On Apr 16, 2013, at 9:30 AM, Richard Gerd Kuesters <rich...@humantech.com.br>
wrote:
> Hello all!
>
> Ok, maybe people asks this a lot, but I wonder if it's possible to perform a
> query using an object as a filter - and I searched for it, didn't found
> anything close to my idea.
>
> Simple dumb example code:
>
>
> class User(Base):
>
> user_id = Column(Integer, Sequence(...), primary_key=True)
> username = Column(Unicode)
>
>
> class Subscription(Base):
>
> subscription_id = Column(Integer, Sequence(...), primary_key=True)
> name = Column(unicode)
> owner_id = Column(Integer, ForeignKey('user.user_id'), nullable=False)
>
> @hybrid_property
> def owner(self):
> if not object_session(self):
> return None
> return object_session(self).query(User).filter(self.owner_id ==
> User.user_id).first()
>
> @owner.setter
> def owner(self, owner):
> self.owner_id = owner.user_id if isinstance(owner, User) else owner
> if object_session(self):
> object_session(self).commit()
>
> # @owner.expression # ???
>
>
> # ok, so far *almost* good
OK, all of that complexity with hybrid_property is not needed at all here.
Just say, "owner = relationship("User")", and you're done. SQLAlchemy manages
one-to-many, many-to-one, and many-to-many automatically with relationship().
>
>
> # but, i would like to do _this_ instead
>
> print session.query(Subscription).filter(Subscription.owner == new_user).all()
>
>
> I've tried it in so many ways that I feel dizzy. The only way I think would
> be using @owner.expression to "return User", but that didn't the trick, it
> only appends "WHERE false" to the query, hehehe.
yeah just use relationship(), and you'd be using the first operator as
described right here in the ORM tutorial (which is a must-read):
http://docs.sqlalchemy.org/en/rel_0_8/orm/tutorial.html#common-relationship-operators
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