I had to define the callable function before it gets called. So in your case I'd put it right under class Product(models.model) Then you should be able to just do: upload_to=get_image_path
Cheers, On Sep 18, 10:20 pm, lingrlongr <[EMAIL PROTECTED]> wrote: > According to the documentation, I can make the upload_to argument a > callable (which must receive 2 args): > > class Product(models.Model): > number = models.CharField(max_length=20) > image = models.ImageField(upload_to='<callable>') > > I want to build a path so it looks like this (with respect to > MEDIA_URL): > images/products/<number>/<original_file_name> > > def get_image_path(instance, filename): > return 'images/products/%s/%s' % (instance.number, filename) > > I tried: > upload_to='get_image_path' > upload_to=get_image_path > upload_to=get_image_path() > upload_to=self.get_image_path > > How can I accomplish that? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
