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donet （MVC）webAPI 的接受json 的操作

news 2025/6/27 23:25:48

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);