MVC4模型错误不正确

Question

Hi I getting this error: The model item passed into the dictionary is of type 'System.Collections.Generic.List`1[DBModel.Telemarketing]', but this dictionary requires a model item of type 'TWeb.Models.LoginModel'

In _Layout.cshtml file i have

@Html.Partial("_LoginPartial") 

this partial login view is rendered in div on _layout page (it`s hides/shows with javaScripts )

@model TWeb.Models.LoginModel

Then I have "Telemarketings" controller having view:

public class TelemarketingController : Controller
{
    private Entities db = new Entities();

    //
    // GET: /Telemarketing/

    public ActionResult Index()
    {
        return View(db.Telemarketings.ToList());
    }

When I click link in _Layout page

@Html.ActionLink("Telemarketingas", "Index", "Telemarketing", new{area="" },new{ })

It throws an error written in top of the post.

I am new in MVC, please help me.

Answer 1

Your "_LoginPartial" expects "LoginModel" model, but since you're not giving it any, Razor engine sets its model to the current view model ("db.Telemarketings.ToList()").

All you have to do is somehow set its model, probably like so:

@Html.Partial("_LoginPartial", new LoginModel()) 

Answer 2

problem 1) Your Partial requires a model, and you're not passing one. proper syntax: @Html.Partial("_LoginPartial", Model.LoginModel)

problem 2) _layout, as far as I know, can't have a Model passed

Solution 1: Use an ActionPartial. AcionPartials are called similarly,

@Html.Action("/Tools/_LoginPartial"). 

The difference is they have an ActionMethod Associated which can return a Model

   public ActionResult _LoginPartial()
    {

          LoginModel Model= new LoginModel();
          //populate Model from whatever

         return View(Model);

    }

Option 2: Pass a LoginModel object to a Viewbag

Viewbag.LoginModel = new LoginModel();

and reference the Viewbag in your _layout's Partial

@Html.Partial("_LoginPartial", Viewbag.LoginModel) 

Answer 3

最简单的方法是从Login Div :)中删除模型声明。

Answer 4

您可以使用此代码

 @Html.Partial("Partial page", new ModelFroLogin())