I have two models
B objects have a foreign key to an
A object. Given a set of
A objects, is there anyway to use the ORM to get a set of
B objects containing the most recent object created for each
Here’s an simplified example:
Class Bakery(models.Model): town = models.CharField() Class Cake(models.Model): bakery = models.ForeignKey(Bakery) baked_at = models.DateTimeField()
So I’m looking for a query that returns the most recent cake baked in each bakery in Anytown, USA.
As far as I know, there is no one-step way of doing this in Django ORM.
But you can split it in two queries:
bakeries = Bakery.objects.annotate(hottest_cake_baked_at=Max('cake__baked_at')) hottest_cakes = Cake.objects.filter(baked_at__in=[b.hottest_cake_baked_at for b in bakeries])
If id’s of cakes are progressing along with bake_at timestamps, you can simplify and disambiguate the above code (in case two cakes arrives at the same time you can get both of them):
hottest_cake_ids = Bakery.objects.annotate(hottest_cake_id=Max('cake__id')).values_list('hottest_cake_id', flat=True) hottest_cakes = Cake.objects.filter(id__in=hottest_cake_ids)
BTW credits for this goes to Daniel Roseman, who once answered similar question of mine:
If the above method is too slow, then I know also second method – you can write custom SQL producing only those Cakes, that are hottest in relevant Bakeries, define it as database VIEW, and then write unmanaged Django model for it. It’s also mentioned in the above django-users thread. Direct link to the original concept is here:
Hope this helps.
If you happen to be using PostGreSQL, you can use Django’s interface to DISTINCT ON:
recent_cakes = Cake.objects.order_by('bakery__id', '-baked_at').distinct('bakery__id')
As the docs say, you must
order by the same fields that you
distinct on. As Simon pointed out below, if you want to do additional sorting, you’ll have to do it in Python-space.
This should do the job:
from django.db.models import Max Bakery.objects.annotate(Max('cake__baked_at'))
hottest_cakes = Cake.objects.filter( baked_at=Subquery( (Cake.objects .filter(bakery=OuterRef('bakery')) .values('bakery') .annotate(last_bake=Max('baked_at')) .values('last_bake')[:1] ) ) ) #BONUS, we can now use this for prefetch_related() bakeries = Bakery.objects.all().prefetch_related( Prefetch('cake_set', queryset=hottest_cakes, to_attr='hottest_cakes' ) ) #usage for bakery in bakeries: print 'Bakery %s has %s hottest_cakes' % (bakery, len(bakery.hottest_cakes))
I was fighting with similar problem and finally come to following solution. It does not rely on
distinct so can be sorted as desired on db-side and also can be used as nested query for filtering. I also believe this implementation is db engine independent, because it’s based on standard sql
HAVING clause. The only drawback is that it will return multiple hottest cakes per bakery, if they are baked in that bakery at exactly same time.
from django.db.models import Max, F Cake.objects.annotate( # annotate with MAX "baked_at" over all cakes in bakery latest_baketime_in_bakery=Max('bakery__cake_set__baked_at') # compare this cake "baked_at" with annotated latest in bakery ).filter(latest_baketime_in_bakery__eq=F('baked_at'))
I haven’t built out the models on my end, but in theory this should work. Broken down:
Cake.objects.filter(bakery__town="Anytown")Should return any cakes whom belong to “Anytown”, assuming the country is not part of the string. The double underscores between
townallow us to access the
.order_by("-created_at")will order the results by their created date, most recent first (take note of the
-(minus) sign in
"-created_at". Without the minus sign, they’d be ordered by oldest to most recent.
[:1]on the end will return only the 1st item in the list which is returned (which would be a list of cakes from Anytown, sorted by most recent first).
Note: This answer is for Django 1.11.
This answer modified from Queries shown here in Django 1.11 Docs.