Invalid Operation Exception: The model item passed into the ViewDataDictionary is of type

Question

When I list the data from the table(Tips) of Type Model as a partialView I get this error .

InvalidOperationException: The model item passed into the ViewDataDictionary is of type 'System.Collections.Generic.List1[RealEstateAspNetCore3._1.Models.Advertisement]', but this ViewDataDictionary instance requires a model item of type 'System.Collections.Generic.IEnumerable1[RealEstateAspNetCore3._1.Models.Tip]'.

The table Tip (Type) contains StatusId as a foreign key.

Tip.cs model class:

    [Key] 
    public int TypeId { get; set; }
    public string TypeName { get; set; }
    public int StatusId { get; set; }
    public virtual Status Status { get; set; }

Status.cs

    [Key] 
    public int StatusId { get; set; }
    public string StatusName { get; set; }
    public List<Tip> Tips { get; set; }

advertisement.cs

    [Key]
    public int AdvId { get; set; }       
    public string Description { get; set; }
    public int StatusId { get; set; }
    public int TypeId { get; set; }
    [ForeignKey("TypeId")]
    public Tip Tip { get; set; }

HomeController

public PartialViewResult StatusName1()
{
    string statusname1 = _context.Tips.Where(i => i.StatusId == 1)
                                      .FirstOrDefault()
                                      .ToString();
    return PartialView(statusname1);
}

Partial view StatusName1:

@model RealEstateAspNetCore3._1.Models.Tip

 <a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button">
  @Model.Status.StatusName <span class="caret"></span> </a>

Index.cshtml in Index I list the advertisement tables and works well and i used model like below

@model IEnumerable<RealEstateAspNetCore3._1.Models.Advertisement>
  

in layout after i call the partial above RenderBoy like this i get error

 <partial name="StatusName1" />

I spent two days trying to solve this, and the error stayed the same.

Thank you

Answer 1

If you're calling the partial view from _Layout.cshtml when you need to reference the full path of the view:

<partial name="~/Views/Home/StatusName1.cshtml" />

Replace Views/Home with whatever the folder names are in your project. See more here: https://learn.microsoft.com/en-us/aspnet/core/mvc/views/partial?view=aspnetcore-5.0#partial-tag-helper

There are model and a whole host of other issues in the snippets you've provided - I think the answer from @pritom-sarkar has addressed most of those

Answer 2

I assumed that your partial view name is StatusName1

change your code like below:-

public IActionResult StatusName1()
{
    var status1 = _context.Tips.Where(i => i.StatusId == 1)
                                      .FirstOrDefault();

    return PartialView("StatusName1",status1);
}

It will resolve your issue.

UPDATE

in MVC core You need to use ViewComponent instead partial view.

create a folder with the name Components The class must inherit from the ViewComponent. I will give this component class name CategoryWiseMenu

namespace DigitalShop.Components
{
    public class CategoryWiseMenu:ViewComponent
    {
       private readonly ApplicationDbContext _db;

        public CategoryWiseMenu(ApplicationDbContext db)
        {
            _db = db;
        }

        public IViewComponentResult Invoke()
        {
            var c = _db.Tips.Where(i => i.StatusId == 1)
                                      .FirstOrDefault();
            return View("Default.cshtml",c);
        }
    }
}

~/Views/Shared/Components/Default.cshtml

Create Default.cshtml in Components folder like above path.

Demo Default.cshtml

<ul class="navbar-nav flex-grow-1">


    @foreach (var rr in Model)
    {
        <li class="nav-item text-dark">
            <a asp-controller="" asp-action=""
                class="nav-link text-dark">@rr.TypeName</a>
        </li>
    }
    
</ul>

now include below code in your layout page.

@await Component.InvokeAsync("CategoryWiseMenu")

now it will works fine which you want.