Alex Brasetvik escribió: > On Dec 9, 2009, at 12:41 , Antonio Beamud Montero wrote: > > >> s.intersect(select([tags.c.channel], tags.c.tag == tag)) >> > > This is returning a new select which is the intersection --- which you are > throwing away. It is *not* modifying s in-place. > > You probably want to do > > s = s.intersect(select([tags.c.channel], tags.c.tag == tag)) > Ops.. thank you... I've solved it using this approach:
lsel = [] for t in ltags: lsel.append(select(.....)) xsec = intersect(*lsel) Now it works perfect. Greetings. > -- > Alex Brasetvik > > -- > > You received this message because you are subscribed to the Google Groups > "sqlalchemy" group. > To post to this group, send email to sqlalch...@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 sqlalch...@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.