Question

我有以下代码段：

public void getForcast()
{
    string url = string.Format("https://samples.openweathermap.org/data/2.5/weather?q=London&appid=b6907d289e10d714a6e88b30761fae22");
    using (WebClient web = new WebClient())
    {
        var json = web.DownloadString(url);
        var obj = JsonConvert.DeserializeObject<WeatherData.WeatherForcast>(json);
        WeatherData.WeatherForcast forcast = obj;
        WeatherMark.Text = string.Format("{0}", forcast.list[1].weathers[0].description);
    }
}

我希望它从预测列表中获取其描述。

但是我收到了这个错误

对象引用未设置为对象的实例

这是我的全部课程清单：

class WeatherForcast
{
    public List<list> list { get; set; }
}
public class weather
{
    public string main { get; set; }
    public string description { get; set; }
}
public class list
{
    public List<weather> weathers { get; set; }
}

有人知道为什么会出现吗？

Answer 1

JSON很可能与提供的类定义不匹配，该类定义在解析时会导致对象为空。

在浏览器中调用显示的URL将提供以下JSON响应

{“ coord”：{“ lon”：-0.13，“ lat”：51.51}，“ weather”：[{“ id”：300，“ main”：“毛毛雨”，“描述”：“光强度下毛毛雨“，” icon“：” 09d“}]，” base“：” stations“，” main“：{” temp“：280.32，” pressure“：1012，”湿度“：81，” temp_min“：279.15， “ temp_max”：281.15}，“可见性”：10000，“风”：{“速度”：4.1，“度”：80}，“云”：{“全部”：90}，“ dt”：1485789600，“ sys“：{” type“：1，” id“：5091，” message“：0.0103，” country“：” GB“，” sunrise“：1485762037，” sunset“：1485794875}，” id“：2643743，” name“：” London“，” cod“：200}

可以将其映射到已显示的经过一些修改的代码。

public class WeatherForcast {
    public List<weather> weather { get; set; }
}

public class weather {
    public string main { get; set; }
    public string description { get; set; }
}

有一些在线工具可用于放置JSON，它将生成JSON的映射类。

Answer 2

您可以使用＆mode = xml在Xml中获取天气，然后使用xml序列化：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
namespace ConsoleApplication1
{
    class Program
    {
        const string URL = "https://samples.openweathermap.org/data/2.5/weather?q=London&mode=xml&appid=b6907d289e10d714a6e88b30761fae22";
        static void Main(string[] args)
        {
            XmlReader reader = XmlReader.Create(URL);

            XmlSerializer serializer = new XmlSerializer(typeof(Weather));
            Weather weather = (Weather)serializer.Deserialize(reader);
        }
    }
    [XmlRoot("current")]
    public class Weather
    {
        public City city { get; set; }
        public Temperature temperature { get; set; }
        public Humidity humidity { get; set; }
        public Pressure pressure { get; set; }
        public Wind wind { get; set; }
    }
    public class City
    {
        [XmlAttribute()]
        public string name { get; set; }

        [XmlAttribute()]
        public string id { get; set; }

        public Coord coord { get; set; }
        public string country { get; set; }
        public Sun sun { get; set; }
    }
    public class Sun
    {
        [XmlAttribute()]
        public DateTime rise { get; set; }

        [XmlAttribute()]
        public DateTime set { get; set; }        
    }
    public class Coord
    {
        [XmlAttribute()]
        public decimal lon { get; set; }

        [XmlAttribute()]
        public decimal lat { get; set; }
    }
    public class Temperature
    {
        [XmlAttribute()]
        public decimal value { get; set; }

        [XmlAttribute()]
        public decimal min { get; set; }

        [XmlAttribute()]
        public decimal max { get; set; }
    }
    public class Humidity
    {
        [XmlAttribute()]
        public decimal value { get; set; }
    }
    public class Pressure
    {
        [XmlAttribute()]
        public decimal value { get; set; }
    }
    public class Wind
    {
        public Speed speed { get; set; }
        public Direction direction { get; set; }
    }
    public class Speed
    {
        [XmlAttribute()]
        public decimal value { get; set; }

        [XmlAttribute()]
        public string name { get; set; }
    }
    public class Direction
    {
        [XmlAttribute()]
        public decimal value { get; set; }

        [XmlAttribute()]
        public string name { get; set; }

        [XmlAttribute()]
        public string code { get; set; }
    }
}

解析开放式天气地图API响应失败

2 个答案: