Thank you, as always, for the quick and detailed response.
With the join to the subquery that's on func.max(A.id), once you use
> that function, the column loses it's "A.id-ness", because SQLA doesn't
> know anything about func.max() and for all it knows it could be turning
> it into anything.
>
I figured as much, and obviously this is the correct behavior. I was hoping
there was some way I could tell SQLAlchemy that subquery.id "possesses
A.id-ness"
(e.g. via .select_from()), but that doesn't appear to be the case.
3. build yourself a function, if you want it to look nice you can use
> with_transformation()
>
This is exactly what I'm doing (minus the with_transformation(), which I'm
about to look up...). The problem is that there are additional classes C, D,
E, F, etc. that all point to A (via a ForeignKey and a relationship), and
the query q could involve any one of them. (Obviously if it involved more
than one, I would need to specify the join explicitly.) So it's not at all
straightforward (at least to me) to figure out on what to join -- unless I
require that it be explicitly provided as an argument to the function, i.e.
join_to_min_a(q,
field_to_join_to_A_id).
On Wednesday, July 13, 2016 at 12:16:52 PM UTC-4, Mike Bayer wrote:
>
>
>
> On 07/13/2016 02:29 AM, Seth P wrote:
> > [Apologies for posting an incomplete version of this post earlier.
> > Please ignore it.]
> >
> > If B has a ForeignKey (and relationship) to A (e.g. B.a_id -> A.id),
> > then I can write query(B.b_num).join(A) without specifying the
> > condition, and SQLAlchemy will figure out the join automatically. [See
> > query 0 in the code below.]
> >
> > It will similarly figure out the join of B with a "direct" query ofA.id,
> > e.g.
> > query(A.id).order_by(A.a_num).limit(1).subquery('first_a_id_by_num').
> > [See query 1 in the code below.]
> >
> > However, it will not work with a more complicated query of A.id, e.g.
> > query(func.min(A.id).label('id')).subquery('max_a_id'). [See query 2 in
> > the code below.]
> >
> > Of course, I can make it work by specifying the join condition B.a_id ==
> > subquery_returning_one_id.c.id. [See query 3 in the code below.]
> >
> > I can get the implicit join to work with such a subquery by joining with
> > a separate A.id and using the subquery to filter this A.id, but this
> > seems more convoluted than necessary. [See query 4 in the code below.]
> >
> > I can also get it to work with
> >
> query(A.id).select_from(subquery_returning_one_id).subquery('a_id_from_max_a_id'),
>
>
> > but like query 4, this also introduces an extra reference to A.id. [See
> > query 5 in the code below.]
> >
> > Is there any way to get an implicit join like query 2 to produce sql as
> > in query 3, without introducing (explicitly as in query 4 or implicitly
> > as in query 5) an extra reference to A.id? Or is the extra copy of A.id
> > in queries 4 and 5 pretty harmless performance-wise, and I should just
> > deal with it as the cost of not providing an explicit join condition?
> >
> > Yes, I realize that I can avoid this problem by providing an explicit
> > join condition, but I'd prefer to avoid that if possible. (Also, in case
> > it matters, my actual subquery is more complicated than the
> > func.min(A.id) example here, but in the end returns a single column
> > labeled id with values from A.id.)
>
> So, when you do a thing like query(B).join(A), it is using the foreign
> keys between B and A to figure that out, but *not* the relationship
> "B.a". If OTOH you do query(B).join(B.a), then you *are* using the
> relationship.
>
> With the join to the subquery that's on func.max(A.id), once you use
> that function, the column loses it's "A.id-ness", because SQLA doesn't
> know anything about func.max() and for all it knows it could be turning
> it into anything. So neither a join on FKs nor on the existing
> relationship can figure that out immediately.
>
> It depends here on where you are OK doing the explicit mentioning of
> A.id and B.a_id. it has to be somewhere. It can be:
>
> 1. in a new relationship() that you put on A or B, that doesn't normally
> load but you can use it here
>
> 2. when you make the subquery, include B.a_id in it somehow, like either
> select from B.a_id instead of A.id (if that applies), or pre-fabricate
> your join condition:
>
> q = session.query(B.b_num)
> subquery_returning_one_id =
> session.query(func.min(A.id).label('id')).subquery('max_a_id')
>
> j = subquery_returning_one_id.join(B, B.a_id ==
> subquery_returning_one_id.c.id)
>
> query = q.select_from(j)
>
> 3. build yourself a function, if you want it to look nice you can use
> with_transformation()
>
> def join_to_min_a(q):
> subquery_returning_one_id =
> session.query(func.min(A.id).label('id')).subquery('max_a_id')
> q = q.join(subquery_returning_one_id,
> subquery_returning_one_id.c.id == B.a_id)
> return q
>
> q = session.query(B.b_num)
>
> q = q.with_transformation(join_to_min_a)
> print(q.one())
>
>
>
>
>
> >
> >
> >
> > from sqlalchemy import create_engine, func, Column, Integer, ForeignKey
> > from sqlalchemy.orm import relationship, sessionmaker
> > from sqlalchemy.ext.declarative import declarative_base
> >
> > sqlite = 'sqlite:///test.db'
> > engine = create_engine(sqlite, echo=True)
> > Base = declarative_base(bind=engine)
> >
> >
> > class A(Base):
> > __tablename__ = 'a'
> > id = Column(Integer, primary_key=True)
> > a_num = Column(Integer)
> >
> >
> > class B(Base):
> > __tablename__ = 'b'
> > id = Column(Integer, primary_key=True)
> > b_num = Column(Integer)
> > a_id = Column(Integer, ForeignKey(A.id))
> > a = relationship(A)
> >
> >
> > if __name__ == '__main__':
> > Base.metadata.drop_all()
> > Base.metadata.create_all()
> > session = sessionmaker(bind=engine)()
> > session.add(B(b_num=2, a=A(a_num=1)))
> > session.commit()
> >
> > q = session.query(B.b_num)
> > subquery_returning_one_A_id =
> session.query(A.id).order_by(A.a_num).limit(1).subquery('first_a_id_by_num')
>
> > subquery_returning_one_id =
> session.query(func.min(A.id).label('id')).subquery('max_a_id')
> >
> > i = 0
> > print("\n%d" % i)
> > try:
> > query = q.join(A)
> > print(query.one())
> > except Exception as e:
> > print("Exception:", e)
> >
> > i = 1
> > print("\n%d" % i)
> > try:
> > query = q.join(subquery_returning_one_A_id)
> > print(query.one())
> > except Exception as e:
> > print("Exception:", e)
> >
> > i = 2
> > print("\n%d" % i)
> > try:
> > query = q.join(subquery_returning_one_id)
> > print(query.one())
> > except Exception as e:
> > print("Exception:", e)
> >
> > i = 3
> > print("\n%d" % i)
> > try:
> > query = q.join(subquery_returning_one_id, B.a_id ==
> subquery_returning_one_id.c.id)
> > print(query.one())
> > except Exception as e:
> > print("Exception:", e)
> >
> > i = 4
> > print("\n%d" % i)
> > try:
> > query = q.join(session.query(A.id).filter(A.id ==
> > subquery_returning_one_id.c.id).subquery('a_id_equal_to_max_a_id'))
> > print(query.one())
> > except Exception as e:
> > print("Exception:", e)
> >
> > i = 5
> > print("\n%d" % i)
> > try:
> > query =
> q.join(session.query(A.id).select_from(subquery_returning_one_id).subquery('a_id_from_max_a_id'))
>
>
> > print(query.one())
> > except Exception as e:
> > print("Exception:", e)
> >
> > session.close_all()
> >
> >
> > Relevant output:
> >
> > 0
> > 2016-07-13 02:17:41,901 INFO sqlalchemy.engine.base.Engine BEGIN
> (implicit)
> > 2016-07-13 02:17:41,902 INFO sqlalchemy.engine.base.Engine SELECT
> > b.b_num AS b_b_num
> > FROM b JOIN a ON a.id = b.a_id
> > 2016-07-13 02:17:41,902 INFO sqlalchemy.engine.base.Engine ()
> > (2,)
> >
> > 1
> > 2016-07-13 02:17:41,902 INFO sqlalchemy.engine.base.Engine SELECT
> > b.b_num AS b_b_num
> > FROM b JOIN (SELECT a.id AS id
> > FROM a ORDER BY a.a_num
> > LIMIT ? OFFSET ?) AS first_a_id_by_num ON first_a_id_by_num.id =
> b.a_id
> > 2016-07-13 02:17:41,908 INFO sqlalchemy.engine.base.Engine (1, 0)
> > (2,)
> >
> > 2
> > Exception: Could not find a FROM clause to join from. Tried joining to
> > SELECT min(a.id) AS id
> > FROM a, but got: Can't find any foreign key relationships between 'b'
> > and 'max_a_id'.
> >
> > 3
> > 2016-07-13 02:17:41,912 INFO sqlalchemy.engine.base.Engine SELECT
> > b.b_num AS b_b_num
> > FROM b JOIN (SELECT min(a.id) AS id
> > FROM a) AS max_a_id ON b.a_id = max_a_id.id
> > 2016-07-13 02:17:41,914 INFO sqlalchemy.engine.base.Engine ()
> > (2,)
> >
> > 4
> > 2016-07-13 02:17:41,917 INFO sqlalchemy.engine.base.Engine SELECT
> > b.b_num AS b_b_num
> > FROM b JOIN (SELECT a.id AS id
> > FROM a, (SELECT min(a.id) AS id
> > FROM a) AS max_a_id
> > WHERE a.id = max_a_id.id) AS a_id_equal_to_max_a_id ON
> > a_id_equal_to_max_a_id.id = b.a_id
> > 2016-07-13 02:17:41,920 INFO sqlalchemy.engine.base.Engine ()
> > (2,)
> >
> > 5
> > 2016-07-13 02:17:41,922 INFO sqlalchemy.engine.base.Engine SELECT
> > b.b_num AS b_b_num
> > FROM b JOIN (SELECT a.id AS id
> > FROM a, (SELECT min(a.id) AS id
> > FROM a) AS max_a_id) AS a_id_from_max_a_id ON a_id_from_max_a_id.id =
> b.a_id
> > 2016-07-13 02:17:41,922 INFO sqlalchemy.engine.base.Engine ()
> > (2,)
> >
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