Felasfaw Felasfaw - 17 days ago 8
ASP.NET (C#) Question

ASP.NET MVC 3 Partial View in layout page

I'm working on setting up a shared content (navigation) for an asp.net MVC layout page.

Here is my partial view "_LayoutPartial.cshtml" with code to pull navigation data from a model.

@model MyApp.Models.ViewModel.LayoutViewModel
<p>

@foreach (var item in Model.navHeader)
{
//Test dump of navigation data
@Html.Encode(item.Name);
@Html.Encode(item.URL);

}
</p>


Here is how the code for my controller "LayoutController.cs" looks like.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using MyApp.Models.ViewModel;

namespace MyApp.Controllers
{
public class LayoutController : Controller
{

//
// GET: /Layout/

LayoutViewModel layout = new LayoutViewModel();

public ActionResult Index()
{
return View(layout);
}

}
}


Here is the code for the "_Layout.cshtml" page. I'm attempting to call the partial view here using Html.RenderAction(Action,Controller) method.

<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
</head>
<body>
<p>
@{Html.RenderAction("Index","Layout");}
</p>

@RenderBody()
</body>
</html>


When the layout page executes the @{Html.RenderAction("Index","Layout");} line, it throws out an error message "Error executing child request for handler 'System.Web.Mvc.HttpHandlerUtil+ServerExecuteHttpHandlerAsyncWrapper'."

What am I missing friends? How can I call a partial view in a layout page?

Thank you all in advance!

Answer

Instead of:

public ActionResult Index()
{
    return View(layout);
}

do:

public ActionResult Index()
{
    return PartialView(layout);
}

If you don't do that when you return a normal view from your child action, this normal view attempts to include the Layout, which in turn attempts to render the child action, which in turn returns a view, which in turn includes the Layout, which in turn attempts to render the child action, ... and we end up with names like the one ported by this very same site.

Also in your partial you don't need to do double encoding. The @ Razor function already does HTML encode:

@model MyApp.Models.ViewModel.LayoutViewModel
<p>

    @foreach (var item in Model.navHeader)
    {
        @item.Name 
        @item.URL
    }
</p>