Create zip archive for instant download

Question

In a web app I am working on, the user can create a zip archive of a folder full of files. Here here's the code:

files = torrent[0].files
 zipfile = z.ZipFile(zipname, 'w')
 output = ""

 for f in files:
  zipfile.write(settings.PYRAT_TRANSMISSION_DOWNLOAD_DIR + "/" + f.name, f.name)

downloadurl = settings.PYRAT_DOWNLOAD_BASE_URL + "/" + settings.PYRAT_ARCHIVE_DIR + "/" + filename
output = "Download <a href=\"" + downloadurl + "\">" + torrent_name + "</a>"
return HttpResponse(output)

But this has the nasty side effect of a long wait (10+ seconds) while the zip archive is being downloaded. Is it possible to skip this? Instead of saving the archive to a file, is it possible to send it straight to the user?

I do beleive that torrentflux provides this excat feature I am talking about. Being able to zip GBs of data and download it within a second.

Answer 1

Check this Serving dynamically generated ZIP archives in Django

Answer 2

Does the zip library you are using allow for output to a stream. You could stream directly to the user instead of temporarily writing to a zip file THEN streaming to the user.

Answer 3

Here's a simple Django view function which zips up (as an example) any readable files in /tmp and returns the zip file.

from django.http import HttpResponse
import zipfile
import os
from cStringIO import StringIO # caveats for Python 3.0 apply

def somezip(request):
    file = StringIO()
    zf = zipfile.ZipFile(file, mode='w', compression=zipfile.ZIP_DEFLATED)
    for fn in os.listdir("/tmp"):
        path = os.path.join("/tmp", fn)
        if os.path.isfile(path):
            try:
                zf.write(path)
            except IOError:
                pass
    zf.close()
    response = HttpResponse(file.getvalue(), mimetype="application/zip")
    response['Content-Disposition'] = 'attachment; filename=yourfiles.zip'
    return response

Of course this approach will only work if the zip files will conveniently fit into memory - if not, you'll have to use a disk file (which you're trying to avoid). In that case, you just replace the file = StringIO() with file = open('/path/to/yourfiles.zip', 'wb') and replace the file.getvalue() with code to read the contents of the disk file.

Answer 4

It is possible to pass an iterator to the constructor of a HttpResponse (see docs). That would allow you to create a custom iterator that generates data as it is being requested. However I don't think that will work with a zip (you would have to send partial zip as it is being created).

The proper way, I think, would be to create the files offline, in a separate process. The user could then monitor the progress and then download the file when its ready (possibly by using the iterator method described above). This would be similar what sites like youtube use when you upload a file and wait for it to be processed.