Create zip file from byte[]
This is a variation of the great accepted answer posted by the OP. However, this is for WebForms instead of MVC. I'm working with the assumption that caseAttachmentModel.Body is a byte[]
Essentially everything is the same except with an additional method that sends the zip out as a Response.
using (var compressedFileStream = new MemoryStream()) {
//Create an archive and store the stream in memory.
using (var zipArchive = new ZipArchive(compressedFileStream, ZipArchiveMode.Update, false)) {
foreach (var caseAttachmentModel in caseAttachmentModels) {
//Create a zip entry for each attachment
var zipEntry = zipArchive.CreateEntry(caseAttachmentModel.Name);
//Get the stream of the attachment
using (var originalFileStream = new MemoryStream(caseAttachmentModel.Body)) {
using (var zipEntryStream = zipEntry.Open()) {
//Copy the attachment stream to the zip entry stream
originalFileStream.CopyTo(zipEntryStream);
}
}
}
}
sendOutZIP(compressedFileStream.ToArray(), "FileName.zip");
}
private void sendOutZIP(byte[] zippedFiles, string filename)
{
Response.Clear();
Response.ClearContent();
Response.ClearHeaders();
Response.ContentType = "application/x-compressed";
Response.Charset = string.Empty;
Response.Cache.SetCacheability(System.Web.HttpCacheability.Public);
Response.AddHeader("Content-Disposition", "attachment; filename=" + filename);
Response.BinaryWrite(zippedFiles);
Response.OutputStream.Flush();
Response.OutputStream.Close();
Response.End();
}
I would also like to point out that advice given by @Levi Fuller on references in the accepted answer is spot on!
After a little more playing around and reading I was able to figure this out. Here is how you can create a zip file (archive) with multiple files without writing any temporary data to disk:
using (var compressedFileStream = new MemoryStream())
{
//Create an archive and store the stream in memory.
using (var zipArchive = new ZipArchive(compressedFileStream, ZipArchiveMode.Create, false)) {
foreach (var caseAttachmentModel in caseAttachmentModels) {
//Create a zip entry for each attachment
var zipEntry = zipArchive.CreateEntry(caseAttachmentModel.Name);
//Get the stream of the attachment
using (var originalFileStream = new MemoryStream(caseAttachmentModel.Body))
using (var zipEntryStream = zipEntry.Open()) {
//Copy the attachment stream to the zip entry stream
originalFileStream.CopyTo(zipEntryStream);
}
}
}
return new FileContentResult(compressedFileStream.ToArray(), "application/zip") { FileDownloadName = "Filename.zip" };
}