zend-framework - Zend 框架 : How to render a different action?

Question

我有两个基本相同的操作，但需要不同的 URL。通常我会使用 _forward()呈现另一个 Action :

class MyController extends Zend_Controller_Action
{
    public function actionOneAction()
    {
        $this->_forward('action-two');
    }

    public function actionTwoAction()
    {
        $this->view->foobar = 'foobar';
    }
}

但是，我有一些代码发生在 preDispatch()我只想执行一次:

class MyController extends Zend_Controller_Action
{
    public function preDispatch()
    {
        //execute this only once before actionOne or actionTwo, but not both
    }

    public function actionOneAction()
    {
        $this->_forward('action-two'); //this won't work because preDispatch() will get called a second time
    }

    public function actionTwoAction()
    {
        $this->view->foobar = 'foobar';
    }
}

所以我想也许我可以直接调用这个函数，像这样:

class MyController extends Zend_Controller_Action
{
    public function preDispatch()
    {
        //execute this only once before actionOne or actionTwo, but not both
    }

    public function actionOneAction()
    {
        $this->actionTwoAction(); //execute the same code as actionTwoAction()
    }

    public function actionTwoAction()
    {
        $this->view->foobar = 'foobar';
    }
}

但是现在 Zend Framework 提示找不到 action-one.phtml查看脚本。我不想渲染 actionOne 的 View 脚本。我想呈现 actionTwo 的 View 脚本。我需要做什么？

Answer 1

使用 render()似乎可以解决问题:

public function actionOneAction()
{
    $this->actionTwoAction(); //execute the same code as actionTwoAction()
    $this->render('action-two'); //renders the same view as actionTwoAction()
}