コード例 #1
0
        private void GetWeatherData(string query)
        {
            var hostDetails = new HostDetails("http", "api.openweathermap.org", "data/2.5/weather");
            var json        = JsonHost.GetJson(hostDetails, query);

            _data = JsonConvert.DeserializeObject <CurrentWeatherData>(json);
        }
コード例 #2
0
        /// <summary>
        /// Initializes this class with a raw JSON call (not recommended for production)
        /// </summary>
        /// <param name="url">URL for API call</param>
        public CurrentWeather(string url)
        {
            var json = JsonHost.GetJson(url);

            _data = JsonConvert.DeserializeObject <CurrentWeatherData>(json);
        }