程序在 FtpWebResponse 上挂起

Question

第一次发帖，老读者。我有一个非常烦人的问题，这让我很紧张。我已经设置了一个程序，所以我在 FTP 服务器上监听新文件，如果有新文件我下载它。从那里我处理文件中的一些信息等。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上，一切都很好，但是一旦检测到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);
        }
    }

它挂在的行是这一行： 使用 (FtpWebResponse partResponse = (FtpWebResponse)request.GetResponse())

我的方法在这里是静态的原因是因为我只是在另一个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上死掉？我已经盯着自己瞎了好几个小时了！

Answer 1

我也遇到了这个问题...尝试先完成您的请求，然后在尝试检索响应之前关闭它。这对我有用（实际上是在阅读 MartinNielsen 的评论后尝试过的）。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;

这是我在编写此代码时使用的一些资源（不会让我发布超过 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# 专家，但我使用此代码从我的 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");
                }
            }
        }
    }

告诉我它是否适合你:)