Question

I have a view that is not strongly typed. However I have in this view a partial view that is strongly typed.

How do I do I pass the model to this strongly typed view?

I tried something like

 public ActionResult Test()
        {
              MyData = new Data();
              MyData.One = 1;
              return View("Test",MyData)
        }

In my TestView

<% Html.RenderPartial("PartialView",Model); %>

This give me a stackoverflow exception. So I am not sure how to pass it on. Of course I don’t want to make the test view strongly typed if possible as what happens if I had like 10 strongly typed partial views in that view I would need like some sort of wrapper.

Answer 1

You should extend your model so that it can provide all necessary fields for the view (this is called ViewModel) or you provide them seperately with ViewData.

 public ActionResult Test()
        {
              MyData = new Data();
              MyData.One = 1;
              ViewData["someData"]=MyData;
              return View();
        }

then:

<% Html.RenderPartial("PartialView",ViewData["someData"]); %>

ViewData is a nice losely typed dictionary