it's a scalar subquery. you want to make the select() and then call as_scalar() on it so that it behaves like a column in a SQL expression.
On Jun 24, 2013, at 9:34 AM, Sebastian Elsner <sebast...@risefx.com> wrote: > Hello, > > I am trying to translate this SQL to a SQLAlchemy query, but failed so far: > > select `users`.`name`, `assignments`.`id`, > ( > select count(*) > from `assignments` > where `assignments`.`user_id` = `users`.`id` > ) as `num_assignments` > from `users` > join `assignments` > on `assignments`.`user_id` = `users`.`id` > > I would like to get results of (user_id, assignment_id, > total_assignments_per_user_id). I have found a similar question on the list > (https://groups.google.com/forum/#!topic/sqlalchemy/LBEyRe3w-8Q), and tried > to assemble a query like so, but I am missing something... > > It would also be nice if there were a faster way to do this. Maybe someone > has a good idea. > > And a question on terminology: is it really called subquery if the "subquery" > is in the "select block" (like above). > > Many thanks > > Sebastian > > -- > You received this message because you are subscribed to the Google Groups > "sqlalchemy" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to sqlalchemy+unsubscr...@googlegroups.com. > To post to this group, send email to sqlalchemy@googlegroups.com. > Visit this group at http://groups.google.com/group/sqlalchemy. > For more options, visit https://groups.google.com/groups/opt_out. > > -- You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/groups/opt_out.
