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How to render ZF2 view within JSON response?

So far, I have figured out how to return a typical JSON response in Zend Framework 2. First, I added the ViewJsonStrategy to the strategies section of the view_manager configuration. Then, instead of returning a ViewModel instance from the controller action, I return a JsonModel instance with all my variables set.

Now that I've figured that piece out, I need to understand how to render a view and return it within that JSON response. In ZF1, I was able to use $this->view->render($scriptName), which returned the HTML as a string. In ZF2, the Zend\View\View::render(...) method returns void.

So... how can I render an HTML view script and return it in a JSON response in one request?

This is what I have right now:

    if ($this->getRequest()->isXmlHttpRequest()) {
        $jsonModel = new JsonModel(...);

        /* @todo Render HTML script into `$html` variable, and add to `JsonModel` */
        return $jsonModel;
    } else {
        return new ViewModel(...);
    }
like image 644
webjawns.com Avatar asked Sep 16 '12 22:09

webjawns.com


3 Answers

OK, i think i finally understood what you're doing. I've found a solution that i think matches your criteria. Though i am sure that there is room for improvement, as there's some nasty handwork to be done...

public function indexAction()
{
  if (!$this->getRequest()->isXmlHttpRequest()) {
    return array();
  }

  $htmlViewPart = new ViewModel();
  $htmlViewPart->setTerminal(true)
               ->setTemplate('module/controller/action')
               ->setVariables(array(
                  'key' => 'value'
               ));

  $htmlOutput = $this->getServiceLocator()
                     ->get('viewrenderer')
                     ->render($htmlViewPart);

  $jsonModel = new JsonModel();
  $jsonModel->setVariables(array(
    'html' => $htmlOutput,
    'jsonVar1' => 'jsonVal2',
    'jsonArray' => array(1,2,3,4,5,6)
  ));

  return $jsonModel;
}

As you can see, the templateMap i create is ... nasty ... it's annoying and i'm sure it can be improved by quite a bit. It's a working solution but just not a clean one. Maybe somehow one would be able to grab the, probably already instantiated, default PhpRenderer from the ServiceLocator with it's template- and path-mapping and then it should be cleaner.

Thanks to the comment ot @DrBeza the work needed to be done could be reduced by a fair amount. Now, as I'd initially wanted, we will grab the viewrenderer with all the template mapping intact and simply render the ViewModel directly. The only important factor is that you need to specify the fully qualified template to render (e.g.: "$module/$controller/$action")

I hope this will get you started though ;)

PS: Response looks like this:

Object:
    html: "<h1>Hello World</h1>"
    jsonArray: Array[6]
    jsonVar1: "jsonVal2"
like image 101
Sam Avatar answered Nov 15 '22 08:11

Sam


You can use more easy way to render view for your JSON response.

public function indexAction() {
    $partial = $this->getServiceLocator()->get('viewhelpermanager')->get('partial');
    $data = array(
            'html' => $partial('MyModule/MyPartView.phtml', array("key" => "value")),
            'jsonVar1' => 'jsonVal2',
            'jsonArray' => array(1, 2, 3, 4, 5, 6));
    $isAjax = $this->getRequest()->isXmlHttpRequest());
    return isAjax?new JsonModel($data):new ViewModel($data);
}

Please note before use JsonModel class you need to config View Manager in module.config.php file of your module.

'view_manager' => array(
        .................
        'strategies' => array(
            'ViewJsonStrategy',
        ),
        .................
    ),

it is work for me and hope it help you.

like image 20
Maksym Kalin Avatar answered Nov 15 '22 06:11

Maksym Kalin


In ZF 3 you can achieve the same result with this code

MyControllerFactory.php

public function __invoke(ContainerInterface $container, $requestedName, array $options = null)
{
    $renderer = $container->get('ViewRenderer');

    return new MyController(
        $renderer
    );
}

MyController.php

private $renderer;

public function __construct($renderer) {
    $this->renderer = $renderer;
}

public function indexAction() {

    $htmlViewPart = new ViewModel();
    $htmlViewPart
            ->setTerminal(true)
            ->setTemplate('module/controller/action')
            ->setVariables(array('key' => 'value'));

    $htmlOutput = $this->renderer->render($htmlViewPart);

    $json = \Zend\Json\Json::encode(
        array(
            'html' => $htmlOutput,
            'jsonVar1' => 'jsonVal2',
            'jsonArray' => array(1, 2, 3, 4, 5, 6)
        )
    );

    $response = $this->getResponse();
    $response->setContent($json);

    $response->getHeaders()->addHeaders(array(
        'Content-Type' => 'application/json',
    ));
    return $this->response;
}
like image 2
halanson Avatar answered Nov 15 '22 07:11

halanson