Just to be painfully complete the solution was:
subquery = select([Taggable.c.id], query._criterion,
from_obj= query._from_obj).as_scalar()
names.select (exists ([names.c.id],
and_(names.c.id == obj_tags.c.name_id,
obj_tags.c.parent_id==Taggable.c.id ,
Taggable.c.id.in_ ( subquery) ))
which works in 0.4.1, but not in 0.3.11
Thanks for the help!
Kris
On Nov 27, 8:57 pm, Michael Bayer <[EMAIL PROTECTED]> wrote:
> On Nov 27, 2007, at 11:51 PM, kris wrote:
>
>
>
> > I tried
>
> > names.select (exists ([names.c.id],
> > and_(names.c.id == obj_tags.c.name_id,
> > obj_tags.c.parent_id==Taggable.c.id ,
> > Taggable.c.id == query.compile() )
> > however, I am getting too many fields back with the error
> > "subquery must return only one column"
>
> > Any suggestion on how to ask for only the id column?
>
> oh, well yeah now it will get slightly more tricky. you can pull out
> the "WHERE" criterion by itself using query._criterion, but that wont
> give you the joins anything else, i.e.
>
> select([mytable.c.somecolumn], query._criterion).as_scalar()
>
> if you have joins in there as well, you can get them like:
>
> select([mytable.c.somecolumn], query._criterion,
> from_obj=query._from_obj).as_scalar()
>
> im not sure at the moment how/if this particular kind of operation
> would be a regular query method....maybe query.generate_select() or
> something.. also as_scalar() wraps the select in a "scalar-holding"
> object that makes it act like a single column in expressions. its not
> entirely needed but can be helpful.
--~--~---------~--~----~------------~-------~--~----~
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
-~----------~----~----~----~------~----~------~--~---
