Serving large files ( with high loads ) in Django

Unless you are going to be serving very very small number of such requests, any solution that requires serving your content via django won't be scalable. For anything to scale in future, you'll probably want to move your content storage and serving to to a separate server and then this won't work.

The recommended way would be to keep static content served through a lighter server (such as nginx). To add security, pass the static server a token from django by setting the cookie or via the get parameters.

The token should have following values: timestamp, filename, userid. It should be signed via some key by the django app.

Next, write a small nginx module which checks the token and that the user has indeed access to the file. It should also check that token isn't old enough by checking the timestamp.

You can use the 'sendfile' method as described in this answer.

Practically you need this (c&p):

response = HttpResponse(mimetype='application/force-download')
response['Content-Disposition'] = 'attachment; filename=%s' % smart_str(file_name)
response['X-Sendfile'] = smart_str(path_to_file)
# It's usually a good idea to set the 'Content-Length' header too.
# You can also set any other required headers: Cache-Control, etc.
return response

This requires mod_xsendfile (which is also supported by nginx or lighty)

Your opening of the image loads it in memory and this is what causes the increase in load under heavy use. As posted by Martin the real solution is to serve the file directly.

Here is another approach, which will stream your file in chunks without loading it in memory.

import os
import mimetypes
from django.http import StreamingHttpResponse
from django.core.servers.basehttp import FileWrapper


def download_file(request):
   the_file = '/some/file/name.png'
   filename = os.path.basename(the_file)
   chunk_size = 8192
   response = StreamingHttpResponse(FileWrapper(open(the_file, 'rb'), chunk_size),
                           content_type=mimetypes.guess_type(the_file)[0])
   response['Content-Length'] = os.path.getsize(the_file)    
   response['Content-Disposition'] = "attachment; filename=%s" % filename
   return response