Sweet, it's working. :) Now let's wait for the 0.8.
2011/5/26 Michael Bayer <mike...@zzzcomputing.com> > > On May 26, 2011, at 5:03 PM, Israel Ben Guilherme Fonseca wrote: > > I saw that, but I'm not using a subquery in the from clause. Maybe I wasn't > clear enough. Example: > > outeruser = aliased(User) > inneruser = aliased(User) > > innerselect = session.query(inneruser.id).filter(inneruser.id == > outeruser.id).subquery() > > At this point I already have a problem, the generated from clause is > something like: > > from user as user_2, user as user_1 > > I didnt want the other user_2, because the filter statement is actually > referencing the user of the outerselect: > > outerselect = session.query(outeruser).filter(outeruser.id == innerselect) > > I expected that the innerselect referenced the id of the outer select. > > > oh then you're looking for correlation: > > innerselect = session.query(inneruser.id).filter(inneruser.id == > outeruser.id).correlate(outeruser) > > outerselect = session.query(outeruser).filter(outeruser.id == > innerselect.as_scalar()) > > for some reason the Query is disabling auto-correlation upon subquery(), > .statement or as_scalar(). Sort of wish I had noticed that before > releasing 0.7. Will add a ticket to possibly change that default for 0.8, > see you in a year. > > > -- > 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. > -- 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.
