2012-10-12 117 views
0

我正在尝试从URL中读取xml(以及它的非URL,它的宁静服务),下面是我的代码。我不知道问题是什么,但我无法从URL读取XML。我得到的错误,我没有得到任何数据..从URL中读取xml

string soap = ""; 
if (fPortalGuid != string.Empty) 
{ 
    HttpWebRequest request = WebRequest.Create("http://cramapp-dt-02s.cable.comcast.com:8158/restfulqueryservice/queryservice/getrequestdetails?api_key=......&request_id=33ebc6e9-9def-4f39-adf9-bba2edef3b54") as HttpWebRequest; 
    //request.Credentials = new NetworkCredential("rolland", "409cleaner"); 
    //request.ContentType = "application/json; charset = utf - 8"; 
    request.Method = "POST"; 

    using (Stream stm = request.GetRequestStream()) 
    { 
     using (StreamWriter stmw = new StreamWriter(stm)) 
     { 
      stmw.Write(soap); 
     } 
    } 
    // Get response 
    using (HttpWebResponse response = request.GetResponse() as HttpWebResponse) 
    { 
     // Get the response stream 
     StreamReader reader = new StreamReader(response.GetResponseStream()); 
     // Console application output 
     Console.WriteLine(reader.ReadToEnd()); 
    } 
} 

和XML看起来像这样xml file image

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你有什么错误? – Bobson

+0

您确定您的界面接受POST请求?尝试做一个GET。 –

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你是如何得到你在截图中显示的XML的? XML上面的消息“The XML file does not appear ...”只是浏览器警告你,它显示的XML是内部样式表,而不是Web服务的错误。如果您在C#代码中遇到异常,请提供。 –

回答

0

这条线可以使错误

HttpWebRequest request = WebRequest.Create("http://cramapp-dt-02s.cable.comcast.com:8158/restfulqueryservice/queryservice/getrequestdetails?api_key=......&request_id=33ebc6e9-9def-4f39-adf9-bba2edef3b54") as HttpWebRequest; 

因为在创建功能的网址(太长,所以有...呼喊显示),请尝试找到并粘贴完整链接或此链接:

http://www.w3schools.com/xml/note.xml 
+1

不要太长;只是隐藏他的API密钥。 –

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是的,我隐藏我的api键:) – Philly