The model item passed into the dictionary is of type 'System.Collections.Generic.List`1[Demo.Models.CustomerData]', but this dictionary requires a model item of type 'Demo.Models.CustomerData'.

Hi Sajan,

I have faced today a similar problem. The Exception was like yours but for a diffrent reasons. In a single View I wanted to place three different controls. All of them strongly typed to class A. First was editor control, second was displaying read only and looked similar to a form and third represented item in a collection of records. These are the files I have created:

Controllers/ActionController.cs
Models/A.cs
Models/ActionViewModel.cs
Views/Shared/DisplayTemplates/A.cshtml - form like display
Views/Shared/DisplayTemplates/AItem.cshtml - item for a table display
Views/Shared/EditorTemplates/A.cshtml - ediatble form
Views/Action/Action.cshtml - main view for the page

Views/Action/Action.cshtml Listing

@model ModelItemType.Models.ActionViewModel

@{
    ViewBag.Title = "Action";
}

<h2>Action</h2>
@if (Model.AInstance.IsEditable)
{
@Html.EditorFor(model => model.AInstance)
}
else
{
@Html.DisplayFor(model => model.AInstance)
}
@if (Model.ACollection.Count > 0)
{
<table>
@Html.DisplayFor(model => model.ACollection, "AItem")
</table>
}

Views/Shared/EditorTemplates/A.cshtml Listing

@model ModelItemType.Models.A
@using (Html.BeginForm())
{
    @Html.ValidationSummary(true)
    <fieldset>
        @Html.EditorFor(model => model.MyProperty)
        @Html.ValidationMessageFor(model => model.MyProperty)
        <p>
            <input type="submit" value="Save" />
        </p>
    </fieldset>
}

Views/Shared/DisplayTemplates/A.cshtml Listing

@model ModelItemType.Models.A

@Html.DisplayFor(model => model.MyProperty)

Views/Shared/DisplayTemplates/AItem.cshtml Listing

@model ModelItemType.Models.A

<tr>
    <td>@Html.DisplayFor(model => model.MyProperty)</td>
</tr>

Controllers/ActionController.cs Listing

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using ModelItemType.Models;

namespace ModelItemType.Controllers
{
    public class ActionController : Controller
    {
        //
        // GET: /Action/

        public ActionResult Action()
        {
            ActionViewModel viewModel = new ActionViewModel
            {
                AInstance = new A { MyProperty = "PropertySet1", IsEditable = true },
                ACollection = new List<a> {
                    new A{ MyProperty = "PropertySet2", IsEditable = true},
                    new A{ MyProperty = "PropertySet3", IsEditable = false}
                }
            };
            return View(viewModel);
        }

    }
}

Models/A.cs Listing

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;

namespace ModelItemType.Models
{
    public class A
    {
        public string MyProperty { get; set; }
        public bool IsEditable { get; set; }
    }
}

Models/ActionViewModel.cs Listing

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;

namespace ModelItemType.Models
{
    public class ActionViewModel
    {
        public A AInstance { get; set; }
        public ICollection<a> ACollection { get; set; }
    }
}

This code throws the similar exception as in the title of this post. Look at the Views/Action/Action.cshtml:

@Html.DisplayFor(model => model.ACollection, "AItem")

Although if we change the line to:

@Html.DisplayFor(model => model.ACollection)

The model class is the same. But I was not able to have two controls bound: one to the single instance one to the collection.

Your edited question was:
"cant we show the list of data from database and fill up form to the same page like 1st view"
Solution is:
Yes we can! Just do the following to the views:
1. Change the following line in Views/Action/Action.cshtml:

@Html.DisplayFor(model => model.ACollection, "AItem")

to:

@Html.DisplayFor(model => model.ACollection)

2. Remove Views/Shared/DisplayTemplates/A.cshtml
3. Rename Views/Shared/DisplayTemplates/AItem.cshtml to Views/Shared/DisplayTemplates/A.cshtml

I think this solves your problem.

Unfortunately I'm left with mine. Why my first version does not want to work?

Kind Regards,
Paweł