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);