On 19 Mar, 15:09, Martin <[email protected]> wrote: > I guess > > q = session.query(A).join('b').order_by(B.code).all() > > should do the trick, at least it did under 0.4.8
It works in 0.5.2 too, thanks :) I was wondering though, maybe 0.5.x has a way to do it without the join? I mean, SQLA should have all the informations to figure it out by itself, I think..? --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
