Hi Michael,
Do you take a break sometimes during the week-end? ;-)
Well, this seems quite fine, and a very quick answer as usual!!!
I also agree that this is a "bad" model, but data is provided by an
external partner and I can't update it :-(
I was just sure that SA was able to handle it anyway :-)
I'll try to test this quickly, but not before monday morning in
fact when being back at work :-/
Many thanks,
Thierry
Le Sat, 16 Jun 2012 11:41:29 -0400, Michael Bayer
<mike...@zzzcomputing.com> a �crit:
>
> It's not a basic question at all as this is a rare edge case, and
> it's not a "foreign key" by definition. You need to relate the two
> tables based on a SQL function, in this case a concatenation. In
> some cases this can be tricky, and there's improvements in 0.8 to
> address that, though in this case it seems to work without too much
> difficulty:
>
> from sqlalchemy import *
> from sqlalchemy.orm import *
> from sqlalchemy.ext.declarative import declarative_base
>
> Base= declarative_base()
>
> class A(Base):
> __tablename__ = "a"
>
> id1 = Column(String, primary_key=True)
> id2 = Column(String, primary_key=True)
> bs = relationship("B",
> primaryjoin="B.a_id == A.id1 + A.id2",
> foreign_keys="B.a_id",
> viewonly=True)
>
> class B(Base):
> __tablename__ = "b"
> id = Column(Integer, primary_key=True)
> a_id = Column(String)
>
> e = create_engine("sqlite://", echo=True)
> Base.metadata.create_all(e)
> s = Session(e)
>
> s.add_all([
> A(id1="x", id2="y", bs=[
> B(a_id="xy"),
> B(a_id="xy")
> ]),
> A(id1="q", id2="p", bs=[
> B(a_id="qp")
> ])
> ])
>
> s.commit()
>
> print s.query(A).first().bs
>
> for a in s.query(A).options(joinedload(A.bs)):
> print a.bs
>
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