简体   繁体   中英

Symfony2 - How to render a view from another controller

I have two controllers, homepage and Security.

In the homepage, I am displaying one view and in the security, I am doing some things, and one of them is the email address validation.

What I would like is that when the email validation code is not valid, display the homepage with a flash message. For that, I will have to render the indexAction of the HomepageController, from the Security controller, by giving him as parameter the flash message.

How can this be done? Can I render a route or an action from another controleller?

Thank you in advance.

I believe the checking should not be done in the Security controller. Right place in my opinion is a separate validator service or right in the entity which uses the email address.

But to your question, you can call another controller's action with $this->forward() method:

public function indexAction($name)
{
    $response = $this->forward('AcmeHelloBundle:Hello:fancy', array(
        'name'  => $name,
        'color' => 'green',
    ));

    return $response;
}

The sample comes from symfony2 documentation on: http://symfony.com/doc/2.0/book/controller.html#forwarding

我找到了解决方案,只需通过指定控制器和动作nanme来使用forward函数:

return $this->forward('MerrinMainBundle:Homepage:Index', array('flash_message'=>$flash_message));

redirectToRoute : Just a recap with current symfony versions (as of 2016/11/25 with v2.3+)

public function genericAction(Request $request)
{
    if ($this->evalSomething())
    {
        $request->getSession()->getFlashBag()
            ->add('warning', 'some.flash.message');
        $response = $this->redirectToRoute('app_index', [
            'flash_message' => $request->getSession()->getFlashBag(),
        ]);
    } else {
        //... other logic
    }
    return $response;
}

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM