I've recreated this behavior with the following examples. (using 1.0) #models.py class Building(models.Model): name = models.CharField(max_length=10) floors = models.IntegerField()
class Unit(models.Model): unit_number = models.IntegerField() bedrooms = models.IntegerField() building = models.ForeignKey(Building) #filter_test.py b1 = Building(name="Bld 1",floors=10) b1.save() u = Unit(unit_number=1,bedrooms=1,building=b1) u.save() u = Unit(unit_number=2,bedrooms=2,building=b1) u.save() u = Unit(unit_number=3,bedrooms=2,building=b1) u.save() b2 = Building(name="Bld 2",floors=20) b2.save() u = Unit(unit_number=1,bedrooms=1,building=b2) u.save() u = Unit(unit_number=2,bedrooms=2,building=b2) u.save() u = Unit(unit_number=3,bedrooms=1,building=b2) u.save() b3 = Building(name="Bld 3",floors=30) b3.save() u = Unit(unit_number=1,bedrooms=1,building=b3) u.save() u = Unit(unit_number=2,bedrooms=2,building=b3) u.save() u = Unit(unit_number=3,bedrooms=1,building=b3) u.save() buildings = Building.objects.filter(unit__bedrooms=1) for bld in buildings: print bld.name ######OUTPUT Bld 1 Bld 2 Bld 2 Bld 3 Bld 3 -------------------------------- So its returning multiple duplicate Building instances. Is there a way to make the query so it only returns 1 Building instance for each match (rather than the dupes) So ideally it would have just returned. Bld 1, Bld 2 and Bld 3 because they all happen to have units with 1 bedroom. It looks like the ORM is returning 1 Building instance for each Unit that has 1 bedrooms. --~--~---------~--~----~------------~-------~--~----~ 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 django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
