In sqlalchemy, how can I combine two queries by having a column entry identical?

Question

Suppose I have a mapped class User , mapped to a tables of the same name, and a column "age" for his age. I'm now interested in the following problem:

In the course of my application, there emerge two queries:

q1 = session.query(User).filter(lots of conditions)
q2 = session.query(User).filter(lots of other conditions)

I now want to "join" q2 onto q1 upon the condition that they have the same age. But I have no idea of how this might work. I tried the following without success:

q1.join(q2.subquery(), q1.age==q2.age) # the query doesn't hold the columns of the queried class
q1.join(Age).join(q2.subquery()) # Works only if age is a relationship with mapped class Age

My closest calls were something like this:

a1 = aliased(User)
a2 = aliased(User)
q1 = session.query(a1)
q2 = session.query(a2)
s = q2.subquery()
q1.join(s, a1.age==a2.age).all()
>>> sqlalchemy.exc.OperationalError: (OperationalError) no such column: user_2.age 'SELECT user_1.id AS user_1_id, user_1.name AS user_1_name, user_1.age AS user_1_age \nFROM user AS user_1 JOIN (SELECT user_2.id AS id, user_2.name AS name, user_2.age AS age \nFROM user AS user_2) AS anon_1 ON user_1.age = user_2.age' ()

Any ideas about how to make this run?

Answer 1

After fiddling around and reading answers to questions about subqueries, I managed to find a solution. Instead of the last, offending line, put:

q1.join(s, a1.age==s.columns.age).all()

That way, the on-clause becomes ON user_1.age = anon_1.age , which is what we want.