在sqlalchemy，数据库，ORM中使用标签

Question

q = session.query(
    label('id1', func.least(Communication.initiator_id, Communication.receiver_id)),
    label('id2', func.greatest(Communication.initiator_id, Communication.receiver_id)),
    label('orga_id1', func.greatest(0, 1)),
    label('orga_id2', func.greatest(0, 1)),
    label('nb', func.count(Communication.id))
).filter(
    or_(
        Communication.initiator_id == people_id,
        Communication.receiver_id == people_id
    )
).order_by("nb").group_by('id1', 'id2').all()

So i have this piece of code, and it's working fine but only one thing i would like to add here, and that thing is the use of labels. For Exemple how do i use the name label "id1" in my other label?

I my code to be something like this:

q = session.query(
    label('id1', func.least(Communication.initiator_id, Communication.receiver_id)),
    label('id2', func.greatest(Communication.initiator_id, Communication.receiver_id)),
    label('orga_id1', func.greatest(session.query(People).filter_by(id = "And here somehow i want the value that was created on first label with name 'id1'").first().orga_id, -1)),
    label('orga_id2', func.greatest(session.query(People).filter_by(id = "And here somehow i want the value that was created on second label with name 'id2'").first().orga_id, -1)),
    label('nb', func.count(Communication.id))
).filter(
    or_(
        Communication.initiator_id == people_id,
        Communication.receiver_id == people_id
    )
).order_by("nb").group_by('id1', 'id2').all()

Can someone pls explain me how do i use the value of labels inside the query !!!

Answer 1

You cannot reference the output columns of the parent query in a subquery, though you could reference the parent's source tables:

select 1 as one, (select one) as two;

will not work, but

select 1 as one, (select s.i) as two from generate_series(2, 2) s(i);

does. You can on the other hand first perform your query for fetching the communications and counts and then add the Persons' organization ids on top. Using scalar subqueries the way you've envisioned originally:

id1 = func.least(Communication.initiator_id, Communication.receiver_id)
id2 = func.greatest(Communication.initiator_id, Communication.receiver_id)

sq = session.query(id1.label('id1'),
                   id2.label('id2'),
                   func.count(Communication.id).label('nb')).\
    filter(or_(Communication.initiator_id == people_id,
               Communication.receiver_id == people_id)).\
    order_by('nb').\
    group_by('id1', 'id2').\
    subquery()

orga_id1 = session.query(People.orga_id).filter_by(id=sq.c.id1).as_scalar()
orga_id2 = session.query(People.orga_id).filter_by(id=sq.c.id2).as_scalar()

q = session.query(sq.c.id1,
                  sq.c.id2,
                  func.greatest(orga_id1, -1).label('orga_id1'),
                  func.greatest(orga_id2, -1).label('orga_id2'),
                  sq.c.nb).\
    all()

With LEFT JOINs:

from sqlalchemy.orm import aliased

id1 = func.least(Communication.initiator_id, Communication.receiver_id)
id2 = func.greatest(Communication.initiator_id, Communication.receiver_id)

people1 = aliased(People)
people2 = aliased(People)

sq = session.query(id1.label('id1'),
                   id2.label('id2'),
                   func.count(Communication.id).label('nb')).\
    filter(or_(Communication.initiator_id == people_id,
               Communication.receiver_id == people_id)).\
    order_by('nb').\
    group_by('id1', 'id2').\
    subquery()

q = session.query(sq.c.id1,
                  sq.c.id2,
                  func.coalesce(people1.orga_id, -1).label('orga_id1'),
                  func.coalesce(people2.orga_id, -1).label('orga_id2'),
                  sq.c.nb).\
    outerjoin(people1, people1.id == sq.c.id1).\
    outerjoin(people2, people2.id == sq.c.id2).\
    all()