Question

I have the following django model with an ImageField:

class Pics(models.Model):
    def file_path(instance, filename):
        return '/'.join([instance.user.username, 'pics', filename])
    #...
    file = models.ImageField(upload_to=file_path)

I know I can call pic.file.url to get the url to the image. however, I’m wondering if Django also implements some functionality to automatically host that image?

The code I have currently works fine but it seems redundant:

#urls.py
    urlpatterns += patterns('views',
         url(r'media/(?P<name>\w{1,30})/pics/(?P<file>[^\\/:*?\"<>|]+)$', 'show_pic'),)

then my view is:

#app.views
import mimetypes
def show_pic(request, name, file):
     path = u'media/%s/graphics/%s' % (name, file)
     mime = mimetypes.guess_type(name)[0]
     image_data = open(path, 'rb').read()
     return HttpResponse(image_data, mimetype=mime)

Answer 1

Instead of creating a view to serve the pictures you can directly point to media/<name>/graphics/<file> and let your webserver itself serve the images. Django is not well suited for serving static files, but your webserver on the other hand is designed to handle them efficiently.