Question

我有以下代码，我尝试发出请求查询yahoo api以返回whoid。但我无法生成XML来查询它，错误不会显示。

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;

    }

是否有更好的方式/更容易从响应中生成XML文档？

非常感谢，

Answer 1

比我想象的容易得多，请参阅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);

Answer 2

您可以通过将XmlTextReader实例传递给XDocument.Load（）方法来消除创建XmlDocument的步骤。

从.net代码消费yahoo webservice

2 个答案: