You should be able to do this: p = Publisher.objects.get(pk=1) authors = Author.objects.filter(book__publisher__exact = p).distinct()
list = ([(a.id, a.name) for a in authors]) I don't have a python shell handy so I do not know for sure that the syntax is correct, but I do things like this all the time with Django. -rfd On Aug 14, 2:42 pm, NickJ <[EMAIL PROTECTED]> wrote: > Sorry, think I've answered my own question: Further searching found a > similar request in developers a couple of months back: > > http://groups.google.com/group/django-developers/browse_thread/thread... > > Which didn't get any love. So I guess I'll need to drop in to SQL for > now. > > On Aug 14, 4:34 pm, NickJ <[EMAIL PROTECTED]> wrote: > > > Take 3 models: > > > class Publisher(models.Model): > > name = models.CharField() > > > class Author(models.Model): > > name = models.CharField() > > > class Book(models.Model): > > publisher = models.ForeignKey(Publisher) > > authors = models.ManyToManyField(Author) > > > i.e. a book must have one or more authors, and must have one > > publisher. > > > What i want is to get a distinct list of all authors for a specific > > publisher. Is this possible without dropping back to raw SQL? > > > I want to do something like: > > > p = Publisher.objects.get(pk=1) > > list = p.book_set.values('publisher__id', 'publisher__name').distinct() --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
