Marcin Krol wrote:
> Hello,
>
>
> rsvs = session.query(Reservation).filter(Reservation.email ==
> em).filter(Reservation.newhosts == [] ).options(eagerload('newhosts')).all()
>
> sqlalchemy.exc.InvalidRequestError: Can't compare a collection to an
> object or collection; use contains() to test for membership.
>
> Well I can always do:
>
> rsvs = session.query(Reservation).filter(Reservation.email ==
> em).options(eagerload('newhosts')).all()
>
>
> for r in rsvs:
> if r.newhosts != []:
> print r.id, 'newhosts', r.newhosts
>
> or
>
> for r in rsvs:
> if r.newhosts == []:
> print r.id, 'newhosts', r.newhosts
>
>
> But, that's ugly like. So, the question is, is it possible to query for
> empty or non-empty collection?
>
> Regards,
> mk
>
If you really do need to peek inside non-empty collections like the
example above, then I think your eagerloading query works best.
Otherwise, you can use this (which generates a NOT EXISTS subquery) to
find reservations without any newhosts:
q = session.query(Reservation)
q = q.filter(Reservation.email == em)
q = q.filter(~Reservation.newhosts.any())
rsvs = q.all()
or this (explicit anti-join, assumes that Reservation.newhosts is a
relation to a Host class):
q = session.query(Reservation)
q = q.outerjoin(Reservation.newhosts)
q = q.filter(Reservation.email == em)
q = q.filter(Host.id == None)
rsvs = q.all()
or, if you need to include both empty and non-empty newhosts, with a
flag indicating which:
q = session.query(Reservation, Reservation.newhosts.any())
q = q.filter(Reservation.email == em)
rsvs_with_flag = q.all()
-Conor
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