assuming you have a relation() from Email to HWRep: session .query(Email).filter(~Email.hwreps.any()).filter(~Email.projects.any())
On May 19, 2009, at 12:21 PM, Marcin Krol wrote: > > Hello everyone, > > I have this and it works: > > hwrepemails = session.query(Email).join(HWRep).all() > > hwrepemailids = [ e.id for e in hwrepemails ] > > unassociatedusers = session.query(Email).filter(Email.projects == > None > ).filter > (not_(Email.id.in_(hwrepemailids))).order_by(Email.email).all() > > The idea is to get all emails that have no projects, but to also > exclude > those emails that are associated with HWRep instances. > > Since this way of doing it requires two queries, I was wondering if > there is a way to do it in a single query? > > Nothing comes to my mind in this regard, or I wouldn't ask here... > > Regards, > mk > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
