That works!
Obviously I need to know that the joining field is called a_id, and I can
live with that, since in practice it's uniform. But I'm just curious if
there's an automated way to figure out which entity/column is related to A.
(There could be more than one entity in q, though just one that has a
(unique) ForeignKey to A.)
On Wednesday, July 13, 2016 at 2:06:55 PM UTC-4, Mike Bayer wrote:
>
>
>
> On 07/13/2016 01:04 PM, Seth P wrote:
> > 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).
>
> OK then how about this:
>
> def join_to_min_a(q):
> subquery_returning_one_id =
> session.query(func.min(A.id).label('id')).subquery('max_a_id')
> joining_to = q.column_descriptions[0]['entity'].a_id
> q = q.join(subquery_returning_one_id,
> subquery_returning_one_id.c.id == joining_to)
> return q
>
>
>
>
>
>
> >
> >
> > 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
> > <http://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 <
> http://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
> > <http://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 <
> http://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
> >
> > <http://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 <http://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 <http://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
> > <http://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 <http://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 <http://a.id>) AS id
> > > FROM a) AS max_a_id ON b.a_id = max_a_id.id <http://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 <http://a.id> AS id
> > > FROM a, (SELECT min(a.id <http://a.id>) AS id
> > > FROM a) AS max_a_id
> > > WHERE a.id <http://a.id> = max_a_id.id <http://max_a_id.id>) AS
> > a_id_equal_to_max_a_id ON
> > > a_id_equal_to_max_a_id.id <http://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 <http://a.id> AS id
> > > FROM a, (SELECT min(a.id <http://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 <http://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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