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);
}
错误:
System.NotSupportedException:&#34;不支持路径格式。&#34;
我不知道在这种情况下我应该如何连接到FTP服务器并压缩ram中的目录并将其直接发送到服务器。
......有人可以帮助或链接到类似或同等问题的解决方案吗?
答案 0 :(得分:2)
在MemoryStream
中创建ZIP存档并上传。
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);
}
}
不幸的是,ZipArchive
需要可搜索的流。如果不是,您将能够直接写入FTP请求流,而不需要将整个ZIP文件保存在内存中。
基于:
答案 1 :(得分:1)
像这样的东西可以使ZIP进入内存:
public static byte[] ZipFolderToMemory(string folder)
{
using (var stream = new MemoryStream())
{
using (var archive = new ZipArchive(stream, ZipArchiveMode.Create))
{
foreach (var filePath in Directory.EnumerateFiles(folder))
{
var entry = archive.CreateEntry(Path.GetFileName(filePath));
using (var zipEntry = entry.Open())
using (var file = new FileStream(filePath, FileMode.Open))
{
file.CopyTo(zipEntry);
}
}
}
return stream.ToArray();
}
}
获得字节数组后,您应该可以随时将其发送到服务器。