2010-12-10 165 views
0

我想从窗户访问令牌直播通过使用此代码远程服务器返回错误:(401)未经授权

串requestUrl =“https://consent.live.com连接API /AccessToken.aspx“;

 // Request the access token. 
     string postData = string.Format("{0}?wrap_client_id={1}&wrap_client_secret={2}&wrap_callback={3}&wrap_verification_code={4}&idtype={5}", 
       requestUrl, 
       "000000004C039809", 
       "l4VJekL1vFL1iFVmcP5qLkWv9ukY4mdl", 
       "http://ewshops.com", 
       "dac5d71d-d640-30d1-ebed-3576b132b3ec", 
       "cid"); 
     byte[] postDataEncoded = System.Text.Encoding.UTF8.GetBytes(postData); 

     WebRequest req = HttpWebRequest.Create(requestUrl); 
     req.Method = "POST"; 
     // req. 
     req.ContentType = "application/x-www-form-urlencoded"; 
     req.ContentLength = postDataEncoded.Length; 


     Stream requestStream = req.GetRequestStream(); 
     requestStream.Write(postDataEncoded, 0, postDataEncoded.Length); 

     WebResponse res = req.GetResponse(); 

     string responseBody = null; 

     using (StreamReader sr = new StreamReader(res.GetResponseStream(), Encoding.UTF8)) 
     { 
      responseBody = sr.ReadToEnd(); 
     } 

     // Process FORM POST. 
     NameValueCollection responseCollection = System.Web.HttpUtility.ParseQueryString(responseBody); 

     return responseCollection["wrap_access_token"]; 

,但我已经收到以下错误

远程服务器返回错误:(401)未经授权。

+4

我可以建议你在这个问题中清理一下代码块的格式(例如删除所有这些不必要的空白行),以便它更易读?请参阅[格式帮助](http://stackoverflow.com/editing-help) – stakx 2010-12-10 11:20:31

+1

我已经重新格式化了这篇文章,谢谢 – Hadad 2010-12-10 12:01:57

回答

2

向我们展示响应正文,它通常包含更多信息。在将其添加到uri之前,您还应该编码http://ewshops.com

+0

返回异常返回试图获得响应,这是堆栈跟踪: – Hadad 2010-12-10 16:51:15

2

我有同样的问题,并修复如下。删除requestUrl(“https://consent.live.com/AccessToken.aspx”)和随后的“?”从你的postData。 POST数据应该采用x-www-form-urlencoded格式,并且不包含请求URL。还有HttpUtility.UrlEncode()的所有参数。

相关问题