未能理想化异构json数据

你需要定制 JsonConverter

public class DPInfoConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return objectType == typeof(DPInfo);
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        JObject obj = JObject.Load(reader);
        JProperty points = obj.Properties().FirstOrDefault(p => p.Name != "last");

        DPInfo info = new DPInfo
        {
            key = points.Name,   // remove this line if you don't need the key
            points = points.Value.ToObject<decimal[][]>(),
            last = (long)obj["last"]
        };

        return info;
    }

    public override bool CanWrite
    {
        get { return false; }
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

[JsonConverter] 属性到您的类:

[JsonConverter(typeof(DPInfoConverter))]
public class DPInfo
{
    public string key { get; set; }   // remove this line if you don't need the key
    public decimal[][] points { get; set; }
    public long last { get; set; }
}

然后,像往常一样反序列化,它应该可以工作:

string json = @"
{
  ""keyXXX"": [[1, 2, 3]],
  ""last"": 12345
}";

DPInfo info = JsonConvert.DeserializeObject<DPInfo>(json);

小提琴: https://dotnetfiddle.net/7S6STp

你可以用 JObject.Parse 将字符串解析为 dynamic

在您的情况下,声明如下所示:

dynamic data = JObject.Parse("{ 'keyXXX' : [[1,2,3,...]...], 'last': 12345 }");