Yes, I got the style from there.
I have a great great great great grandchild that I need to be able to
access by the great great great great grandparent id. How would you
recommend doing that, then? I don't want to have to write f_instances =
F.query.join(F.e).join(E.d).join(D.c).join(C.b).join(B.a).filter(A.id ==
1).all() just to find the F's that have A.id 1.
On Tuesday, March 24, 2015 at 6:28:31 AM UTC-7, Michael Bayer wrote:
>
>
>
> Andrew Millspaugh <millspau...@gmail.com <javascript:>> wrote:
>
> > I've got a class hierarchy that looks something like this:
> >
> > [ A ] 1--------* [ B ] 1---------* [ C ] 1----------* [ D ]
> 1--------0..1 [ E ] 1..*----------0..1 [ F ]
> >
> > org proj ticket snap
> bidlimit ticketset
> >
> > And I'm trying to add a relationship from A to F with a backref. The
> relationship definition (on the A model) looks like:
> >
> > f = DB.relationship('F',
> > secondary=(
> > 'join(F, E, F.id == E.f_id)'
> > '.join(D, E.d_id == D.id)'
> > '.join(C, D.c_id == C.id)'
> > '.join(B, C.b_id == B.id)'
> > ),
> > primaryjoin='A.id == B.a_id',
> > secondaryjoin='E.f_id == F.id',
> > backref=DB.backref('a', uselist=False), viewonly=True
> > )
> >
> > Now, if I query A.f, I get all of the F's, instead of just the ones
> which have a relationship with A. I'm sure I'm missing something simple,
> but I can't seem to find it... Any help out there?
>
> what does your SQL output say? Is this query(A).join(A.f) or the “f”
> attribute on an existing “A”? I’m assuming you got this style from
>
> http://docs.sqlalchemy.org/en/rel_0_9/orm/join_conditions.html#composite-secondary-joins,
>
>
> note that its experimental and not well supported. In most cases you
> should
> deal with individual relationships between each class.
>
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