我必须在亚马逊S3中创建一个文件夹。现在必须在zip文件中转换该文件夹。我使用DotNetZip Liberary转换为.zip文件。这是
的链接http://dotnetzip.codeplex.com/wikipage?title=CS-Examples
public void ConvertToZip(string directoryToZip, string zipFileName)
{
try
{
using (client = DisposableAmazonClient())
{
var sourDir = new S3DirectoryInfo(client, bucket, directoryToZip);
var destDir = new S3DirectoryInfo(client, bucket, CCUrlHelper.BackupRootFolderPhysicalPath);
using (var zip = new ZipFile())
{
zip.AddDirectory(sourDir.FullName); // recurses subdirectories
zip.Save(Path.Combine(destDir.FullName, zipFileName));
}
}
logger.Fatal("Successfully converted to Zip.");
}
catch (Exception ex)
{
logger.Error("Error while converting to zip. Error : " + ex.Message);
}
}
当我运行代码时,它显示错误“不支持给定路径的格式。”
答案 0 :(得分:3)
S3DirectoryInfo模拟目录结构,但它不是实际的目录结构,并且DotNetZip不知道如何处理指向S3中的对象的字符串。为了让您这样做,您将不得不下载文件,压缩它们,然后上传生成的zip文件。以下是一些示例代码,说明了如何做到这一点。
class Program { static void Main(string[] args) { var zipFilename = @"c:\temp\data.zip"; var client = new AmazonS3Client(); S3DirectoryInfo rootDir = new S3DirectoryInfo(client, "norm-ziptest"); using (var zip = new ZipFile()) { zip.Name = zipFilename; addFiles(zip, rootDir, ""); }// Move local zip file to S3 var fileInfo = rootDir.GetFile("data.zip"); fileInfo.MoveFromLocal(zipFilename); } static void addFiles(ZipFile zip, S3DirectoryInfo dirInfo, string archiveDirectory) { foreach (var childDirs in dirInfo.GetDirectories()) { var entry = zip.AddDirectoryByName(childDirs.Name); addFiles(zip, childDirs, archiveDirectory + entry.FileName); } foreach (var file in dirInfo.GetFiles()) { using (var stream = file.OpenRead()) { zip.AddEntry(archiveDirectory + file.Name, stream); // Save after adding the file because to force the // immediate read from the S3 Stream since // we don't want to keep that stream open. zip.Save(); } } }
}