MVC AJAX Search results in Partial View [duplicate] - asp.net-mvc

Question

This question and community wiki answer has been added to assist in closing out numerous unanswered questions as discussed in this meta post.
I have some code and when it executes, it throws an exception saying:
The model item passed into the dictionary is of type Bar but this dictionary requires a model item of type Foo
What does this mean, and how do I fix it?

Answer 1

The error means that you're navigating to a view whose model is declared as typeof Foo (by using #model Foo), but you actually passed it a model which is typeof Bar (note the term dictionary is used because a model is passed to the view via a ViewDataDictionary).
The error can be caused by
Passing the wrong model from a controller method to a view (or partial view)
Common examples include using a query that creates an anonymous object (or collection of anonymous objects) and passing it to the view
var model = db.Foos.Select(x => new
{
ID = x.ID,
Name = x.Name
};
return View(model); // passes an anonymous object to a view declared with #model Foo
or passing a collection of objects to a view that expect a single object
var model = db.Foos.Where(x => x.ID == id);
return View(model); // passes IEnumerable<Foo> to a view declared with #model Foo
The error can be easily identified at compile time by explicitly declaring the model type in the controller to match the model in the view rather than using var.
Passing the wrong model from a view to a partial view
Given the following model
public class Foo
{
public Bar MyBar { get; set; }
}
and a main view declared with #model Foo and a partial view declared with #model Bar, then
Foo model = db.Foos.Where(x => x.ID == id).Include(x => x.Bar).FirstOrDefault();
return View(model);
will return the correct model to the main view. However the exception will be thrown if the view includes
#Html.Partial("_Bar") // or #{ Html.RenderPartial("_Bar"); }
By default, the model passed to the partial view is the model declared in the main view and you need to use
#Html.Partial("_Bar", Model.MyBar) // or #{ Html.RenderPartial("_Bar", Model.MyBar); }
to pass the instance of Bar to the partial view. Note also that if the value of MyBar is null (has not been initialized), then by default Foo will be passed to the partial, in which case, it needs to be
#Html.Partial("_Bar", new Bar())
Declaring a model in a layout
If a layout file includes a model declaration, then all views that use that layout must declare the same model, or a model that derives from that model.
If you want to include the html for a separate model in a Layout, then in the Layout, use #Html.Action(...) to call a [ChildActionOnly] method initializes that model and returns a partial view for it.

Answer 2

This question already has a great answer, but I ran into the same error, in a different scenario: displaying a List in an EditorTemplate.
I have a model like this:
public class Foo
{
public string FooName { get; set; }
public List<Bar> Bars { get; set; }
}
public class Bar
{
public string BarName { get; set; }
}
And this is my main view:
#model Foo
#Html.TextBoxFor(m => m.Name, new { #class = "form-control" })
#Html.EditorFor(m => m.Bars)
And this is my Bar EditorTemplate (Bar.cshtml)
#model List<Bar>
<div class="some-style">
#foreach (var item in Model)
{
<label>#item.BarName</label>
}
</div>
And I got this error:
The model item passed into the dictionary is of type 'Bar', but this
dictionary requires a model item of type
'System.Collections.Generic.List`1[Bar]
The reason for this error is that EditorFor already iterates the List for you, so if you pass a collection to it, it would display the editor template once for each item in the collection.
This is how I fixed this problem:
Brought the styles outside of the editor template, and into the main view:
#model Foo
#Html.TextBoxFor(m => m.Name, new { #class = "form-control" })
<div class="some-style">
#Html.EditorFor(m => m.Bars)
</div>
And changed the EditorTemplate (Bar.cshtml) to this:
#model Bar
<label>#Model.BarName</label>

Answer 3

Observe if the view has the model required:
View
#model IEnumerable<WFAccess.Models.ViewModels.SiteViewModel>
<div class="row">
<table class="table table-striped table-hover table-width-custom">
<thead>
<tr>
....
Controller
[HttpGet]
public ActionResult ListItems()
{
SiteStore site = new SiteStore();
site.GetSites();
IEnumerable<SiteViewModel> sites =
site.SitesList.Select(s => new SiteViewModel
{
Id = s.Id,
Type = s.Type
});
return PartialView("_ListItems", sites);
}
In my case I Use a partial view but runs in normal views

Answer 4

Consider the partial map.cshtml at Partials/Map.cshtml. This can be called from the Page where the partial is to be rendered, simply by using the <partial> tag:
<partial name="Partials/Map" model="new Pages.Partials.MapModel()" />
This is one of the easiest methods I encountered (although I am using razor pages, I am sure same is for MVC too)

Answer 5

First you need to return an IEnumerable version of your model to the list view.
#model IEnumerable<IdentityManager.Models.MerchantDetail>
Second, you need to return a list from the database. I am doing it via SQL Server, so this is code I got working.
public IActionResult Merchant_Boarding_List()
List<MerchantDetail> merchList = new List<MerchantDetail>();
var model = new MerchantDetail();
try
{
using (var con = new SqlConnection(Common.DB_CONNECTION_STRING_BOARDING))
{
con.Open();
using (var command = new SqlCommand("select * from MerchantDetail md where md.UserGUID = '" + UserGUID + "'", con))
{
using (SqlDataReader reader = command.ExecuteReader())
{
while (reader.Read())
{
model.biz_dbaBusinessName = reader["biz_dbaBusinessName"].ToString();
merchList.Add(model);
}
}
}
}
}
catch (Exception ex)
{
}
return View(merchList);

Answer 6

Passing the model value that is populated from a controller method to a view
public async Task<IActionResult> Index()
{
//Getting Data from Database
var model= await _context.GetData();
//Selecting Populated Data from the Model and passing to view
return View(model.Value);
}

Answer 7

one more thing.
if your view is a partial/sub page and the model for that partial view is null for some reason (e.g no data) you will get this error. Just need to handle the null partial view model