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,) > > > > -- > > You received this message because you are subscribed to the Google > > Groups "sqlalchemy" group. > > To unsubscribe from this group and stop receiving emails from it, send > > an email to sqlalchemy+...@googlegroups.com <javascript:> > > <mailto:sqlalchemy+unsubscr...@googlegroups.com <javascript:>>. > > To post to this group, send email to sqlal...@googlegroups.com > <javascript:> > > <mailto:sqlal...@googlegroups.com <javascript:>>. > > Visit this group at https://groups.google.com/group/sqlalchemy. > > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Groups "sqlalchemy" group. 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