压缩目录并上传到FTP服务器，而不用在C＃中本地保存.zip文件 [英] Zip a directory and upload to FTP server without saving the .zip file locally in C#

问题描述

我的问题是标题。我试过这个：

pre $ public void UploadToFtp（List< strucProduktdaten> ProductData）
{
ProductData。 ForEach（delegate（strucProduktdaten data）
{
ZipFile.CreateFromDirectory（data.Quellpfad，data.Zielpfad，CompressionLevel.Fastest，true）;
}）;
}


static void Main（string [] args）
{
List< strucProduktdaten> ProductDataList = new List< strucProduktdaten>（）;
strucProduktdaten ProduktData = new strucProduktdaten（）;
ProduktData.Quellpfad = @Path\to\zip;
ProduktData.Zielpfad = @链接到ftp; //< - 我知道如果没有用uname和密码连接到ftp，链接就没有意义了

ProductDataList.Add（ProduktData）;

ftpClient.UploadToFtp（ProductDataList）;



错误：

System.NotSupportedException：路径格式不受支持。

我不知道我如何应该在这种情况下连接到FTP服务器，并在ram中压缩目录，并直接发送到服务器。

...有人可以帮忙或有链接到一个相似或相同的问题解决了什么？

解决方案

在 MemoryStream中创建ZIP压缩文件 {
使用（流memoryStream =新的MemoryStream（））



  （var archive = new ZipArchive（memoryStream，ZipArchiveMode.Create，true））
 {
 foreach（在Directory.EnumerateFiles（@C：\source\directory）中的字符串路径）
 {
 ZipArchiveEntry entry = archive.CreateEntry（Path.GetFileName（path））;使用（Stream entryStream = entry.Open（））
 using（Stream fileStream = File.OpenRead（path））
 
 fileStream.CopyTo（entryStream） ; 
} 
} 
} 
 
 memoryStream.Seek（0，SeekOrigin.Begin）; 
 
 FtpWebRequest请求= 
（FtpWebRequest）WebRequest.Create（ftp://ftp.example.com/remote/path/archive.zip）; 
 request.Credentials = new NetworkCredential（username，password）; 
 request.Method = WebRequestMethods.Ftp.UploadFile; 
 using（Stream ftpStream = request.GetRequestStream（））
 {
 memoryStream.CopyTo（ftpStream）; 
 
 
  

不幸的是， ZipArchive code>需要可搜索的流。如果没有，您可以直接写入FTP请求流，而不需要将整个ZIP文件保存在内存中。

---

基于：

• 使用System.IO.Compression在内存中创建ZIP存档


• 上传一个文件从一个字符串或流到一个FTP服务器

Hey guys my question is the title. I have tried this:

public void UploadToFtp(List<strucProduktdaten> ProductData)
{
    ProductData.ForEach(delegate( strucProduktdaten data )
    {
        ZipFile.CreateFromDirectory(data.Quellpfad, data.Zielpfad, CompressionLevel.Fastest, true);
    });
}


static void Main(string[] args)
{
    List<strucProduktdaten> ProductDataList = new List<strucProduktdaten>();
    strucProduktdaten ProduktData = new strucProduktdaten();
    ProduktData.Quellpfad = @"Path\to\zip";
    ProduktData.Zielpfad = @"Link to the ftp"; // <- i know the link makes no sense without a connect to the ftp with uname and password

    ProductDataList.Add(ProduktData);

    ftpClient.UploadToFtp(ProductDataList);
}

Error:

System.NotSupportedException:"The Path format is not supported."

I have no idea how I should connect in this case to the FTP server and zipping the directory in ram and send it directly to the server.

... can someone help or have a link to a similar or equal problem what was solved?

解决方案

Create the ZIP archive in MemoryStream and upload it.

using (Stream memoryStream = new MemoryStream())
{
    using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
    {
        foreach (string path in Directory.EnumerateFiles(@"C:\source\directory"))
        {
            ZipArchiveEntry entry = archive.CreateEntry(Path.GetFileName(path));

            using (Stream entryStream = entry.Open())
            using (Stream fileStream = File.OpenRead(path))
            {
                fileStream.CopyTo(entryStream);
            }
        }
    }

    memoryStream.Seek(0, SeekOrigin.Begin);

    FtpWebRequest request =
        (FtpWebRequest)WebRequest.Create("ftp://ftp.example.com/remote/path/archive.zip");
    request.Credentials = new NetworkCredential("username", "password");
    request.Method = WebRequestMethods.Ftp.UploadFile;
    using (Stream ftpStream = request.GetRequestStream())
    {
        memoryStream.CopyTo(ftpStream);
    }
}

Unfortunately the ZipArchive requires a seekable stream. Were it not, you would be able to write directly to the FTP request stream and won't need to keep a whole ZIP file in a memory.

---

Based on:

• Creating a ZIP Archive in Memory Using System.IO.Compression
• Upload a file to an FTP server from a string or stream

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