SqlAlchemy: filter to match all instead of any values in list?

Question

I want to query a junction table for the value of column aID that matches all values of a list of ids ids=[3,5] in column bID.

This is my junction table (JT):

 aID    bID
   1      1
   1      2
   2      5
   2      3
   1      3
   3      5

I have this query: session.query(JT.aID).filter(JT.bID.in_(ids)).all()

This query returns the aID values 1, 2 and 3 because they all have rows with either 3 or 5 in the bID column. What I want the query to return is 2 because that is the only aID value that has all values of the ids list in its bID column.

Don't know how to explain the problem better, but how can I get to the result?

Answer 1

Based on @Gordon Linoff answer and with two tables A and B where A has a relation one- to-many towards B called A.bs the SqlAlchemy equivalent would be:

from sqlalchemy import func  
session.query(A).join(B).filter(B.id.in_(<your_list>)).group_by(A.id).having(func.count(A.bs) == len(<your_list>)).all()

Answer 2

You are looking for a query that works on sets of rows. I think a group by with having clause is the best approach:

select aid
from jt
where bid in (<your list>)
group by aid
having count(distinct bid) = 2

If you can put the ids that you desire in a table, you can do the following more generic approach:

select aid
from jt join
     bids
     on jf.bid = bids.bid
group by aid
having count(distinct jt.bid) = (select count(*) from bids)