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);
}
`
这是将一个实体转换为XML文件，然后压缩它的方法:
private void downloadFile(EntityXML xml) {
string nameDownloadXml = "File_1.xml";
string nameDownloadZip = "File_1.zip";
var serializer = new XmlSerializer(typeof(EntityXML));
Response.Clear();
Response.ClearContent();
Response.ClearHeaders();
Response.AddHeader("content-disposition", "attachment;filename=" + nameDownloadZip);
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var demoFile = archive.CreateEntry(nameDownloadXml);
using (var entryStream = demoFile.Open())
using (StreamWriter writer = new StreamWriter(entryStream, System.Text.Encoding.UTF8))
{
serializer.Serialize(writer, xml);
}
}
using (var fileStream = Response.OutputStream)
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
Response.End();
}
感谢ZipArchive创建无效的ZIP文件，我得到:
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
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))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
这表明我们需要在使用ZipArchive之前调用Dispose，正如Amir所建议的，这可能是因为它将最后的字节(如校验和)写入存档，使其完整。但是为了不关闭流，这样我们就可以重用它，你需要将true作为第三个参数传递给ZipArchive。
