I can create a folder, I can rename the file, but I can not upload and download a file to ftp. The exception that is shown is System.Net.WebException: The remote server returned an error: (500) Syntax error, command not recognized.
Anyone have an idea why?
log file:
System.Net Information: 0 : [6112] FtpControlStream#7746814 - Resposta recebida [500 PORT/EPRT (Active Mode/Extended Active Mode) is not supported. Use PASV/EPSV instead of this]
System.Net Information: 0 : [6112] FtpWebRequest#30923613::(Liberando a conexão de FTP#7746814.)
System.Net Error: 0 : [6112] Exceção em FtpWebRequest#30923613::GetRequestStream - O servidor remoto retornou um erro: (500) Erro de sintaxe, comando não reconhecido.
Code:
/* Upload File */
public void upload(string remoteFile, string localFile)
{
try
{
/* Create an FTP Request */
ftpRequest = (FtpWebRequest)FtpWebRequest.Create(host + remoteFile);
/* Log in to the FTP Server with the User Name and Password Provided */
ftpRequest.Credentials = new NetworkCredential(user, pass);
/* When in doubt, use these options */
ftpRequest.UseBinary = true;
ftpRequest.UsePassive = false;
ftpRequest.KeepAlive = true;
/* Specify the Type of FTP Request */
ftpRequest.Method = WebRequestMethods.Ftp.UploadFile;
/* Establish Return Communication with the FTP Server */
Stream ftpStream = ftpRequest.GetRequestStream();
/* Open a File Stream to Read the File for Upload */
FileStream localFileStream = new FileStream(localFile, FileMode.Create);
/* Buffer for the Downloaded Data */
byte[] byteBuffer = new byte[bufferSize];
int bytesSent = localFileStream.Read(byteBuffer, 0, bufferSize);
/* Upload the File by Sending the Buffered Data Until the Transfer is Complete */
try
{
while (bytesSent != 0)
{
ftpStream.Write(byteBuffer, 0, bytesSent);
bytesSent = localFileStream.Read(byteBuffer, 0, bufferSize);
}
}
catch (Exception ex) { Console.WriteLine(ex.ToString()); }
/* Resource Cleanup */
localFileStream.Close();
ftpStream.Close();
ftpRequest = null;
}
catch (Exception ex) { Console.WriteLine(ex.ToString()); }
return;
}