如何从休息Web服务中获取特定部分数据？

Question

我是休息webservices的新手,我试着了解如何使用它们.

我在C#中定义一个http get请求,如下所示:

static string HttpGet(string url)
        {
            HttpWebRequest req = WebRequest.Create(url)
                                 as HttpWebRequest;
            string result = null;
            using (HttpWebResponse resp = req.GetResponse()
                                          as HttpWebResponse)
            {
                StreamReader reader =
                    new StreamReader(resp.GetResponseStream());
                result = reader.ReadToEnd();
            }
            return result;
        }

然后我在一个按钮中使用HttpGet从这个天气webservice 获取xml数据(我把txtOut发现我的代码工作).

private void btnGet_Click(object sender, RoutedEventArgs e)
        {
            string test = HttpGet("http://api.openweathermap.org/data/2.5/weather?q=London&mode=xml");

            txtOut.Text = test;
        }

它将整个xml从发送获取请求发送到该URL:http://api.openweathermap.org/data/2.5/weather？q = London&mode = xml

所以我的问题是如何在变量中保存该xml的特定部分？像卡尔文的最低温度,所以我可以将其转换为华氏温度或摄氏温度.

请帮帮我.

Answer 1

你api也返回json,所以你可以像下面一样使用它(使用Json.Net)

using (WebClient wc = new Webclient())
{
    var json = wc.DownloadString("http://api.openweathermap.org/data/2.5/weather?q=London&mode=json");
    var obj = JsonConvert.DeserializeObject<OpenWeatherMap.Root>(json);
}

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

    public class Sys
    {
        public int type { get; set; }
        public int id { get; set; }
        public double message { get; set; }
        public string country { get; set; }
        public int sunrise { get; set; }
        public int sunset { 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 Main
    {
        public double temp { get; set; }
        public int humidity { get; set; }
        public double pressure { get; set; }
        public double temp_min { get; set; }
        public double temp_max { get; set; }
    }

    public class Wind
    {
        public double speed { get; set; }
        public double gust { get; set; }
        public int deg { get; set; }
    }



    public class Clouds
    {
        public int all { get; set; }
    }

    public class Root
    {
        public Coord coord { get; set; }
        public Sys sys { get; set; }
        public List<Weather> weather { get; set; }
        public string @base { get; set; }
        public Main main { get; set; }
        public Wind wind { get; set; }
        public Dictionary<string,double> rain { get; set; }
        public Clouds clouds { get; set; }
        public int dt { get; set; }
        public int id { get; set; }
        public string name { get; set; }
        public int cod { get; set; }
    }
}

如果你坚持使用xml,那么你可以使用Linq2Xml + XPath

var xDoc = XDocument.Load("http://api.openweathermap.org/data/2.5/weather?q=London&mode=xml");

var windSpeed  = (double)xDoc.XPathSelectElement("//wind/speed").Attribute("value");

要么

var temp = (double)xDoc.Root.Element("temperature").Attribute("value");