使用ID作为名称，从json对象创建强类型的c#对象

Question

使用ID作为名称，从json对象创建强类型的c#对象我试图利用API为一个著名的在线会议提供商。其中一个API调用返回的对象如下所示。{     "5234592":{     "pollsAndSurveys":{         "questionsAsked":1,         "surveyCount":0,         "percentageSurveysCompleted":0,         "percentagePollsCompleted":100,         "pollCount":2},     "attendance":{         "averageAttendanceTimeSeconds":253,         "averageInterestRating":0,         "averageAttentiveness":0,         "registrantCount":1,         "percentageAttendance":100}     },     "5235291":{     "pollsAndSurveys":{         "questionsAsked":2,         "surveyCount":0,         "percentageSurveysCompleted":0,         "percentagePollsCompleted":0,         "pollCount":0},     "attendance":{         "averageAttendanceTimeSeconds":83,         "averageInterestRating":0,         "averageAttentiveness":0,         "registrantCount":1,         "percentageAttendance":100}     }}我试图在C#中创建一个强类型的对象，以便处理这些数据。我可以为pollsAndSurvey位和考勤位创建对象，但是我不知道如何处理id号，在本例中是5234592&5235291，这是会话的标识符。public class AttendanceStatistics{     [JsonProperty(PropertyName = "registrantCount")]     public int RegistrantCount { get; set; }     [JsonProperty(PropertyName = "percentageAttendance")]     public float PercentageAttendance{ get; set; }     [JsonProperty(PropertyName = "averageInterestRating")]     public float AverageInterestRating { get; set; }     [JsonProperty(PropertyName = "averageAttentiveness")]     public float AverageAttentiveness { get; set; }     [JsonProperty(PropertyName = "averageAttendanceTimeSeconds")]     public float AverageAttendanceTimeSeconds { get; set; }}public class PollsAndSurveysStatistics{     [JsonProperty(PropertyName = "pollCount")]     public int PollCount { get; set; }     [JsonProperty(PropertyName = "surveyCount")]     public float SurveyCount { get; set; }     [JsonProperty(PropertyName = "questionsAsked")]     public int QuestionsAsked { get; set; }我很确定WebinarPerformanceStats是问题所在，但我不知道从哪里出发。我要改变什么才能得到NewtonSoft.Json.JsonConvert.DeserializeObject<WebinarPerformanceStats>(theJsonResponse)去工作？