如何在C＃中解析OpenWeather一个电话？

Question

我已经成功地使用JSON解析了OpenWeather的“当前天气数据”，但我对“一个电话”有疑问。它返回排列的数据有所不同。除了“天气”列表之外，我可以拉出所有内容。我为其创建的类返回零值。

{
"lat": 41.9585,
"lon": -78.1892,
"timezone": "America/New_York",
"timezone_offset": -14400,
"current": {
    "dt": 1652559634,
    "sunrise": 1652522008,
    "sunset": 1652574293,
    "temp": 74.53,
    "feels_like": 75.11,
    "pressure": 1014,
    "humidity": 73,
    "dew_point": 65.32,
    "uvi": 2.37,
    "clouds": 100,
    "visibility": 10000,
    "wind_speed": 7.47,
    "wind_deg": 173,
    "wind_gust": 10.63,
    "weather": [
        {
            "id": 804,
            "main": "Clouds",
            "description": "overcast clouds",
            "icon": "04d"
        }
    ]
}
}

string url = string.Format("https://api.openweathermap.org/data/2.5/onecall?lat={0}&lon={1}&exclude=minutely,hourly,daily&appid={2}&units=imperial", 41.958496166, -78.18916591, APIKey);

string jason = web.DownloadString(url);
WeatherInfo.root info = JsonConvert.DeserializeObject<WeatherInfo.root>(jason);


public static class WeatherInfo
{
    public class current
    {
        public double dt { get; set; }
        public double sunrise { get; set; }
        public double sunset { get; set; }
        public double temp { get; set; }
        public double pressure { get; set; }
        public double humidity { get; set; }
        public double dew_point { get; set; }
        public double uvi { get; set; }
        public double clouds { get; set; }
        public double visibility { get; set; }
        public double wind_speed { get; set; }
        public double wind_deg { get; set; }
        public double wind_gust { get; set; }
    }

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

    public class root
    {
        public current current { get; set; }
        public List<weather> weather { get; set; }
    }
}