statuscodes=Member.objects.values('status').order_by('status').distinct() It gives me the expected results. Three items in the dictionary from a database table of about 10,000 records: [{'status': u'ACTIVE'}, {'status': u'RESIGNED'}, {'status': u'TRANSFER'}]
it's what i both expected and want. On Jul 21, 7:33 pm, Subhranath Chunder <subhran...@gmail.com> wrote: > I thought you were trying to get: > > I'm expecting only the four distinct records. > > You'll get a ValuesQuerySet or list of dictionaries with redundant keys this > way. > > Thanks, > Subhranath Chunder. > > > > On Wed, Jul 21, 2010 at 11:40 PM, rmschne <rmsc...@gmail.com> wrote: > > Member.objects.values('status').order_by('status').distinct() from > > jaymzcd works perfectly. Thanks! > > > -- > > You received this message because you are subscribed to the Google Groups > > "Django users" group. > > To post to this group, send email to django-us...@googlegroups.com. > > To unsubscribe from this group, send email to > > django-users+unsubscr...@googlegroups.com<django-users%2bunsubscr...@google > > groups.com> > > . > > For more options, visit this group at > >http://groups.google.com/group/django-users?hl=en. -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@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.