从ID为名称的json对象创建一个强类型的c＃对象

  • Z时代
  • 2024-01-10
  • 分类：问答

我正在尝试为一个著名的在线会议提供者使用该API。他们的API调用之一返回一个看起来像这样的对象。

{
    "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＃中创建一个强类型对象，以便我可以处理这些数据。我可以为pollsAndSurveys位和出席者位创建对象，但是我不知道如何处理ID号，在这种情况下为5234592和5235291，即会话的标识符。

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; }
    [JsonProperty(PropertyName = "percentagePollsCompleted")]
    public float PercentagePollsCompleted { get; set; }
    [JsonProperty(PropertyName = "percentageSurveysCompleted")]
    public float PercentageSurveysCompleted { get; set; }
}
public class SessionPerformanceStats
{
    [JsonProperty(PropertyName = "attendance")]
    public AttendanceStatistics Attendance { get; set; }
    [JsonProperty(PropertyName = "pollsAndSurveys")]
    public PollsAndSurveysStatistics PollsAndSurveys { get; set; }
}
public class WebinarPerformanceStats
{
    public List<SessionPerformanceStats> Stats { get; set; }
}

我很确定WebinarPerformanceStats是问题所在，但我不知道从这里出发。我要改变什么才能得到

NewtonSoft.Json.JsonConvert.DeserializeObject<WebinarPerformanceStats>(theJsonResponse)

上班？

回答：

使您的根对象成为字典：

var dictionary = JsonConvert.DeserializeObject<Dictionary<string, SessionPerformanceStats>>(theJsonResponse);

Json.NET将字典与JSON对象之间的序列化，并将键转换为属性名称。在您的情况下，ID号将反序列化为字典键。如果您确定它们始终是数字，则可以这样声明它们：

var dictionary = JsonConvert.DeserializeObject<Dictionary<long, SessionPerformanceStats>>(theJsonResponse);

请参见序列化字典和反序列化字典

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