Django order_by() filter with distinct()

Question

How can I make an order_by like this ....

p = Product.objects.filter(vendornumber='403516006')\                    .order_by('-created').distinct('vendor__name') 

The problem is that I have multiple vendors with the same name, and I only want the latest product by the vendor ..

Hope it makes sense?

I got this DB error:

SELECT DISTINCT ON expressions must match initial ORDER BY expressions LINE 1: SELECT DISTINCT ON ("search_vendor"."name") "search_product"...

Answer 1

Based on your error message and this other question, it seems to me this would fix it:

p = Product.objects.filter(vendornumber='403516006')\                .order_by('vendor__name', '-created').distinct('vendor__name') 

That is, it seems that the DISTINCT ON expression(s) must match the leftmost ORDER BY expression(s). So by making the column you use in distinct as the first column in the order_by, I think it should work.

Answer 2

Just matching leftmost order_by() arg and distinct() did not work for me, producing the same error (Django 1.8.7 bug or a feature)?

qs.order_by('project').distinct('project') 

however it worked when I changed to:

qs.order_by('project__id').distinct('project') 

and I do not even have multiple order_by args.