pardon my sql-ignorancy, but cant u express this in just one
expression? it should be possible, it is a graph/set arithmetics
after all...
mmh, (could be very wrong!) something like
- get all rows that has some b_id from the looked list
- group(?) somehow by a_id, and then finger the a_id
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
Cool. I wasn't sure if it was ready for filter(A.bs == list_of_bs).
When I tried to do that before, I must have let some silly syntax
error keep me from realizing that it was a workable construction.
Thanks!
On Mar 5, 8:20 am, Michael Bayer [EMAIL PROTECTED] wrote:
On Mar 5, 2008, at 10:50
