ASP.NET MVC 3 Partial View in layout page

问题

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!

回答1:

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>

回答2:

First verify that your child view is inside the Shared directory

@Html.Partial("_LayoutPartial")

 @{Html.RenderAction("actionname", "controller name");}

And don't use @Html.Encode(), Razor is already doing for u. Just use

@item.Name 
@item.URL

回答3:

I have solved this error getting on Layout page

System.Web.Mvc.HttpHandlerUtil+ServerExecuteHttpHandlerAsyncWrapper

Important ! First create partial view inside shared folder

In Controller ,

public PartialViewResult Userdetails()
{
   ....
   return PartialView("PartialViewName", obj);  
}

In Layout Page,

@{Html.RenderAction("action","controller");}

回答4:

I know this is an old question but I thought I would throw this in here. You can use either Html.Action or Html.RenderAction. They both technically do the same thing but depending on how much content you're returning back can have an impact on which one you should really use for best efficiency.

Both of the methods allow you to call into an action method from a view and output the results of the action in place within the view. The difference between the two is that Html.RenderAction will render the result directly to the Response (which is more efficient if the action returns a large amount of HTML) whereas Html.Action returns a string with the result.

Source

来源：https://stackoverflow.com/questions/11459390/asp-net-mvc-3-partial-view-in-layout-page

标签

ASP.NET

asp.net-mvc-2

partial-views