string result ="{"AppointmentID":463236,"Message":"Successfully Appointment Booked","Success":true,"MessageCode":200,"isError":false,"Exception":null,"ReturnedValue":null}"
dynamic d = JsonConvert.DeserializeObject<dynamic>(result);
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d.GetType()为Newtonsoft.Json.Linq.JObject
那么如何将其反序列化为动态对象而不是 JObject
目前尚不清楚什么对您不起作用以及为什么您关心返回类型,但您可以直接访问反序列化对象的属性,如下所示:
string result = @"{""AppointmentID"":463236,""Message"":""Successfully Appointment Booked"",""Success"":true,""MessageCode"":200,""isError"":false,""Exception"":null,""ReturnedValue"":null}";
dynamic d = JsonConvert.DeserializeObject<dynamic>(result);
string message = d.Message;
int code = d.MessageCode;
...
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