SqlAlchemy closure in subquery - sqlalchemy

Question

I have searched many topics and didn't find the answer, or question was too complex. So okay. This is my first question.
Here is the SQL
SELECT parent.*,
(
SELECT COUNT(*)
FROM child
WHERE parent.id = child.parent_id
)
FROM parent
How to do this clause in sqlalchemy?
WHERE ui.invited_by = u.id
Can it be reproduced in collections ? sql expressions ?
P.S. I know that it can be done by group_by. But i need by subquery.
Thank you.

Answer 1

The SA query (using subquery) will give you the results you want:
sq = session.query(Child.parent_id, func.count(Child.id).label("child_num")).group_by(Child.parent_id)
sq = sq.subquery()
# use outerjoin to have also those Parents with 0 (zero) children
q = session.query(Parent, sq.c.child_num).outerjoin(sq)
q = q.filter(Parent.id == 1) # add your filter here: ui.invited_by = u.id
for x in q.all():
print x
although the subquery is not exactly as you described, but rather something like:
SELECT parent.*,
anon_1.child_num AS anon_1_child_num
FROM parent
LEFT JOIN (SELECT child.parent_id AS parent_id,
count(child.id) AS child_num
FROM child
GROUP BY child.parent_id
) AS anon_1
ON parent.id = anon_1.parent_id
Still do not understand why you need a sub-query the way you described though.

Answer 2

I find here really awesome answer. But too also too complicated. First of all i want to tell that closure in sql world is CORRELATION.
This is NOT the same but helps me.
pparent = Parent.__table__.alias('pparent') # using table directly to alias.
subquery = s.query(count(Child.id)).join(pparent) # usual thing but use aliased table.
s.query(Parent, subquery.filter(Parent.id == pparent.id).correlate(Parent).as_scalar()) #magic