Hm.... I'm lost. This is where I am now:
session.query(User.id, Assignment.id,
session.query(func.count(Assignment.id)).filter(Assignment.user_id ==
User.id).as_scalar()).join(Assignment).all()
On 06/24/2013 04:26 PM, Michael Bayer wrote:
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
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