Zend Framework: Render multiple Views in one Layou

2019-02-01 03:13发布

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问题:

I want to generate a dynamic site using Zend_Layout.

My layout (/application/layouts/scripts/layout.phtml) contains the following lines:

...        
<body>

        <?php echo $this->render('header.phtml') ?>

        <div id="content"><?php echo $this->layout()->content ?></div>

        <?php echo $this->render('footer.phtml') ?>

    </body>
...

If i browse to my index controller index action - Zend renders the index view (application/views/scripts/index/index.phtml) inside of $this->layout()->content automatically.

Now i want to render the views of to different controller actions in the layout. So i generate a new controller auth with an action login that shows a login form.

I change my layout to:

  ...        
    <body>

            <?php echo $this->render('header.phtml') ?>

            <div id="content"><?php echo $this->layout()->content ?></div>
            <div id="login"><?php echo $this->layout()->login ?></div>

            <?php echo $this->render('footer.phtml') ?>

        </body>
    ...

When i browse to index/index, i want to define in this action that zend should render auth/login view inside $this->layout()->login and for example news/list inside of $this->layout()->content.

index/index is than a kind of a page layout - and auth/login and news/list a kind of widget

How to do this?

回答1:

First advice is to avoid the Action view helper at all costs, it will probably be removed in ZF 2.0 anyway. (ZF-5840) (Why the actionstack is evil)

This is related to a question I asked - and bittarman's answer is pretty useful. The best way to implement something like that is to have a view helper that can generate your "login" area. My_View_Helper_Login for instance. Then your layout can call $this->login(), and so can the view script for user/login. As far as having index/index render the content from news/list just forward the request to the other controller/action from the controller. $this->_forward('list', 'news');



回答2:

I would also advise against using the action view helper. Unless you have a bunch of logic in your controller, you probably don't need to dispatch another request to another controller just to render a view partial.

I would recommend simply using a view partial just like you have done with your header.phtml and footer.phtml:

<body>

        <?php echo $this->render('header.phtml') ?>

        <div id="content"><?php echo $this->layout()->content ?></div>

        <div id="login"><?php echo $this->render('auth/login.phtml') ?></div>

        <?php echo $this->render('footer.phtml') ?>

</body>

And maybe your auth/login.phtml view script looks like this:

<div id="login_box">
    <?php if (empty($this->user)): ?>
        Please log in
    <?php else: ?>
        Hello <?php echo $this->user->name ?>
    <?php endif; ?>
</div>

As long as you set your view variables at some point in your controller, you can call the render view helper from within a view (or even a controller if you wanted to).

#index controller
public function indexAction()
{
    $this->view->user = Model_User::getUserFromSession();
}


回答3:

You can use the not so speed performant

$this->action()

or you try it with 

$this->partial()

(see http://framework.zend.com/manual/en/zend.view.helpers.html#zend.view.helpers.initial.partial )



回答4:

I got to this page when looking for an answer to my own problem, which was that my layout got called multiple times. This happened because of multiple reasons:

1) I made a bad ajax call, had /help instead of /module/help

2) I called an action, exampleAction(), I had to put a 

$this->_helper->layout->disableLayout();

To prevent the layout from being rendered again.

3) I did a redirect, try using a forward or route.



标签： zend-framework zend-view
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