How to transfer paramenters to Partial View in ASP.NET MVC?

Question

On the Details view I want to display a grid from other table. How to implement it better? I'm trying to do it in a such a way (error: Cannot implicitly convert type 'void' to 'object'):

<%= Html.RenderPartial("~/Views/Appartament/Index.ascx", new { id  = Model.blockhouse_id })%>

Here is the code from details view:

<%@ Page Title="" Language="C#" MasterPageFile="~/Views/Shared/Site.Master" Inherits="System.Web.Mvc.ViewPage<testMVC.Models.Blockhouse>" %>

<asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server">

    <h2>Details</h2>

        <p>
            blockhouse_id:
            <%= Html.Encode(Model.blockhouse_id) %>
        </p>
        <p>
            name:
            <%= Html.Encode(Model.name) %>
        </p>    

         <p>
            Appartaments:
            <%= Html.RenderPartial("~/Views/Appartament/Index.ascx", new { id  = Model.blockhouse_id })%>
        </p>       

</asp:Content>

And the code of the Index.ascx controller:

public ActionResult Index(int blockhouse_id)
        {
            var _appartament = apt.GetBlockAppartaments(blockhouse_id).ToList();
            return View("Index", _appartament);
        }

Answer 1

RenderPartial doesn't have a return value.

Instead of

<%= %>

Use

<% %>

• ASP.NET MVC RenderPartial Compiler Error

As for Razor, instead of

@

Use

@{ }