在一個地方處理所有Ajax異常，並返回錯誤以查看

Question

我有一個相當大的應用程序，有數十個Ajax調用。我想在單個地方記錄出現在ajax調用中的任何錯誤。我在我的_Layout.cshmtl視圖上放置了一個jquery警報，以便可以將異常傳遞給警報。如何從我的HandleExceptionAttribute類中返回錯誤字符串到我的視圖？

控制器動作：

[HandleExceptionAttribute] 
    [HttpPost] 
    [Authorize (Roles = "View Only,Admin,A-Team Manager,A-Team Analyst")] 
    public JsonResult GetEntitySorMapTable (Decimal entityId) 
    { 
     //Added this line to hit my HandleExceptionAttribute 
     throw new DivideByZeroException(); 

     List<EntitySorMapView> entitySorMaps = null; 

     if (entityId == 0) 
     { 
      entitySorMaps = new List<EntitySorMapView> (); 
     } 

     entitySorMaps = RealmsModel.RealmsDataInterface ().SelectEntitySorMapByEntityId (entityId); 

     String data = HtmlHelpers.BuildEntitySorMapTable (entitySorMaps); 

     return new JsonResult () 
     { 
      Data = data, 
      MaxJsonLength = Int32.MaxValue 
     };    
    } 

錯誤屬性：

public class HandleExceptionAttribute : HandleErrorAttribute 
{ 
    public override void OnException(ExceptionContext filterContext) 
    { 
     if (filterContext.HttpContext.Request.IsAjaxRequest() && filterContext.Exception != null) 
     { 
      filterContext.HttpContext.Response.StatusCode = (int)HttpStatusCode.InternalServerError; 
      filterContext.Result = new JsonResult 
      { 
       JsonRequestBehavior = JsonRequestBehavior.AllowGet, 
       Data = new 
       { 
        filterContext.Exception.Message, 
        filterContext.Exception.StackTrace 
       } 
      }; 
      filterContext.ExceptionHandled = true; 
     } 
     else 
     { 
      base.OnException(filterContext); 
     } 
    } 
} 

_layout查看Ajax錯誤的腳本：

<script type="text/javascript"> 
    $(document).ajaxError(function (xhr, status, error) { 
     e.stopPropagation(); 
     if (xhr.error != null) 
      alert('Error: ' + xhr.responseText + ' status: ' + status + ' Exception: ' + error); 
    }); 
</script> 

Answer 1

您可以使用

filterContext.Result = new JsonResult 
{ 
    Data = new { errorMessage = "Your custom error message" }, 
    JsonRequestBehavior = JsonRequestBehavior.AllowGet 
}; 

和您的_layout.cshtml文件中

$(document).ajaxError(function (event, jqxhr, settings, thrownError) { 
    if (jqxhr.error != null) { 
     var result = JSON.parse(jqxhr.responseText); 
      console.log(result.errorMessage) 
    } 
});