我有下面的代码,我试图提出一个查询yahoo api的请求,以便返回whoid。但是我无法生成XML来查询它,错误不会显示出来。

private string getWOEID()
{
 string woeID = "";

  String reqUrl = "http://query.yahooapis.com/v1/public/yql?q=select%20woeid%20from%20geo.places%20where%20text%3D%22farnborough%2Champshire%2Cuk%22&format=xml";
  HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(reqUrl);
  //load the response into a response object
  WebResponse resp = wr.GetResponse();
  // create a new stream that can be placed into an XmlTextReader
  Stream str = resp.GetResponseStream();
  XmlTextReader reader = new XmlTextReader(str);
  reader.XmlResolver = null;
  // create a new Xml document and loading feed data in to it
  XmlDocument xmldoc = new XmlDocument();
  xmldoc.Load(reader);

  //query the woeid with using linq
  XDocument doc = XDocument.Parse(xmldoc.ToString());
  woeID = doc.Descendants()
                .Where(element => element.Name == "woeid")
                .FirstOrDefault().Value;
  return woeID;

    }

有没有更好的方法/更容易从响应生成xml文档?
多谢,

最佳答案

比我想象的容易多了,看http://developer.yahoo.com/dotnet/howto-xml_cs.html

    String reqUrl = "http://query.yahooapis.com/v1/public/yql?q=select%20woeid%20from%20geo.places%20where%20text%3D%22farnborough%2Champshire%2Cuk%22&format=xml";

XmlDocument doc = new XmlDocument();
 doc.Load(reqUrl);

10-08 20:09