程序在FtpWebResponse上挂起

Question

第一次的海报，长期的读者。我有一个非常烦人的问题，一直让我心烦意乱。我设置了一个程序，这样我就可以在FTP服务器上监听新文件，如果有新文件，我就下载它。从那里我开始处理文件中的一些信息，等等。当我第二次运行我的序列时，我的问题就出现了。也就是说，在我下载的第一个文件中，一切都是完全正常的，但是一旦检测到一个新文件并且我的程序尝试下载它，我的程序就会挂起。

 private static void DownloadFile(string s)
    {
        try
        {
            FtpWebRequest request = (FtpWebRequest)WebRequest.Create("ftp://blabla.com/"+s);
            request.Method = WebRequestMethods.Ftp.DownloadFile;
            request.Credentials = new NetworkCredential("xxx" ,"zzz");

            using (FtpWebResponse partResponse = (FtpWebResponse)request.GetResponse())
            {
                Stream partReader = partResponse.GetResponseStream();

                byte[] buffer = new byte[1024];
                FileInfo fi = new FileInfo(path);
                FileStream memStream = fi.Create();
                while (true)
                {
                    int bytesRead = partReader.Read(buffer, 0, buffer.Length - 1);
                    if (bytesRead == 0)
                        break;

                    memStream.Write(buffer, 0, bytesRead);
                }
                partResponse.Close();
                memStream.Close();
            }
            Console.WriteLine(DateTime.Now + " file downloaded");
            MoveFileToInProgress(s);
        }
        catch (Exception e)
        {
            Console.WriteLine(e.Message);
        }
    }

它挂起的代码行是这样的:使用(FtpWebResponse partResponse = (FtpWebResponse)request.GetResponse())

这里我的方法是静态的，因为我只是在一个不同的项目中运行它来测试它。我的问题是，为什么它只在第二个文件中消失？我已经盯着自己看了好几个小时了！

Answer 1

我也遇到了这个问题...尝试先完成请求，然后将其关闭，然后再尝试检索响应。这对我很有效(实际上在阅读了MartinNielsen的评论后尝试了一下)。这就是我所做的。

        // connect to the ftp site
        FtpWebRequest ftpRequest = (FtpWebRequest)WebRequest.Create(ftpUri);
        ftpRequest.Method = WebRequestMethods.Ftp.UploadFile;
        ftpRequest.Credentials = new NetworkCredential(ftpUser, ftpPassword);

        // setting proxy to null so that it does not go through the proxy
        ftpRequest.Proxy = null;

        // get file information
        StreamReader fileStream = new StreamReader(filePath);
        byte[] fileBytes = Encoding.UTF8.GetBytes(fileStream.ReadToEnd());
        ftpRequest.ContentLength = fileBytes.Length;
        fileStream.Close();

        // open connection to ftp site
        Stream ftpRequestStream = ftpRequest.GetRequestStream();

        // write the file to the stream
        ftpRequestStream.Write(fileBytes, 0, fileBytes.Length);

        // close the stream
        ftpRequestStream.Close();

        // get the response from the server
        FtpWebResponse ftpUploadResponse = (FtpWebResponse)ftpRequest.GetResponse();
        string result = ftpUploadResponse.StatusDescription;

        // close response
        ftpUploadResponse.Close();

        // return response to calling code
        return result;

下面是我在编写这段代码时使用的几个资源(我不会发布超过2个，还有更多)

How to: Upload Files with FTP

Uploading a file -- "The requested URI is invalid for this FTP command"

Answer 2

我不是C#方面的专家，但我使用以下代码从我的ftp下载文件：

public void Download(string filename)
    {
        // I try to download five times before crash
        for (int i = 1; i < 5; i++)
        {
            try
            {
                FtpWebRequest ftp = (FtpWebRequest)FtpWebRequest.Create(Global.Path + "/" + filename);
                ftp.Credentials = new NetworkCredential(User, Pass);
                ftp.KeepAlive = false;
                ftp.Method = WebRequestMethods.Ftp.DownloadFile;
                ftp.UseBinary = true;
                ftp.Proxy = null;

                int buffLength = 2048;
                byte[] buff = new byte[buffLength];
                int contentLen;

                string LocalDirectory = Application.StartupPath.ToString() + "/downloads/" + filename;
                using (FileStream fs = new FileStream(LocalDirectory, FileMode.Create, FileAccess.Write, FileShare.None))
                using (Stream strm = ftp.GetResponse().GetResponseStream())
                {
                    contentLen = strm.Read(buff, 0, buffLength);
                    while (contentLen != 0)
                    {
                        fs.Write(buff, 0, contentLen);
                        contentLen = strm.Read(buff, 0, buffLength);
                    }
                }

                Process.Start(LocalDirectory);
                break;
            }
            catch (Exception exc)
            {
                if (i == 5)
                {
                    MessageBox.Show("Can't download, try number: " + i + "/5 \n\n Error: " + exc.Message,
                        "Problem downloading the file");
                }
            }
        }
    }

告诉我它是否适用于您:)