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webrequest get throws an exception "The remote server returned an error: (501) Not Implemented."

开发者 https://www.devze.com 2023-02-05 14:54 出处:网络
I\'m trying to load a file using webrequest method.First I need to log in then get a file or a directory listing. https:///xxx.yyy.zzz/login_template

I'm trying to load a file using webrequest method. First I need to log in then get a file or a directory listing. https:///xxx.yyy.zzz/login_template

When I look at the website source in firefox, I see

  <META http-equiv="Content-Type" content="text/html">
  ....
  <form method="post" action="/template/login" enctype="application/x-www-form- urlencoded">
  ....
  <input name="user" type="text">
  <input name="password" type="password">
  <input type=hidden name="switch" value="Log In">
  <input type="submit" value="Accept">

So, I wrote this code:

  public static string DownloadFile()
  {
     CookieContainer cookieJar = new CookieContainer();
     HttpWebRequest request = (HttpWebRequest)WebRequest.Create("https:///xxx.yyy.zzz/login_template");
     request.CookieContainer = cookieJar;

     // Set the credentials.
     request.Proxy.Credentials = CredentialCache.DefaultNetworkCredentials;
     request.Credentials = new NetworkCredential(userName, pass);
     request.KeepAlive = true;
     request.UserAgent = "SecureTransport";
     request.ContentType = @"application/x-www-开发者_StackOverflowform-urlencoded";
     request.Method = WebRequestMethods.Http.Post;

     bool loggedin = false;
     try
     {
         // first need to log in
         string postData = "user=" + userName + "&Password=" + pass;
         byte[] postBuffer = System.Text.Encoding.GetEncoding(1252).GetBytes(postData);
         request.ContentLength = postBuffer.Length;
         Stream newStream = request.GetRequestStream();
         // Send the data.
         newStream.Write(postBuffer, 0, postBuffer.Length);
         newStream.Close();

         using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
         {
             // Get the stream containing content returned by the server.
              if (response.StatusCode == HttpStatusCode.OK)
              {
                  loggedin = true;
              }
              response.Close();
          }

The response that I get is OK - so it seems that I successfully logged in. But then, I need to go to another url to get a file https:///xxx.yyy.zzz/myfile.zip

 HttpWebRequest requestToGetFile = (HttpWebRequest)WebRequest.Create("https:///xxx.yyy.zzz/myfile.zip");
 requestToGetFile.Method = WebRequestMethods.Http.Get;
 requestToGetFile.CookieContainer = cookieJar;

 requestToGetFile.UserAgent = "SecureTransport";
 requestToGetFile.ContentType = @"application/octet-stream";
 using (HttpWebResponse responseToGetFile = (HttpWebResponse)requestToGetFile.GetResponse())
 {
    if (responseToGetDir.StatusCode != HttpStatusCode.OK)
    {    
       ...
    }
 }

I always get an exception System.Exception: System.Net.WebException: The remote server returned an error: (501) Not Implemented. at System.Net.HttpWebRequest.GetResponse()

I've read all I could find about this error - and it seems that the problem should be in the way I do get - there is something in the request that it's not being handled properly on the server - but I don't see what is wrong.


The "not implemented" is because you are specifying a ContentType to a GET request. It does not mean anything to the server (It is mostly used during a POST request and you want to submit a payload such as XML). You need to check the response for the right content type to make sure you are getting a zip file but to make the request, you have to remove that ContentType specification.

I think this is what MisterZimbu is pointing out in the comment as well. :)


The problem was in postData. Apparently, the remote server didn't like "user=" + userName + "&Password=" + pass; sting - it was expecting exactly this sting "user=ID&Password=Pass". Which doesn't make sense, but that's what I was told and it works. Thank you very body for your suggestions. Jenny

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