ajax 如何在 Zend 框架 2 中禁用渲染视图？

How to disable render view in zend framework 2?

提问by Tai T

I want to use some ajax, but I don't know how to use function as the same as setNoRender() in zend framework 2 to disable for render view.

我想使用一些 ajax，但我不知道如何使用与 zend 框架 2 中的 setNoRender() 相同的函数来禁用渲染视图。

How to disable rendering view in zend framework 2?

如何在 Zend 框架 2 中禁用渲染视图？

回答by Blanchon Vincent

• To disable your view :

  public function myactionAction()
  {
      // your code here ...
      return false;
  }
  

• 要禁用您的视图：

  public function myactionAction()
  {
      // your code here ...
      return false;
  }
  

"return false" disables the view and not the layout! why? because the accepted types are:

“返回假”禁用视图而不是布局！为什么？因为接受的类型是：

• ViewModel
• array
• null

• 视图模型
• 大批
• 空值

so "false" disable the view.

所以“false”禁用视图。

• To disable layout and view, return a response object:

  public function myactionAction()
  {
      // your code here ...
      return $this->response;
  }
  

• To disable layout:

  public function myactionAction()
  {
      // your code here ...
      $view = new ViewModel();
      $view->setTerminal(true);
      return $view;
  }
  

• 要禁用布局和视图，请返回一个响应对象：

  public function myactionAction()
  {
      // your code here ...
      return $this->response;
  }
  

• 要禁用布局：

  public function myactionAction()
  {
      // your code here ...
      $view = new ViewModel();
      $view->setTerminal(true);
      return $view;
  }
  

回答by Rob Allen

If you're using JSON, then look at the view's JsonStrategyand return a JsonModelfrom you controller. See this article.

如果您使用的是 JSON，请查看视图JsonStrategy并JsonModel从控制器返回 a 。请参阅这篇文章。

Alternatively, you can return an Responsefrom your controller and the whole view layer is skipped:

或者，您可以Response从控制器返回 an并跳过整个视图层：

public function testAction()
{
    $response = $this->getResponse();
    $response->setStatusCode(200);
    $response->setContent('foo');
    return $response;
}   

回答by Cmyker

Properand simple solution to do this

正确和简单的解决方案来做到这一点

public function testAction()
{
    $data = array(
        'result' => true,
        'data' => array()
    );
    return $this->getResponse()->setContent(Json::encode($data));
}

Details: http://cmyker.blogspot.com/2012/11/zend-framework-2-ajax-return-json.html

详情：http: //cmyker.blogspot.com/2012/11/zend-framework-2-ajax-return-json.html

回答by Svyatoslav Tretyak

I found some answer.

我找到了一些答案。

Though $this->layout()->getLayout()returns the name/path of the newly selected layout... The layout does not change with any of the following commands...

虽然$this->layout()->getLayout()返回新选择的布局的名称/路径...布局不会随着以下任何命令而改变...

within a controller

在控制器内

$this->getLocator()->get('view')->layout()->setLayout('layouts/ajax.phtml');
$this->getLocator()->get('view')->layout()->setLayout('ajax');
$this->getLocator()->get('view')->layout()->disableLayout();

within a view PHTML file

在视图 PHTML 文件中

$this->layout()->setLayout('layouts/ajax.phtml');
$this->layout()->setLayout('ajax');
$this->layout()->disableLayout();

回答by Killan

...
use Zend\View\Model\JsonModel;

public function myAction() {
    ...

    $view = new JsonModel($myArray);
    $view->setTerminal(true);
    return $view;
}

回答by Kdecom

$view = new ViewModel(); $view->setTerminate(true);

$view = new ViewModel(); $view->setTerminate(true);