asp.net-mvc - 在ASP.NET MVC中，如何重构try/catch并返回JSON错误消息？

Question

如何在新的ASP.NET MVC应用程序中最好地创建基本 Controller ，该应用程序将包含样板代码来处理派生 Controller 中每个 Action 的所有try/catch例程，并在登录Nlog后返回标准JSON错误消息。我还想处理404错误以重定向到自定义404 View 。我的目标是使用OWIN Cookie身份验证，并希望在Cookie过期时退出。我所有的操作都将返回JsonResult并通过jQuery Ajax调用。

在以前的项目中，我使用了以下方法:

用Global.asax.cs

  protected void Application_Error(object sender, EventArgs e)
    {
        Exception lastError = Server.GetLastError();
        Server.ClearError();

        var statusCode = 0;

        statusCode = lastError.GetType() == typeof(HttpException) ? ((HttpException)lastError).GetHttpCode() : 500;

        var routeData = new RouteData();
        routeData.Values.Add("controller", "Error");
        routeData.Values.Add("statusCode", statusCode);
        routeData.Values.Add("exception", lastError);

        if (new HttpRequestWrapper(HttpContext.Current.Request).IsAjaxRequest())
        {
            routeData.Values.Add("action", "Ajax");     
        }
        else
        {
            routeData.Values.Add("action", "Index");
        }

        var requestContext = new RequestContext(new HttpContextWrapper(Context), routeData);

        IController controller = new ErrorController();
        controller.Execute(requestContext);

        Response.End();
    }

用ErrorController.cs

 public class ErrorController : Controller
{
    public ActionResult Index(int statusCode, Exception exception)
    {
        var model = new ErrorModel { HttpStatusCode = statusCode, Exception = exception.Message };

        Response.StatusCode = statusCode;

        return View(model);
    }

    public JsonResult Ajax(int statusCode, Exception exception, Dictionary<string,string> validationErrors = null)
    {
        var model = new ErrorModel { HttpStatusCode = statusCode, Exception = exception.Message };

        if (exception.GetType() == typeof (DbEntityValidationException))
        {
            model.ValidationErrors = null;
        }
        else
        {
            model.ValidationErrors = validationErrors;
        }

        Response.StatusCode = statusCode;

        return Json(model, JsonRequestBehavior.AllowGet);
    }
}

Answer 1

最好的方法是通过扩展HandleErrorAttribute像这样:

public class AjaxAwareHandleErrorAttribute : HandleErrorAttribute
{
    public override void OnException(ExceptionContext filterContext)
    {
        // Execute the normal exception handling routine...
        base.OnException(filterContext);            

        // Verify if AJAX request...
        if (filterContext.HttpContext.Request.IsAjaxRequest())
        {
            // Use Json in case of AJAX request...
            var result = new JsonResult();                
            result.Data = new { Error = filterContext.Exception.Message };

            filterContext.Result = result;
        }
    }
}

然后，您需要做的就是在FilterConfig文件夹下的App_Start类中注册该属性，如下所示:

public class FilterConfig
{
    public static void RegisterGlobalFilters(GlobalFilterCollection filters)
    {
        filters.Add(new AjaxAwareHandleErrorAttribute());
    }
}