On 19 Mar, 15:09, Martin <mar...@chinasoftinc.com> wrote:
> I guess
>
> q = session.query(A).join('b').order_by(B.code).all()
>
> should do the trick, at least it did under 0.4.8It 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 sqlalchemy@googlegroups.com To unsubscribe from this group, send email to sqlalchemy+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~----------~----~----~----~------~----~------~--~---
