在ASP.NET MVC中如何重构try / catch并返回JSON错误消息?

时间:2014-12-01 20:10:37

标签: asp.net-mvc json error-handling refactoring

如何在新的ASP.NET MVC应用程序中创建基本控制器,该应用程序将包含样板代码,以处理派生控制器中每个操作的所有try / catch例程,并在登录到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视图。我的目标是使用OWIN cookie身份验证,并希望在Cookie过期时注销。我的所有操作都将返回JsonResult并通过jQuery Ajax调用。

在之前的项目中,我使用了以下方法:

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();
    }

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);
    }
}

1 个答案:

答案 0 :(得分:0)

最好的方法是将HandleErrorAttribute扩展为:

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类的App_Start类下注册该属性,如下所示:

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