C#使用实体类获取JSON参数

Question

在C#后台，这种JSON数据要怎么获取里面的所有参数内容呢？

{
    "result":"success",
    "root":{
        "track":[
            {
                "orderno":"A210600870",
                "billcode":"400039547",
                "detail":[
                    {
                        "memo":"快件已发往深圳",
                        "time":"2022-06-09 13:14:00",
                        "scantype":"发往"
                    },
                    {
                        "memo":"联运通 已收运(10件)",
                        "time":"2022-06-09 13:12:35",
                        "scantype":"开单"
                    },
                    {
                        "memo":"快件到达深圳中转中心",
                        "time":"2022-06-09 13:14:00",
                        "scantype":"到达"
                    },
                    {
                        "memo":"快件开始派送",
                        "time":"2022-06-09 13:15:00",
                        "scantype":"派送"
                    },
                    {
                        "memo":"蔡杰杰已签收已签收",
                        "time":"2022-06-09 13:15:00",
                        "scantype":"签收"
                    }
                ]
            }
        ]
    },
    "message":""
}

想通过实体类来获取里面的内容。谢谢！！！

Answer 1

按照这个结构建实体类，然后用
Newtonsoft.Json 反序列化 就可以了

    public class Result
    {
        public string result { get; set; }

        public RootClass root { get; set; }
    }

    public class RootClass
    {
        public List<TrackClass> track { get; set; }
    }
    public class TrackClass
    {
        public string orderno { get; set; }
        public string billcode { get; set; }
        public List<Detail> detail { get; set; }
    }
    public class Detail
    {
        public string memo { get; set; }
        public string time { get; set; }
        public string scantype { get; set; }
    }

序列化代码

        var model = Newtonsoft.Json.JsonConvert.DeserializeObject<Result>(jsonStr);