HttpWebResponse wresp = (HttpWebResponse)wreq.GetResponse();无法连接远程服务器
如题，想获取网页的“源代码”，用这个方法
public String GetPageCode(String PageURL, String Charset)
{
try
{
String strHtml = "";
HttpWebRequest wreq = (HttpWebRequest)WebRequest.Create(PageURL);
CookieContainer cookieCon = new CookieContainer();
wreq.CookieContainer = cookieCon;
HttpWebResponse wresp = (HttpWebResponse)wreq.GetResponse();
Stream s = wresp.GetResponseStream();
StreamReader objReader = new StreamReader(s, System.Text.Encoding.GetEncoding(Charset));
string strLine = "";
while (strLine != null)
{
strLine = objReader.ReadLine();
if (strLine != null)
{
strHtml += strLine.Trim();
}
}
strHtml = strHtml.Replace("<br />", "\r\n");

return strHtml;
}
catch (Exception n)
{
return n.Message;
}
}
传入的网址，是正确的，但总提示：无法连接远程服务器，请问怎么解决，谢谢了

[解决办法]

C# code

/// <summary>          /// 创建GET方式的HTTP请求          /// </summary>          /// <param name="url">请求的URL</param>          /// <param name="timeout">请求的超时时间</param>          /// <param name="userAgent">请求的客户端浏览器信息，可以为空</param>          /// <param name="cookies">随同HTTP请求发送的Cookie信息，如果不需要身份验证可以为空</param>          /// <returns></returns>          public static HttpWebResponse CreateGetHttpResponse(string url,int? timeout, string userAgent,CookieCollection cookies)          {              if (string.IsNullOrEmpty(url))              {                  throw new ArgumentNullException("url");              }              HttpWebRequest request = WebRequest.Create(url) as HttpWebRequest;              request.Method = "GET";              request.UserAgent = DefaultUserAgent;              if (!string.IsNullOrEmpty(userAgent))              {                  request.UserAgent = userAgent;              }              if (timeout.HasValue)              {                  request.Timeout = timeout.Value;              }              if (cookies != null)              {                  request.CookieContainer = new CookieContainer();                  request.CookieContainer.Add(cookies);              }              return request.GetResponse() as HttpWebResponse;          }