我试图使用MemoryStream创建一个简单的演示文本文件的ZIP存档,如下所示:
using (var memoryStream = new MemoryStream())
using (var archive = new ZipArchive(memoryStream , ZipArchiveMode.Create))
{
var demoFile = archive.CreateEntry("foo.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write("Bar!");
}
using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
{
stream.CopyTo(fileStream);
}
}
如果我运行这段代码,就会创建归档文件本身,但foo.txt不会。
然而,如果我直接用文件流替换MemoryStream,存档将被正确创建:
using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
using (var archive = new ZipArchive(fileStream, FileMode.Create))
{
// ...
}
是否可以使用MemoryStream来创建没有FileStream的ZIP存档?
函数返回包含zip文件的流
public static Stream ZipGenerator(List<string> files)
{
ZipArchiveEntry fileInArchive;
Stream entryStream;
int i = 0;
List<byte[]> byteArray = new List<byte[]>();
foreach (var file in files)
{
byteArray.Add(File.ReadAllBytes(file));
}
var outStream = new MemoryStream();
using (var archive = new ZipArchive(outStream, ZipArchiveMode.Create, true))
{
foreach (var file in files)
{
fileInArchive=(archive.CreateEntry(Path.GetFileName(file), CompressionLevel.Optimal));
using (entryStream = fileInArchive.Open())
{
using (var fileToCompressStream = new MemoryStream(byteArray[i]))
{
fileToCompressStream.CopyTo(entryStream);
}
i++;
}
}
}
outStream.Position = 0;
return outStream;
}
如果你想,写zip文件流。
using (var fileStream = new FileStream(@"D:\Tools\DBExtractor\DBExtractor\bin\Debug\test.zip", FileMode.Create))
{
outStream.Position = 0;
outStream.WriteTo(fileStream);
}
`