Having Django serve downloadable files

For the "best of both worlds" you could combine S.Lott's solution with the xsendfile module: django generates the path to the file (or the file itself), but the actual file serving is handled by Apache/Lighttpd. Once you've set up mod_xsendfile, integrating with your view takes a few lines of code:

from django.utils.encoding import smart_strresponse = HttpResponse(mimetype='application/force-download') # mimetype is replaced by content_type for django 1.7response['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

Of course, this will only work if you have control over your server, or your hosting company has mod_xsendfile already set up.

EDIT:

mimetype is replaced by content_type for django 1.7

response = HttpResponse(content_type='application/force-download')  

EDIT: For nginx check this, it uses X-Accel-Redirect instead of apache X-Sendfile header.

A "download" is simply an HTTP header change.

See http://docs.djangoproject.com/en/dev/ref/request-response/#telling-the-browser-to-treat-the-response-as-a-file-attachment for how to respond with a download.

You only need one URL definition for "/download".

The request's GET or POST dictionary will have the "f=somefile.txt" information.

Your view function will simply merge the base path with the "f" value, open the file, create and return a response object. It should be less than 12 lines of code.

For a very simple but not efficient or scalable solution, you can just use the built in django serve view. This is excellent for quick prototypes or one-off work, but as has been mentioned throughout this question, you should use something like apache or nginx in production.

from django.views.static import servefilepath = '/some/path/to/local/file.txt'return serve(request, os.path.basename(filepath), os.path.dirname(filepath))