Question

我有一个复杂的JSON响应，我需要解析和替换数据。

这只是一天的剪辑，但需要在回复中做好所有日子。

我如何解析这个并且只是改变值。我需要修改并发送回主机。

  {
  "schedulePeriods":` [
    {
      "day": "Monday",
      "periodType": "WakeOcc1",
      "startTime": 26,
      "isCancelled": false,
      "heatSetpoint": 70.0,
      "coolSetpoint": 75.0,
      "fanMode": "Auto"
    },
    {
      "day": "Monday",
      "periodType": "LeaveUnocc1",
      "startTime": 32,
      "isCancelled": false,
      "heatSetpoint": 65.0,
      "coolSetpoint": 75.0,
      "fanMode": "Auto"
    },
    {
      "day": "Monday",
      "periodType": "ReturnOcc2",
      "startTime": 66,
      "isCancelled": false,
      "heatSetpoint": 72.0,
      "coolSetpoint": 75.0,
      "fanMode": "Auto"
    },
    {
      "day": "Monday",
      "periodType": "SleepUnocc2",
      "startTime": 0,
      "isCancelled": false,
      "heatSetpoint": 65.0,
      "coolSetpoint": 75.0,
      "fanMode": "Auto"
    },
    {
      "day": "Tuesday",
      "periodType": "WakeOcc1",
      "startTime": 26,
      "isCancelled": false,
      "heatSetpoint": 70.0,
      "coolSetpoint": 75.0,
      "fanMode": "Auto"
    },

所以我现在拥有的是：`

class Day
    {
        public string day { get; set; }
        public string periodType { get; set; }
        public int startTime { get; set; }
        public double heatSetpoint { get; set; }
        public double coolSetpoint { get; set; }


        public override string ToString()
        {
            return string.Format("{0}\n{1}\n{2}\n{3}\n{4}", day, periodType, startTime, heatSetpoint, coolSetpoint);
        }
    }

    static void Main(string[] args)
    {

         String JSONstring = File.ReadAllText("json.json");
         JObject o = JObject.Parse(JSONstring);


         string day = (string)o.SelectToken("schedulePeriods[0].day");
         string periodType = (string)o.SelectToken("schedulePeriods[0].periodType");
         int startTime = (int)o.SelectToken("schedulePeriods[0].startTime");
         double heatSetpoint = (double)o.SelectToken("schedulePeriods[0].heatSetpoint");
         double coolSetpoint = (double)o.SelectToken("schedulePeriods[0].coolSetpoint");
         Console.WriteLine("{0}\n{1}\n{2}\n{3}\n{4}",day,periodType,startTime,heatSetpoint,coolSetpoint);


         Console.WriteLine("Change Value to:");
         string answer = Console.ReadLine();

         day = answer;


         Console.WriteLine("new value {0}", day);
              string data = JsonConvert.SerializeObject(JSONstring,Formatting.Indented);
             File.WriteAllText("newfile.json", data);
              Console.ReadLine();

}

我可以从第一个大括号中获取数据，但没有更多。我需要从每个阵列中获取。我还想修改一个给定的条目，然后用适当的json格式保存编辑的文件（这根本不起作用）`

解决。感谢您的反馈

Answer 1

在您的应用程序中创建以下类

public class RootObject
{
    public List<SchedulePeriod> schedulePeriods { get; set; }
}

public class SchedulePeriod
{
    public string day { get; set; }
    public string periodType { get; set; }
    public int startTime { get; set; }
    public bool isCancelled { get; set; }
    public double heatSetpoint { get; set; }
    public double coolSetpoint { get; set; }
    public string fanMode { get; set; }
}

然后使用Newtonsoft.Json反序列化你的json，如

public static T DeserializeObject<T>(string jsonObj)
{
    return JsonConvert.DeserializeObject<T>(jsonObj,
        new JsonSerializerSettings() { ReferenceLoopHandling = ReferenceLoopHandling.Ignore, Formatting = Formatting.Indented });
}

并称之为

RootObject obj = DeserializeObject<RootObject>(your_json);

然后访问您的日程安排。

希望它有所帮助。

解析嵌套的JSON并替换C＃中的数据

1 个答案: