Asked  7 Months ago    Answers:  5   Viewed   35 times

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 ZendViewView::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(...);



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();
                  'key' => 'value'

  $htmlOutput = $this->getServiceLocator()

  $jsonModel = new JsonModel();
    '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:

    html: "<h1>Hello World</h1>"
    jsonArray: Array[6]
    jsonVar1: "jsonVal2"
Wednesday, March 31, 2021
answered 7 Months ago

ZendLdap is not included any more in a default installation. As the ZendFramework-Team has splitted the zendframework into multiple components not all of them are needed any more. And ZendLdap is one of the "not needed" ones as it has a hard dependency on ext/ldap as you can't use it without the LDAP-extension of PHP. For details on that have a look at

You should therefore run composer require zendframework/zend-ldap from the base-directory of your project and you should be up and running. By now you get that as a suggestions after the installation.

zendframework/zendframework suggests installing zendframework/zend-ldap (zend-ldap component ~2.5.0, if you need LDAP features)
Saturday, May 29, 2021
answered 5 Months ago

Apparently, this is a known bug in the Zend Framework:

Saturday, May 29, 2021
answered 5 Months ago

Express' res.render() is still accessible in your res object.

function registerAction(req, res) {
   // handle user registration

  res.render('/path/to/view', function (err, html) {
    // email user
        subject: "Welcome",
        text: html

  // render view to the user
  return res.view({
     user: newUser
Saturday, August 7, 2021
answered 3 Months ago

RenderPartialView is a custom extension method which renders a view as a string.

It wasn't mentioned in the article (what you have referred originally) but you find it in the sample code attached to the article. It can be found under the HelpersReders.cs

Here is code of the method in question:

public static string RenderPartialView(this Controller controller, 
    string viewName, object model)
    if (string.IsNullOrEmpty(viewName))
        viewName = controller.ControllerContext.RouteData

    controller.ViewData.Model = model;
    using (var sw = new StringWriter())
        ViewEngineResult viewResult = ViewEngines.Engines
            .FindPartialView(controller.ControllerContext, viewName);
        var viewContext = new ViewContext(controller.ControllerContext, 
            viewResult.View, controller.ViewData, controller.TempData, sw);
        viewResult.View.Render(viewContext, sw);

        return sw.GetStringBuilder().ToString();
Saturday, August 28, 2021
answered 2 Months ago
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