RE: problem WebRequest and WebResponse

Tech-Archive recommends: Fix windows errors by optimizing your registry

You should be able to do this very simply without alll that code:

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

-- Peter
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"Juan Pablo MV" wrote:

Hi,

I have a problem using HttpWebRequest so I wish somebody can help me.

1. Application: I have a windows service that gets a XML through a
http (URL) from my provider (feeder). Once I get the XML I process it
and store the XML's information in a database.

2. The following method is the one that I have been using to get the
XML:

public XmlDocument GetXMLHttp(string strURL)
{

// URL request
HttpWebRequest request =
(HttpWebRequest)WebRequest.Create(strURL); ;
request.Method = "POST";
request.Timeout = 10000; // 10 secs
request.ContentType = "application/x-www-form-urlencoded";
request.KeepAlive = false;
request.UserAgent = ".NET Framework Example Client";

XmlDocument rssDoc = new XmlDocument();
WebResponse response = null;
Stream dataStream = null;
try
{
// I use the default credentials
request.Credentials =
CredentialCache.DefaultCredentials;
//' Getting answer.
response = request.GetResponse();
//' Opening the answer stram
dataStream = response.GetResponseStream();
//' Read content
rssDoc.Load(dataStream);
//' Close stream.
dataStream.Close();
dataStream.Dispose();
//' Close WebResponse.
response.Close();
response = null;
//' Close HttpWebRequest.
request.Abort();
request = null;


}
catch
{
//' Close stream.
if (dataStream != null) dataStream.Close();
if (dataStream != null) dataStream.Dispose();

//' Close WebResponse.
if (response != null) response.Close();
response = null;

if (request != null) request.Abort();
request = null;


WebProxy proxyObject = new WebProxy("192.168.0.2",
8080);
proxyObject.Credentials = new
NetworkCredential("usuario", "clave", "dominio");
proxyObject.BypassProxyOnLocal = true;
WebRequest.DefaultWebProxy = proxyObject;
request = (HttpWebRequest)WebRequest.Create(strURL);
request.KeepAlive = false;
Stream rssStream = null;
// Getting answer from the URL
try
{
response = request.GetResponse();
rssStream = response.GetResponseStream();
rssDoc.Load(rssStream);

//' Close stream.
rssStream.Close();
rssStream.Dispose();
//' Close WebResponse.
response.Close();
response = null;
//' Close HttpWebRequest.
request.Abort();
request = null;

}
catch (WebException e)
{
//' Close stream.
if (rssStream != null) rssStream.Close();
if (rssStream != null) rssStream.Dispose();

//' Close WebResponse.
if (response != null) response.Close();
response = null;

if (request != null) request.Abort();
request = null;

if
(Convert.ToBoolean(Dictionary(AppDomain.CurrentDomain.BaseDirectory.ToString()
+ "\\XML\\Config.xml", "Log", "/Config/AppConfig")))
{
// Sending email with the error found
System.IO.StreamWriter myFile = new
System.IO.StreamWriter("C:\\error\\errorLectura" +
DateTime.Now.ToString("MMddyyyy HHmmssfff") + ".txt");
myFile.WriteLine(e);
myFile.Close();
}
}
}
return rssDoc;
}


3. Now, this service is working fine for a short period of time. It
gets data from the URL and process it fine, but after a couple of
hours I am getting the following error:

System.Net.WebException: Unable to connect to the remote server --->
System.Net.Sockets.SocketException: A connection attempt failed
because the connected party did not properly respond after a period
of
time, or established connection failed because connected host has
failed to respond
at System.Net.Sockets.Socket.DoConnect(EndPoint endPointSnapshot,
SocketAddress socketAddress)
at System.Net.Sockets.Socket.InternalConnect(EndPoint remoteEP)
at System.Net.ServicePoint.ConnectSocketInternal(Boolean
connectFailure, Socket s4, Socket s6, Socket& socket, IPAddress&
address, ConnectSocketState state, IAsyncResult asyncResult, Int32
timeout, Exception& exception)
--- End of inner exception stack trace ---
at System.Net.HttpWebRequest.GetResponse()
at MarcadorCORE.MarcadorXMLUtils.GetXMLHttp(String strURL)


4. I am trying everything I can do but for some reason I have been
getting the same error so I will appreciate your help in this matter.

Best regards,

Juan


.

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