将控制器渲染到CakePHP中的不同视图 [英] Rendering controller to a different view in CakePHP

问题描述

有一种方法来渲染控制器到不同的视图，然后正常吗？我试图将一些数据从控制器传递到非默认视图。意思是我的控制器被调用：

  class StocksRealtimeController extends AppController {
 var $ uses ='StockRealtime' 
 function index（）{
 $ action ='/ TestView'; 
 $ this-> set（'stocksRT'，$ this-> StockRealtime-> find（'all'））; 
 // $ this  - > viewPath ='Pages'; 
 $ this  - > render（'/ TestView / index'）; 
} 
} 
  

...和我的视图在views->另一个问题是，如何将该值传递给PHP而不是CakePHP框架外的ctp文件？

我已尝试过这里的一切没有运气。

解决方案

正确的方式：

$ this - >正如上面提到的答案，你可以使用 $ this->（$ test $ / index'）; ; set 将传递变量传递给View。

但是，如果这不会给你想要的，我猜你还想要显示另一个布局（非默认布局）。你可以尝试 $ this - > layout ='layoutname'; （布局在布局文件夹中，默认为default.ctp）

strong> CakePHP的控制器不能够将数据传递到非视图文件（如.php）。 CakePHP的视图应以.ctp

结尾

Is there a way to render a controller to a different view then normal? I'm trying to pass some data from the controller to a non-default view. Meaning my controller is called:

class StocksRealtimeController extends AppController {
    var $uses               = 'StockRealtime';
    function index(){
        $action = '/TestView';
        $this->set('stocksRT', $this->StockRealtime->find('all'));
        //$this -> viewPath = 'Pages';
        $this -> render('/TestView/index');
    }
}

... and My view is in views->TestView->index.ctp

Another question I have is, how to pass that value to a PHP and not a ctp file outside of the CakePHP framework?

I have tried everything from here with no luck.

解决方案

The right way:

$this -> render('TestView/index');

As the answer above mention, you can use $this->set to pass variable to the View.

However, if that doesn't give you what you want I'm guessing that you also want to action to display another layout (non-default layout). You can try doing $this -> layout = 'layoutname'; (Layouts are in the layout folder, default on is default.ctp)

Note: CakePHP's controller isn't designed to be able to pass data to a non-view file (like .php). CakePHP's views should be ending with .ctp

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