Question

Everything works fine out of the box, but after adding a new controller that returns a strongly typed view using the model ‘MySite.Models.Cars’, i get the an error when calling Html.Partial(“LogOnPartial”) in SiteLayout.cshtml. I call the new view like this:

return View(db.Cars.FirstOrDefault());

This is the error:

The model item passed into the
dictionary is of type
‘MySite.Models.Cars’, but this
dictionary requires a model item of
type ‘MySite.Models.LogOnModel’.

Very frustrating. When i use the new view without supplying a model it works again.

return View();

I setup the exact same conditions in MVC 2/ASPX and it worked just fine. I am not sure whether this is an MVC 2/MVC 3 or ASPX/Razor issue. In fact it might be a late night issue…

Very thankful for any reply.

EDIT:

First line from LogOnPartial.cshtml (i removed the underscore)

@model MySite.Models.LogOnModel

First line from Index.cshtml (new view):

@model MySite.Models.Cars

Answer 1

By default the _LogOnModel.cshtml partial as created by the template is not strongly typed and doesn’t require any model. If you modified it so that it requires a model you will need to pass this model when calling it:

@Html.Partial("_LogOnPartial", SomeLogonModelInstance)