从 json 对象创建一个以 ID 为名称的强类型 c# 对象答案

【问题标题】：Create a strongly typed c# object from json object with ID as the name从 json 对象创建一个以 ID 为名称的强类型 c# 对象
【发布时间】：2015-12-10 23:19:59
【问题描述】：

我正在尝试将 API 用于知名的在线会议提供商。他们的其中一个 API 调用返回一个如下所示的对象。

{
    "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}
    }
}

我正在尝试在 C# 中创建一个强类型对象，以便处理这些数据。我可以为 pollsAndSurveys 位和出勤位创建对象，但我不知道如何处理 id 号，在本例中为 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，但我不知道从哪里开始。我必须改变什么才能得到 p>

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

上班？

【问题讨论】：

标签： c# json json.net

【解决方案1】：

让你的根对象成为字典：

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

Json.NET 将字典从 JSON 对象序列化到 JSON 对象，并将键转换为属性名称。在您的情况下，ID 号将被反序列化为字典键。如果您确定它们将始终是数字，则可以这样声明它们：

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

见Serialize a Dictionary和Deserialize a Dictionary

【讨论】：