执行控制器方法之前，Ajax发布到控制器失败

Question

我正在尝试使用Ajax发布保存更改。我似乎在执行控制器方法之前就已经碰到了Ajax方法的fail事件。我想要成功返回“ 0”或“ 1”，或者将其他任何错误消息作为字符串返回给我的Ajax方法。我在这里想念什么？ 下面是我的控制器和Ajax post方法的摘录。

Assets.Launch = {
    ProductService: function ($stockItemId, $stockItemName, successCallback, failureCallback) {
        var data = {
            "stockItemId": $stockItemId,
            "stockItemName": $stockItemName
        };

        $.ajax({
            url: "/Purchasing/AddEditProductService",
            type: "POST",
            data: data
        })

        .done(function (result) {
            if (result === "0") {
                successCallback($stockItemName + " Successfully added");
            }
            else if (result === "1") {
                successCallback($stockItemName + " Successfully updated");
            }
            else {
                failureCallback(result);
            }
        })
            .fail(function (jqxhr, textStatus, errorThrown) {                    
            failureCallback(errorThrown);
        });


    }
};


public JsonResult AddEditProductService(int? stockItemId, string stockItemName)
    {
        string errorToReport = "";
        bool isDuplicate = false;
        bool isAdding = (stockItemId == 0);
        StockItems entity;

        using (var ctx = iuow)
        {
            if (isAdding)
            {
                entity = new StockItems();
            }
            else
            {
                entity = ctx.StockItems.Find(e => e.StockItemId == stockItemId).FirstOrDefault();
            }

            ................

            try
            {
                if (isAdding)
                {
                    ctx.StockItems.Add(entity);
                    ctx.SaveChanges();
                    return Json("0");
                }
                else
                {
                    ctx.StockItems.Update(entity);
                    ctx.SaveChanges();
                    return Json("1");
                }
            }

            catch (DbUpdateException ex)
            {
                if (ex.InnerException != null)
                {
                    var sqlException = ex.InnerException as SqlException;
                    isDuplicate = IsDuplicate(sqlException);
                    if (isDuplicate)
                    {
                        errorToReport = "Cannot insert duplicate key [" + stockItemName + "]";
                    }
                    else
                    {
                        errorToReport = sqlException.Message;
                    }
                }
                return Json(errorToReport);
            }

            catch (Exception exception)
            {
                errorToReport = (exception.InnerException == null) ? exception.Message : exception.InnerException.Message;
                _logger.LogError(errorToReport);
                return Json(errorToReport);
            }
        }
    }