SQLAlchemy filter query by related object

Question

Using SQLAlchemy, I have a one to many relation with two tables - users and scores. I am trying to query the top 10 users sorted by their aggregate score over the past X amo

Answer 1

I am assuming the column (not the relation) you're using for the join is called Score.user_id, so change it if this is not the case.

You will need to do something like this:

DBSession.query(Score.user_id, func.sum(Score.score_amount).label('total_score')).group_by(Score.user_id).filter(Score.created > somedate).order_by('total_score DESC')[:10]

However this will result in tuples of (user_id, total_score). I'm not sure if the computed score is actually important to you, but if it is, you will probably want to do something like this:

users_scores = []
q = DBSession.query(Score.user_id, func.sum(Score.score_amount).label('total_score')).group_by(Score.user_id).filter(Score.created > somedate).order_by('total_score DESC')[:10]
for user_id, total_score in q:
    user = DBSession.query(User)
    users_scores.append((user, total_score))

This will result in 11 queries being executed, however. It is possible to do it all in a single query, but due to various limitations in SQLAlchemy, it will likely create a very ugly multi-join query or subquery (dependent on engine) and it won't be very performant.

If you plan on doing something like this often and you have a large amount of scores, consider denormalizing the current score onto the user table. It's more work to upkeep, but will result in a single non-join query like:

DBSession.query(User).order_by(User.computed_score.desc())

Hope that helps.