33👍
✅
You can try the following code, assuming that object_name
is an object of that model:
filename = object_name.file.name.split('/')[-1]
response = HttpResponse(object_name.file, content_type='text/plain')
response['Content-Disposition'] = 'attachment; filename=%s' % filename
return response
See the following part of the Django documentation on sending files directly: https://docs.djangoproject.com/en/dev/ref/request-response/#telling-the-browser-to-treat-the-response-as-a-file-attachment
6👍
https://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.FileField.storage
All that will be stored in your database is a path to the file (relative to MEDIA_ROOT). You’ll most likely want to use the convenience url function provided by Django. For example, if your ImageField is called mug_shot, you can get the absolute path to your image in a template with {{ object.mug_shot.url }}.
👤jpic
Source:stackexchange.com