如何将此JSON结果解析为对象？

Question

我需要将此JSON字符串解析为我的"WeatherJson"类型的对象.但是我不知道如何解析字符串中的阵列,如"'天气’: [{"id":802,"main":"Clouds","description":"scattered clouds","icon":"03d"}].实体类会是什么样子？

JSON字符串:

{
  "coord": {"lon":79.85,"lat":6.93},
  "sys": {
    "type": 1, 
    "id": 7864, 
    "message": 0.0145,
    "country": "LK",
    "sunrise": 1435883361,
    "sunset": 1435928421
  },
  "weather": [
     {"id":802, "main":"Clouds", "description":"scattered clouds", "icon":"03d"}
  ],
  "base": "stations",
  "main": {
    "temp": 302.15,
    "pressure": 1013,
    "humidity": 79,
    "temp_min": 302.15,
    "temp_max": 302.15
  },
  "visibility":10000,
  "wind": { "speed": 4.1, "deg": 220 },
  "clouds": { "all": 40 },
  "dt": 1435893000,
  "id":1248991,
  "name":"Colombo",
  "cod":200
}

编辑

我需要从代码中检索以下值:

WeatherJson w = new WeatherJson();
Console.WriteLine(w.weather.description);
//that above line was retrieved and stored from the JSONArray named 'weather' in the main json response

Answer 1

您应该只使JSON中的数组与POCO中的列表或数组类型匹配.这是使用您提供的JSON的简短但完整的示例:

using System;
using System.Collections.Generic;
using System.IO;
using Newtonsoft.Json;

class Test
{ 
    static void Main(string[] args) 
    {
        var json = File.ReadAllText("weather.json");
        var root = JsonConvert.DeserializeObject<Root>(json);
        Console.WriteLine(root.Weather[0].Description);
    }
}

public class Root
{
    // Just a few of the properties
    public Coord Coord { get; set; }
    public List<Weather> Weather { get; set; }
    public int Visibility { get; set; }
    public string Name { get; set; }
}

public class Weather
{
    public int Id { get; set; }
    public string Description { get; set; }
}

public class Coord
{
    public double Lon { get; set; }
    public double Lat { get; set; }
}