C# 使用自定义类将json解析到字典中

我有这个json。 我可以将其解析为下一个对象

字典

如何从该对象生成类似于
字典
，其中

{
  "data" : {
                 "TTyTo3qHkFq8b-KsIHZOlQ" : {
                       "2016-05-01" : {
                             "value1" : 11,
                             "value2" : 87
                       },
                       "2016-05-02" : {
                             "value1" : 2,
                             "value2" : 34
                       }
                 }
    }
}

classcustomclass1{
日期时间日期{get；set；}
字典值{get；set；}
}

?

对于反序列化，我使用的是Json.NET。
您需要实现自定义以实现您想要的

首先，您需要创建表示整个JSON的
Root
类：

class CustomClass1 {
    DateTime Date {get;set;}
    Dictionary<string, int> Value {get;set;}
}

最后将其传递到
反序列化对象
方法重载：

public class CustomClassConverter : JsonConverter
{
    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        //Implement if needed
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        var obj = (JObject)JObject.ReadFrom(reader);

        var root = new Root();
        root.Data = new Dictionary<string, List<CustomClass1>>();

        JObject value = (JObject)obj.GetValue("data");
        foreach (JProperty prop1 in value.Properties())
        {
            var customObjects = new List<CustomClass1>();
            foreach (JProperty prop2 in prop1.Values())
            {
                var customObject = new CustomClass1();
                customObject.Date = DateTime.Parse(prop2.Name);
                customObject.Values = new Dictionary<string, int>();
                foreach (JProperty prop3 in prop2.Values())
                {
                    customObject.Values.Add(prop3.Name, prop3.ToObject<int>());
                }
                customObjects.Add(customObject);
            }
            root.Data.Add(prop1.Name, customObjects);
        }

        return root;
    }

    public override bool CanConvert(Type t)
    {
        return t.Equals(typeof(Root)); //Converter which handles only Root types
    }

    public override bool CanRead
    {
        get { return true; }
    }
}

var root=JsonConvert.DeserializeObject（json，新的CustomClassConverter（））；
您发布的内容不是有效的JSON。用于验证您的JSON。@rufer7，小补丁关于
数据
属性，您需要忽略它吗？@DarjanBogdan，不
数据
应该存在
public class CustomClassConverter : JsonConverter
{
    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        //Implement if needed
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        var obj = (JObject)JObject.ReadFrom(reader);

        var root = new Root();
        root.Data = new Dictionary<string, List<CustomClass1>>();

        JObject value = (JObject)obj.GetValue("data");
        foreach (JProperty prop1 in value.Properties())
        {
            var customObjects = new List<CustomClass1>();
            foreach (JProperty prop2 in prop1.Values())
            {
                var customObject = new CustomClass1();
                customObject.Date = DateTime.Parse(prop2.Name);
                customObject.Values = new Dictionary<string, int>();
                foreach (JProperty prop3 in prop2.Values())
                {
                    customObject.Values.Add(prop3.Name, prop3.ToObject<int>());
                }
                customObjects.Add(customObject);
            }
            root.Data.Add(prop1.Name, customObjects);
        }

        return root;
    }

    public override bool CanConvert(Type t)
    {
        return t.Equals(typeof(Root)); //Converter which handles only Root types
    }

    public override bool CanRead
    {
        get { return true; }
    }
}

var root = JsonConvert.DeserializeObject<Root>(json, new CustomClassConverter());