asp.net-mvc – ASP.NET MVC用户友好401错误

我在ASP.NET MVC site in a way like suggests this post中实现了错误处理.

404错误都可以正常工作.但是如何正确显示用户友好的屏幕401错误？他们通常不会抛出可以在Application_Error()中处理的异常,而是返回HttpUnauthorizedResult.一种可能的方法是将以下代码添加到Application_EndRequest()方法的末尾

if (Context.Response.StatusCode == 401)
{
    throw new HttpException(401,"You are not authorised");
    // or UserFriendlyErrorRedirect(new HttpException(401,"You are not authorised")),witout exception
}

但在Application_EndRequest()Context.Session == null,errorController.Execute()失败,因为它不能使用默认的TempDataProvider.

// Call target Controller and pass the routeData.
  IController errorController = new ErrorController();
  errorController.Execute(new RequestContext(    
       new HttpContextwrapper(Context),routeData)); // Additional information: The SessionStateTempDataProvider requires SessionState to be enabled.

那么你能建议一些最佳实践,如何在ASP.NET MVC应用程序中“用户友好的处理”401？

谢谢.

解决方法

看看HandleErrorAttribute.从它的子类或添加您自己的实现,将处理您感兴趣的所有状态代码.您可以使它为每个错误类型返回单独的错误视图.

这是一个如何创建你的句柄错误异常过滤器的想法.我抛出了大部分的东西,只关注我们的要领.绝对看一下原来的实现,添加参数检查和其他重要的事情.

public class HandleManyErrorsAttribute : Filterattribute,IExceptionFilter
{
    public virtual void OnException(ExceptionContext filterContext)
    {
        if (filterContext.ExceptionHandled || !filterContext.HttpContext.IsCustomErrorEnabled)
            return;

        Exception exception = filterContext.Exception;

        string viewName = string.Empty;
        object viewmodel = null;
        int httpCode = new HttpException(null,exception).GetHttpCode();
        if (httpCode == 500)
        {
            viewName = "Error500View";
            viewmodel = new Error500Model();
        }
        else if (httpCode == 404)
        {
            viewName = "Error404View";
            viewmodel = new Error404Model();
        }
        else if (httpCode == 401)
        {
            viewName = "Error401View";
            viewmodel = new Error401Model();
        }

        string controllerName = (string)filterContext.RouteData.Values["controller"];
        string actionName = (string)filterContext.RouteData.Values["action"];
        filterContext.Result = new ViewResult
        {
            ViewName = viewName,MasterName = Master,ViewData = viewmodel,TempData = filterContext.Controller.TempData
        };
        filterContext.ExceptionHandled = true;
        filterContext.HttpContext.Response.Clear();
        filterContext.HttpContext.Response.StatusCode = httpCode;

        filterContext.HttpContext.Response.TrySkipIisCustomErrors = true;
    }
}

然后,使用此属性“装饰”控制器操作：

[HandleManyErrors]
public ActionResult DoSomethingBuggy ()
{
    // ...
}

原文地址：https://www.jb51.cc/aspnet/250535.html