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  • David Dupont 61 posts 115 karma points
    Nov 29, 2012 @ 20:32
    David Dupont

    Pass strongly type model in razor view

    Hi everyone,

    I'm trying to implement strongly type model in controller using Umbraco 4.10 and I can't figure what I'm doing wrong.

    Here is my razor view:

    @model Umbraco.Web.Mvc.UmbracoViewPage<Emakina.Umbraco.DocumentTypes.TextPage>
        this.Layout = "SiteRoot.cshtml";

    And here is my custom controller :

        public class TextPageController : RenderMvcController
            public override ActionResult Index(RenderModel model)
                //we will create a custom model
                var customModel = ContentHelper.GetCurrentContent() as TextPage;
                //Example for 4.10:
                //get the template name from the route values:
                var template = ControllerContext.RouteData.Values["action"].ToString();
                //return an empty content result if the template doesn't physically 
                //exist on the file system
                if (!EnsurePhsyicalViewExists(template))
                    return Content("");
                //return the current template with an instance of MyCustomModel
                return View(template, customModel);

    When I call the page I get this ysod and I wonder why since the model returned in controller is TextPage and the model declared in razor view is TextPage.

    Thanks per advance.


  • David Dupont 61 posts 115 karma points
    Nov 29, 2012 @ 20:57
    David Dupont

    Here is the error :

    [InvalidOperationException: The model item passed into the dictionary is of type 'Emakina.Umbraco.DocumentTypes.TextPage', but this dictionary requires a model item of type 'Umbraco.Web.Mvc.UmbracoViewPage`1[Emakina.Umbraco.DocumentTypes.TextPage]'.]
       System.Web.Mvc.ViewDataDictionary`1.SetModel(Object value) +403
       System.Web.Mvc.ViewDataDictionary..ctor(ViewDataDictionary dictionary) +708
       System.Web.Mvc.WebViewPage`1.SetViewData(ViewDataDictionary viewData) +74
       System.Web.Mvc.RazorView.RenderView(ViewContext viewContext, TextWriter writer, Object instance) +139
       System.Web.Mvc.ViewResultBase.ExecuteResult(ControllerContext context) +384
       System.Web.Mvc.<>c__DisplayClass1c.<InvokeActionResultWithFilters>b__19() +33
       System.Web.Mvc.ControllerActionInvoker.InvokeActionResultFilter(IResultFilter filter, ResultExecutingContext preContext, Func`1 continuation) +818804
       System.Web.Mvc.ControllerActionInvoker.InvokeActionResultWithFilters(ControllerContext controllerContext, IList`1 filters, ActionResult actionResult) +265
       System.Web.Mvc.ControllerActionInvoker.InvokeAction(ControllerContext controllerContext, String actionName) +818320
       System.Web.Mvc.Controller.ExecuteCore() +159
       System.Web.Mvc.ControllerBase.Execute(RequestContext requestContext) +335
       System.Web.Mvc.<>c__DisplayClassb.<BeginProcessRequest>b__5() +62
       System.Web.Mvc.Async.<>c__DisplayClass1.<MakeVoidDelegate>b__0() +20
       System.Web.Mvc.<>c__DisplayClasse.<EndProcessRequest>b__d() +54
       System.Web.CallHandlerExecutionStep.System.Web.HttpApplication.IExecutionStep.Execute() +469
       System.Web.HttpApplication.ExecuteStep(IExecutionStep step, Boolean& completedSynchronously) +375
  • David Dupont 61 posts 115 karma points
    Nov 29, 2012 @ 21:14
    David Dupont

    Forget about it, the following line was the culprit :

  • Edwin van Koppen 156 posts 270 karma points
    Dec 04, 2012 @ 16:44
    Edwin van Koppen

    You give the wrong object back, i stopped using the index to override and use my own controller like this

    using System;
    using System.Collections.Generic;
    using System.Linq;
    using System.Web;
    using System.Web.Mvc;
    using Umbraco.Web.Models;
    using Umbraco.Web.Mvc;
    using umbraco.cms.businesslogic.template;
    using ViewModel;

    namespace W3S.Controllers
        public class HomeController : W3S.Models.Master
            public ActionResult Home(RenderModel Model) {
                HomeViewModel HVM = new HomeViewModel();
                HVM.UmbracoModel = Model;

                return View(HVM);

    namespace ViewModel {
        public class HomeViewModel {
            public RenderModel UmbracoModel { get; set; }

    @model ViewModel.HomeViewModel
        ViewBag.Title = "Test";
        Layout = "~/Views/Shared/_Layout.cshtml";




  • Edwin van Koppen 156 posts 270 karma points
    Dec 05, 2012 @ 08:51
    Edwin van Koppen

    I'm busy with the same problem but i stumbled upon this one

    So strong typed objects you cannot get a url... pretty big problem for me.

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