Question

我们如何从C＃中的JSON结构中提取或检索子节点值。

我的应用正在使用OpenWeatherMap，我需要从 city ， temp 中检索名称来自列表和描述来自天气节点，我的JSON和类结构如下

{
  "cod": "200",
  "message": 0.0284,
  "city": {
    "id": 2643743,
    "name": "London",
    "coord": {
      "lon": -0.12574,
      "lat": 51.50853
    },
    "country": "GB",
    "population": 0,
    "sys": {
      "population": 0
     }
  },
  "cnt": 1,
  "list": [
    {
      "dt": 1429268400,
      "temp": {
        "day": 12.21,
        "min": 4.86,
        "max": 13.18,
        "night": 4.86,
        "eve": 11.76,
        "morn": 12.21
      },
      "pressure": 1028.8,
      "humidity": 66,
      "weather": [
         {
           "id": 803,
           "main": "Clouds",
           "description": "broken clouds",
           "icon": "04d"
        }
      ],
      "speed": 5.9,
      "deg": 67,
      "clouds": 80
    }
  ]
}

C＃Class

public class WeatherForeCast
{
    public string City { get; set; }
    public decimal Day { get; set; }
    public decimal Min { get; set; }
    public decimal Max { get; set; }
    public decimal Night { get; set; }
    public string Description { get; set; }
}

到目前为止，我熟悉使用JSON.net将C＃对象序列化和反序列化为具有完全相同结构的JSON。

Answer 1

如果您只想填充WeatherForecast的实例，则可以在普通SelectToken上使用几个JObject来电：

var parsed = JObject.Parse(json);
var forecast = new WeatherForeCast();

forecast.City = parsed.SelectToken("city.name").Value<string>();
forecast.Day = parsed.SelectToken("list[0].temp.day").Value<decimal>();
forecast.Description = parsed.SelectToken("list[0].weather[0].description").Value<string>();
forecast.Min = parsed.SelectToken("list[0].temp.min").Value<decimal>();
forecast.Max = parsed.SelectToken("list[0].temp.max").Value<decimal>();
forecast.Night = parsed.SelectToken("list[0].temp.night").Value<decimal>();

请注意，这是非常脆弱的，它正在对JSON的内容做出假设。如果JSON发生更改，SelectToken中各种属性的路径将不正确，此代码将引发异常。

Answer 2

使用json2csharp.com生成您的课程。

public class Coord
{
    public double lon { get; set; }
    public double lat { get; set; }
}

public class Sys
{
    public int population { get; set; }
}

public class City
{
    public int id { get; set; }
    public string name { get; set; }
    public Coord coord { get; set; }
    public string country { get; set; }
    public int population { get; set; }
    public Sys sys { get; set; }
}

public class Temp
{
    public double day { get; set; }
    public double min { get; set; }
    public double max { get; set; }
    public double night { get; set; }
    public double eve { get; set; }
    public double morn { get; set; }
}

public class Weather
{
    public int id { get; set; }
    public string main { get; set; }
    public string description { get; set; }
    public string icon { get; set; }
}

public class List
{
    public int dt { get; set; }
    public Temp temp { get; set; }
    public double pressure { get; set; }
    public int humidity { get; set; }
    public List<Weather> weather { get; set; }
    public double speed { get; set; }
    public int deg { get; set; }
    public int clouds { get; set; }
}

public class RootObject
{
    public string cod { get; set; }
    public double message { get; set; }
    public City city { get; set; }
    public int cnt { get; set; }
    public List<List> list { get; set; }
}

然后使用JSON.NET反序列化到类结构中并提取所需的属性。

var jsonObject = JsonConvert.DeserializeObject<RootObject>(jsonString);

您现在有一个RootObject实例，您可以根据需要遍历它以提取您需要的特定值。

C＃：从JSON结构中提取/检索子节点

2 个答案: