In sqlalchemy, how can I combine two queries by having a column entry identical?
join, python, sqlalchemy
Solution
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.
Problem
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?