Django Model - Get distinct value list

I try to get a list of distinct foreign keys and I wrote this:

my_ids = Entity.objects.values('foreign_key').distinct()

But I get just a list of UNDISTINCT foreign keys... What am I missing?

Thanks!

Passing an argument to distinct doesn't work for MySQL-databases (AFAIK)

This one works and returns just one object:

Entity.objects.order_by('foreign_key').values('foreign_key').distinct()

Perhaps you might want to go with this:

Entity.objects.order_by().values_list('foreign_key', flat=True).distinct()

Entity.objects.values_list('foreign_key', flat=True).distinct().order_by()

distinct not work without order_by, as explained in Django documentation:

> Any fields used in an order_by() call are included in the SQL SELECT > columns. This can sometimes lead to unexpected results when used in > conjunction with distinct(). If you order by fields from a related > model, those fields will be added to the selected columns and they may > make otherwise duplicate rows appear to be distinct. Since the extra > columns don’t appear in the returned results (they are only there to > support ordering), it sometimes looks like non-distinct results are > being returned. > > Similarly, if you use a values() query to restrict the columns > selected, the columns used in any order_by() (or default model > ordering) will still be involved and may affect uniqueness of the > results. > > The moral here is that if you are using distinct() be careful about > ordering by related models. Similarly, when using distinct() and > values() together, be careful when ordering by fields not in the > values() call.

Entity.objects.order_by('foreign_key').distinct('foreign_key')

If you already have them as a list, then convert it to a set() to get the distinct values.