donet （MVC）webAPI 的接受json 的操作

news2025/1/11 18:51:01

直接用对象来进行接收，这个方法还不错的。

    public class BangdingWeiguiJiluController : ApiController
    {

        /// <summary>
        /// Json数据录入错误信息
        /// </summary>
        /// <param name="WeiguiInfos"></param>
        /// <returns></returns>

        [System.Web.Http.HttpPost]
        public JsonResult<ResultMsgWeb> WeiguiJiluAdd([FromBody]BangdingWeiguiJilu WeiguiInfos )
        {
            //BangdingWeiguiJilu modelInf = new BangdingWeiguiJilu() { SetDate= DateTime.Now,YuangongId= 2756,SetType=1,Creator = 9999999 };
            //string Infr = JsonConvert.SerializeObject(modelInf);
            ResultMsgWeb Msg = new ResultMsgWeb();  
            try
            {
                ResultMsg MsgInsert = new DAL.DALBangdingWeiguiJilu().Insert(WeiguiInfos);
                if (MsgInsert.Success)
                {
                    Msg.Code = 200;
                    Msg.Datas = "处理完成" ;
                    Msg.Msg = $"处理完成";
                    Msg.RecordCount = 0;
                }
                else
                {
                    Msg.Code = 400;
                    Msg.Datas = "数据处理失败";
                    Msg.Msg = $"数据处理失败";
                    Msg.RecordCount = 0;
                }
            }
            catch (System.Exception e)
            {
                Msg.Code = 404;
                Msg.Datas = "数据处理失败";
                Msg.Msg = $"数据处理失败" + e.Message;
                Msg.RecordCount = 0;
            }
            return Json<ResultMsgWeb>(Msg);
        }
  }

在这里插入图片描述

调用方法就是：

var client = new RestClient("http://localhost:56727/api/BangdingWeiguiJilu/WeiguiJiluAdd");
client.Timeout = -1;
var request = new RestRequest(Method.POST);
client.UserAgent = "Apifox/1.0.0 (https://apifox.com)";
request.AddHeader("Content-Type", "application/json");
var body = @"<body data here>";
request.AddParameter("application/json", body,  ParameterType.RequestBody);
IRestResponse response = client.Execute(request);
Console.WriteLine(response.Content);