从 postDispatch() 传递一个变量以查看 Zend Framework 中的实例 [英] Pass a variable from postDispatch() to view instance in Zend Framework

问题描述

我有一个带有 postDispatch() 钩子的控制器插件，还有一个 $variable.

I have a controller plugin with postDispatch() hook, and there I have a $variable.

如何将这个变量传递给 view 实例?

How to pass this variable to the view instance?

我尝试了 Zend_Layout::getMvcInstance()->getView()，但这会返回新的视图实例(不是应用程序资源).与 $bootstrap->getResource('view') 相同.

I tried Zend_Layout::getMvcInstance()->getView(), but this returns new view instance (not the application resource). The same with $bootstrap->getResource('view').

我不想将它作为请求参数传递.
现在，作为一种解决方法，我使用 Zend_Registry.

I don't want to pass it as a request param.
Now, as a workaround I do it using Zend_Registry.

但是，这是最好的方法吗?

But, is it the best way?

推荐答案

我一直在使用 ViewRenderer 操作助手在需要时获取视图.这似乎是 Zend 类访问视图对象的最常见方式.

I've been using the ViewRenderer action helper to get the view when I need it. It seems to be the most common way that Zend classes access the view object.

所以，在控制器插件中:

class App_Controller_Plugin_ViewSetup extends Zend_Controller_Plugin_Abstract {

  public function postDispatch() {

    $view = Zend_Controller_Action_HelperBroker::getStaticHelper('viewRenderer')->view;

    echo $view->variable;

    $view->variable = 'Hello, World';

  }

}

在控制器中:

class IndexController extends Zend_Controller_Action {

  public function indexAction() {

    $this->view->variable = 'Go away, World';

  }

}

在视图脚本中:

<?php echo $this->variable; ?>

输出是:走开，世界走，世界

问题似乎在于视图脚本在 postDispatch() 方法被调用之前呈现，因为这确实返回了主视图对象.

It seems like the problem is that the view script renders before the postDispatch() method is called, because this does return the main view object.

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