Я пытаюсь использовать следующий код: я получаю поврежденный zip файл. Зачем? Имена файлов выглядят нормально. Возможно, они не являются относительными именами, и что проблема?
private void trySharpZipLib(ArrayList filesToInclude)
{
// Response header
Response.Clear();
Response.ClearHeaders();
Response.Cache.SetCacheability(HttpCacheability.NoCache);
Response.StatusCode = 200; // http://community.icsharpcode.net/forums/p/6946/20138.aspx
long zipSize = calculateZipSize(filesToInclude);
string contentValue =
string.Format("attachment; filename=moshe.zip;"
); // + " size={0}", zipSize);
Response.ContentType = "application/octet-stream"; //"application/zip";
Response.AddHeader("Content-Disposition", contentValue);
Response.Flush();
using (ZipOutputStream zipOutputStream = new ZipOutputStream(Response.OutputStream) )
{
zipOutputStream.SetLevel(0);
foreach (string f in filesToInclude)
{
string filename = Path.Combine(Server.MapPath("."), f);
using (FileStream fs = File.OpenRead(filename))
{
ZipEntry entry =
new ZipEntry(ZipEntry.CleanName(filename))
{
DateTime = File.GetCreationTime(filename),
CompressionMethod = CompressionMethod.Stored,
Size = fs.Length
};
zipOutputStream.PutNextEntry(entry);
byte[] buffer = new byte[fs.Length];
// write to zipoutStream via buffer.
// The zipoutStream is directly connected to Response.Output (in the constructor)
ICSharpCode.SharpZipLib.Core.StreamUtils.Copy(fs, zipOutputStream, buffer);
Response.Flush(); // for immediate response to user
} // .. using file stream
}// .. each file
}
Response.Flush();
Response.End();
}