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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