Now what about the fact that ([1,2,3] == [2,3,1]) is False in Python? Is it the same for the == in the filter() shown above? I want this query to succeed regardless of order. Do I have to keep my list_of_bs in a known order, both in the collection and in my filter list, for it to work? It just so happens that I do have a criterion by which I *could* order them consistently, but I wasn't planning to have to do that unless necessary.
On Mar 5, 8:20 am, Michael Bayer <[EMAIL PROTECTED]> wrote: > On Mar 5, 2008, at 10:50 AM, Eric Ongerth wrote: > > > Anyway -- so what would really clean it all up would be: > > > session.query(A).filter(A.bs.contains(list_of_bs_being_sought)).all(). > > > THAT would do exactly what I'm trying to accomplish. But it would > > require contains() to accept a list and know what to do with it. My > > proposal would be that the expected behavior is for contains() to > > construct an intersect of selects where each select is like the one it > > creates in its simpler case where the argument to contains() is a > > scalar instead of a list. Does that make sense? > > Well i think we'd call the operator intersect(). However I think this > will do what you want right now if you were to say: > > session.query(A).filter(A.bs == list_of_bs).all() > > since it will generate individual EXISTS predicates for each element > in the list. --~--~---------~--~----~------------~-------~--~----~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~----------~----~----~----~------~----~------~--~---
