Question

我想创建一个自定义异常过滤器，它将捕获返回JSON结果的控制器操作中抛出的异常。

我想重构以下操作方法：

        public JsonResult ShowContent()
    {
        try
        {
            // Do some business logic work that might throw a business logic exception ...
            //throw new ApplicationException("this is a business exception");

            var viewModel = new DialogModel
                                {
                                    FirstName = "John",
                                    LastName = "Doe"
                                };

            // Other exceptions that might happen:
            //throw new SqlException(...);
            //throw new OtherException(...);
            //throw new ArgumentException("this is an unhandeled exception");

            return
                Json(
                    new
                        {
                            Status = DialogResultStatusEnum.Success.ToString(),
                            Page = this.RenderPartialViewToString("ShowContent", viewModel)
                        });
        }
        catch (ApplicationException exception)
        {
            return Json(new { Status = DialogResultStatusEnum.Error.ToString(), Page = exception.Message });
        }
        catch (Exception exception)
        {
            return Json(new { Status = DialogResultStatusEnum.Exception.ToString(), Page = "<h2>PROBLEM!</h2>" });
        }
    }
}

我想要做的是创建一个自定义异常过滤器属性，该属性将捕获操作中引发的任何异常，遵循以下逻辑：

检查是否有异常
- 否：返回
- 是：
  - 如果是BusinessLogic异常 - 返回JSON结果
  - 如果有其他未处理的异常：
    - 日志
    - 使用不同的结果代码

Answer 1

我发现可以使用this article中的代码解决此问题（对其进行微小更改。）

public class HandleJsonExceptionAttribute : ActionFilterAttribute
{
    #region Instance Methods

    public override void OnActionExecuted(ActionExecutedContext filterContext)
    {
        if (filterContext.Exception != null)
        {
            filterContext.HttpContext.Response.StatusCode = (int)System.Net.HttpStatusCode.InternalServerError;
            filterContext.Result = new JsonResult()
            {
                JsonRequestBehavior = JsonRequestBehavior.AllowGet,
                Data = new
                {
                    Error = filterContext.Exception.Message
                }
            };
            filterContext.ExceptionHandled = true;
        }
    }

    #endregion
}

Answer 2

  public class YourController : BaseController
    {
        public JsonResult showcontent()
        {
            // your logic here to return foo json
            return Json ( new { "dfd" }); // just a dummy return text replace it wil your code
        }
    }

    public class BaseController : Controller
    {
 // Catch block for all exceptions in your controller
        protected override void OnException(ExceptionContext filterContext)
        {
            base.OnException(filterContext);
            if (filterContext.Exception.Equals(typeof(ApplicationException)))
            {
                //do  json and return
            }
            else
            {
                // non applictaion exception
// redirect to generic error controllor and error action which will return json result

            }
        }

    }

请参阅此链接，了解如何创建和使用HandleError attribute

* 编辑 for HandleAttribute for Actions *

//not tested code
public class HandleErrorfilter : HandleErrorAttribute
    {
         public string ErrorMessage { get; set; }

        public override void OnException(ExceptionContext filterContext)
        {
                string message = string.Empty;
                //if application exception
                // do  something
                 else
                    message = "An error occured while attemting to perform the last action.  Sorry for the inconvenience.";
            }
            else
            {
                base.OnException(filterContext);
            }
        }

      public class YourController : BaseController
        {
            [HandleErrorfilter]
            public JsonResult showcontent()
            {
                // your logic here to return foo json
                return Json ( new { "dfd" }); // just a dummy return text replace it wil your code
            }
        }

如何从自定义异常过滤器返回JSON结果？

2 个答案: