the Single-Responsibility Principle)
• A view is tightly coupled to HTML. Because of this, you cannot re-use the same view for another stuff, like XML, JSON.
• If you do require "views/" . $file . ".php"; in render() method - you again tighly couple it. What if you change the location of views? Then you would have to slightly rewrite your method. This approach merely kills reuse-ability.

要刷新您的基本知识，请执行以下操作:

To refresh your basic knowledge:

仅用于单一目的.它会更改模型状态-也就是说，它应该接受来自$_POST，$_GET，$_FILES，$_COOKIE的输入.在控制器中，仅应完成变量分配，并且仅此.

Serves only singular purpose. It changes model state - that is, it should take an input that comes from $_POST, $_GET, $_FILES, $_COOKIE. In controller only variable assignment should be done and nothing more.

class Controller
{
   public function indexAction()
   {
        $this->view->setVar('age', $this->request->getPostParam('age'));
        $this->view->setVar('user', $this->request->getPostParam('user'));
        //...
   }
}

查看

视图可直接访问模型.为了使视图更可重用和可维护，您最好将必需的内容作为函数参数(或通过设置器)传递

View

A view has a direct access to a model. In order to make make views more re-usable and maintainable you'd better pass required things as function parameters (or via setters)

class View
{
   public function render($templateFile, array $vars = array())
   {
      ob_start();
      extract($vars);
      require($templateFile);

      return ob_get_clean();
   }
}

应如何初始化视图以及应如何将变量传递给视图?

首先-一个视图应该在MVC-triad外部实例化.由于控制器写入视图或模型，因此您将通过控制器传递变量以进行查看.

How the view should be initialized and how the variables should be passed to it?

First of all - a view should be instantiated outside MVC-triad. Since a controller writes either to view or model - you'd pass variables to view via controller.

$model = new Model();
$view = new View($model);

$controller = new Controller($view);

// This will assign variables to view
$controller->indexAction();

echo $view->render();

注意:在现实世界中，模型不是类，而是抽象层.出于演示目的，我将其称为Model.

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