naktinis wrote: > I tried calling .subquery() method on each union subquery like this: > q1 = Thing.query().filter(...).order_by(Thing.a.desc()).limit > (1).subquery() > q2 = Thing.query().filter(...).order_by(Thing.a.asc()).limit > (1).subquery() > q = q1.union(q2).order_by(Thing.id)
I know you're not doing that since the alias object returned by subquery() does not have a union() method. you'll have to wrap each subquery in a SELECT like this: q1 = sess.query(C).order_by(C.data).limit(2).subquery().select() q2 = sess.query(C).order_by(C.data.desc()).limit(2).subquery().select() print sess.query(C).select_from(union(q1, q2)).order_by(C.data) --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
