ASP.NET MVC PartialView with List

Question

I am at my first MVC project. I want to create a page with a header and in this header to be placed a partial view with category list.

This is what I did so far: I created the master page (_Home.cshtml). Than in Shared folder I created a View (Category.cshtml). See my picture. enter image description here

My Category.cshtml content:

@model IEnumerable<ArtSchool.Models.Category>
<p>
    @Html.ActionLink("Create New", "Create")
</p>
<table>
<tr>
    <th>
        @Html.DisplayNameFor(model => model.Name)
    </th>
    <th>
        @Html.DisplayNameFor(model => model.Visible)
    </th>
    <th></th>
</tr>

@foreach (var item in Model) {
<tr>
    <td>
        @Html.DisplayFor(modelItem => item.Name)
    </td>
    <td>
        @Html.DisplayFor(modelItem => item.Visible)
    </td>
    <td>
        @Html.ActionLink("Edit", "Edit", new { id=item.ID }) |
        @Html.ActionLink("Details", "Details", new { id=item.ID }) |
        @Html.ActionLink("Delete", "Delete", new { id=item.ID })
    </td>
</tr>

}

My master page file:

@{
Layout = null;
}

<!DOCTYPE html>

 <html>
 <head>
<meta name="viewport" content="width=device-width" />
<title>_Home</title>
</head>
<body>
<div>

    @Html.Partial("ASCX/Header")

    @Html.Partial("Category")

    @RenderBody()
</div>

When I run the project I got the error:

enter image description here

I know that is a newbie question but it's my first MVC project. Thanks!

You need to pass a Model with this call @Html.Partial("Category"). You should call a Controller/Action in which this model gets generated and then return the partial. — Andrei V
– Andrei V, Commented Feb 12, 2014 at 13:39
what have you done in your actions?? you should render partial view in your controller — Ehsan
– Ehsan, Commented Feb 12, 2014 at 13:46
Are you sure? In my project I have nothing more he showed. I'm not passing any additional data,calling controler,returning partial and everything works — szpic
– szpic, Commented Feb 12, 2014 at 13:47
@szpic, is your partial view strongly typed? If so, then any reference to the Model will yield the same error. — Andrei V
– Andrei V, Commented Feb 12, 2014 at 13:47
@AndreiV Yes. In my vs strongly typed is default. There is no checkbox create strongly Typed etc — szpic
– szpic, Commented Feb 12, 2014 at 13:53

xurca · Accepted Answer · 2017-02-20 14:04:42Z

7

Solution 1

If you want to use partial view you need to pass model to this helper this way

@Html.Partial("Category", CategoryModel)

before you pass this model you have to fill it with some data.

Solution 2

also you can use @Html.Action() Method with name of ActionResult method which will return partial view for you.

for example:

  @Html.Action("GetCategories", "ControllerName")

  public ActionResult GetCategories() {
    // fill some data for your model here
    return PartialView("Category", model);
  }

edited Feb 20, 2017 at 14:04

answered Feb 12, 2014 at 13:49

xurca

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Comments

Cristian E. · Accepted Answer · 2014-02-12 13:47:59Z

2

If you want to interpret those partials as some static sections inside your HTML, then I would suggest you to call Html.Action() which returns your partials:

@Html.Action("GetPageHeader","Home")
@Html.Action("GetPageCategories","Home")

HomeController

[HttpGet]
public ActionResult GetPageHeader()
{
   return PartialView(@"~/Views/Shared/_PageHeader.cshtml");
}

[HttpGet]
public ActionResult GetPageCategories()
{
   var categories = databaseContext.GetAllCategories(); //Get your categs
   return PartialView(@"~/Views/Shared/_Categories.cshtml",categories);
}

answered Feb 12, 2014 at 13:47

Cristian E.

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