asp.net-mvc – 在ASP.NET MVC控制器结果中设置HTTP状态不呈现视

I have a custom ActionResult for
returning certain HTTP Errors,like
NotFoundResult and ForbiddenResult,
they all derive from ViewResult.

请允许我建议您替代错误处理：

首先创建一个错误控制器和相应的视图：

public class ErrorController : Controller
{
    public ActionResult General()
    {
        return View();
    }

    public ActionResult HttpError404()
    {
        return View();
    }

    public ActionResult HttpError500()
    {
        return View();
    }
}

在Global.asax中定义Application_Error方法：

protected void Application_Error(object sender,EventArgs e)
{
    var exception = Server.GetLastError();
    // TODO: Log the exception with your favorite logging framework

    Response.Clear();
    var httpException = exception as HttpException;

    var routeData = new RouteData();
    // Take the ErrorController
    routeData.Values.Add("controller","error");

    if (httpException == null)
    {
        // Use the General action for any unhandled error
        routeData.Values.Add("action","general");
    }
    else
    {
        switch (httpException.GetHttpCode())
        {
            case 404:
                routeData.Values.Add("action","httpError404");
                break;
            case 500:
                routeData.Values.Add("action","httpError500");
                break;
            default:
                routeData.Values.Add("action","general");
                break;
        }
    }

    // Add the exception to route data so that the error controller 
    // could use it with RouteData.Values["error"]
    routeData.Values.Add("error",exception);

    Server.ClearError();
    IController errorController = new ErrorController();
    errorController.Execute(new RequestContext(new HttpContextWrapper(Context),routeData));
}

最后抛出适当的例外：

public class HomeController: Controller
{
    public ActionResult Index(int id)
    {
        var model = _repository.GetModel(id);
        if (model == null)
        {
            throw new HttpException(404,"Model not found with id = " + id);
        }
        return View(model);
    }
}