It's more of a convenience thing. I want a nice way to be able to do
something like:
A.fs
or
F.a
For example, if a user is only allowed to modify an F object if they have a
key for the corresponding a object, I'd like to write something like:
if current_user.authorized(F.a):
# do something
On the other side, I'd like to be able to get the list of all fs on an A
without having to do a query. I want to be able to write something like:
for f_instance in A.fs:
# do something
If you think the best way to do this is to simply switch to using queries,
I'm fine with that. I just felt that it was awkward. It will change my
procedures to something like:
if current_user.authorized(F.query.join(*F.join_to_a()).one()):
# do something
for f_instance in F.query.join(F.a).all():
# do something
I'm pretty inexperienced with SQLAlchemy. I mostly want to know the best
way to deal with a relationship like this. I am trying to avoid adding a
fake relationship directly between F and A, as it could get out of sync
with the actual nested relationship. In general, if I have a sqlalchemy
class for which I know the instances will always be uniquely identified by
a (great)^n grandparent, how should I be accessing the grandparents from
the grandchild and vice versa?
On Tuesday, March 24, 2015 at 1:12:24 AM UTC-7, Andrew Millspaugh 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?
>
>
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