I want users on the site to be able to download files whose paths are obscured so they cannot be directly downloaded.
For instance, I'd like the URL to be something like this, "http://example.com/download/?f=somefile.txt
And on the server, I know that all downloadable files reside in a folder "/home/user/files/".
Is there a way to make Django serve that file for download as opposed to trying to find a URL and View to display it?
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_str response = HttpResponse(mimetype='application/force-download') # mimetype is replaced by content_type for django 1.7 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
Of course, this will only work if you have control over your server, or your hosting company has mod_xsendfile already set up.
mimetype is replaced by content_type for django 1.7
response = HttpResponse(content_type='application/force-download'
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
POST dictionary will have the
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.