Thanks for your quick answer. However, I still don't understand why another relation is necessary.
> relation ItemC.b takes care of the assignment of table_c.b_id, but there > is nothing here for SQLA to know about table_c.a_id. table_c.a_id is *not* the problem - table_c.b_id is not assigned and that value has nothing to do with itemA at all. In my opinion the ForeignKeyConstraint on Table C contains all information SQLA should need in order to populate the primary key of C. The relation from B->A works essentially the same as C->B with the only difference that B has one autoincremented PK column and one assigned from A, while C has one autoinc column and two values assigned from B. In both cases, a FK constraint is present, specifying the relation between the tables. Looking at the FK constraints, SQLA should be able to determine the order in which the models are to be persisted (A->B- >C) and then assign all autoincremented values correctly. > You need another relation() on ItemC which expresses this, and you need to > assign itemA1 to > ItemC explicitly, independently of the ItemB stuff. This might work, but this doesn't make sense to me, because ItemC shouldn't really have to know about itemA because the PrimaryKey of C only depends on the primarykey of B. The fact, that the PK on B is in turn dependent on A should not be a factor. In fact, I modified the testcase and removed a_id from table_c altogether - The autoincremented id of itemB is still not set on the C- Model, so I'm still thinking this is a bug with nested relations - the fact that the primary key is composed of multiple columns does not seem to contribute to the problem. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~----------~----~----~----~------~----~------~--~---
