Django: Serving a Download in a Generic View

So I want to serve a couple of mp3s from a folder in /home/username/music. I didn't think this would be such a big deal but I am a bit confused on how to do it using generic views and my own url.

urls.py

url(r'^song/(?P<song_id>\d+)/download/$', song_down开发者_如何学JAVAload, name='song_download'),

The example I am following is found in the generic view section of the Django documentations: http://docs.djangoproject.com/en/dev/topics/generic-views/ (It's all the way at the bottom)

I am not 100% sure on how to tailor this to my needs. Here is my views.py

def song_download(request, song_id):
    song = Song.objects.get(id=song_id)

    response = object_detail(
        request,
        object_id = song_id,
        mimetype = "audio/mpeg",
    )
    response['Content-Disposition'= "attachment; filename=%s - %s.mp3" % (song.artist, song.title)

    return response

I am actually at a loss of how to convey that I want it to spit out my mp3 instead of what it does now which is to output a .mp3 with all of the current pages html contained. Should my template be my mp3? Do I need to setup apache to serve the files or is Django able to retrieve the mp3 from the filesystem(proper permissions of course) and serve that? If it do need to configure Apache how do I tell Django that?

Thanks in advance. These files are all on the HD so I don't need to "generate" anything on the spot and I'd like to prevent revealing the location of these files if at all possible. A simple /song/1234/download would be fantastic.

Why do you want to do this with a generic view? It's very easy to do this without generic views:

from django.http import HttpResponse


def song_download(request, song_id):
    song = Song.objects.get(id=song_id)
    fsock = open('/path/to/file.mp3', 'rb')
    response = HttpResponse(fsock, content_type='audio/mpeg')
    response['Content-Disposition'] = "attachment; filename=%s - %s.mp3" % \
                                     (song.artist, song.title)
    return response

I'm not sure if it's possible to make this work somehow with a generic view. But either way, using one is redundant here. With no template to render, the context that is automatically provided by the generic view is useless.

To wrap my comment to Tomasz Zielinski into a real answer:

For several reasons it is indeed better to let apache/nginx/etc do the work of sending files. Most servers have mechanisms to help in that usecase: Apache and lighttpd have xsendfile, nginx has X-Accel-Redirect.

The idea is that you can use all the features of django like nice urls, authentification methods, etc, but let the server do the work of serving files. What your django view has to do, is to return a response with a special header. The server will then replace the response with the actual file.

Example for apache:

def song_download(request):
    path = '/path/to/file.mp3'
    response = HttpResponse()
    response['X-Sendfile'] = smart_str(path)
    response['Content-Type'] = "audio/mpeg"
    response['Content-Length'] = os.stat(path).st_size
    return response

• install mode_xsendfile
• add XSendFileOn on and (depending on the version) XSendFileAllowAbove on or XSendFilePath the/path/to/serve/from to your apache configuration.

This way you don't reveale the file location, and keep all the url management in django.

Serving static files with Django is a bad idea, use Apache, nginx etc.

https://docs.djangoproject.com/en/dev/howto/static-files/deployment/

To answer the original question how to use a generic view, you could do the following:

from django.views.generic import DetailView
from django.http.response import FileResponse

class DownloadSong(DetailView):

    model = Song

    def get(self, request, *args, **kwargs):
        super().get(request, *args, **kwargs)
        song = self.object
        return FileResponse(open(song, 'rb'), 
                            as_attachment=True, 
                            filename=f'{song.artist} - {song.title}.mp3')

Docs:

• Detailview: https://docs.djangoproject.com/en/3.2/ref/class-based-views/generic-display/#detailview
• FileResponse: https://docs.djangoproject.com/en/3.2/ref/request-response/#fileresponse-objects

If your Django version does not have the FileResponse object, use the HttpResponse as shown in the other answers.