I read part of a book called "Getting Things Done", which made the point, "if
it takes less than two minutes, do it now". So I respond to these really fast
strictly just to keep my inbox empty. Good luck !
On Jun 16, 2012, at 11:55 AM, Thierry Florac wrote:
>
> 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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