C# (CSharp) BlogResponse.SerializeToJson示例

示例#1

        public void OnException(ExceptionContext context)
        {
            if (context.Exception != null)
            {
                var result = new BlogResponse();
                result.IsFailed(context.Exception.Message);

                context.Result = new ContentResult()
                {
                    Content    = result.SerializeToJson(),
                    StatusCode = StatusCodes.Status200OK
                };

                context.ExceptionHandled = true;
            }
        }