oh sorry, there is a new flag that I have not documented yet called
"correlated", you need to set that to false (although it shouldnt be
needed for this example, thats slightly buggy):
subq = users.select(order_by=[users.c.username], limit=2,
correlate=False).alias('users')
s = select(
[subq.c.username, roles.c.role],
from_obj = [
subq.join(roles, subq.c.username==roles.c.username)
]
)
the query I get is:
SELECT users.username, roles.role
FROM (SELECT users.username AS username
FROM users ORDER BY users.username
LIMIT 2) AS users JOIN roles ON users.username = roles.username
On Mar 17, 2006, at 1:54 AM, sifu wrote:
hi!
On 3/16/06, Michael Bayer <[EMAIL PROTECTED]> wrote:
I need to document the "from_obj" parameter better:
subq = users.select(order_by=[users.c.username], limit=2).alias
('users')
select(
[subq.c.username, roles.c.role],
from_obj = [
subq.join(roles, subq.c.username==roles.c.username)
]
)
i tried this already once before, but postgres isn't happy with the
resulting query:
SQLError: (ProgrammingError) missing FROM-clause entry in subquery for
table "users"
here is the query sqlalchemy generates:
SELECT users.username, roles.role
FROM (SELECT users.username AS username, users.password AS password,
users.email AS email ORDER BY users.username
LIMIT 2) AS users JOIN roles ON users.username = roles.username
thank you,
sifu
On Mar 16, 2006, at 4:43 AM, sifu wrote:
hi!
i'm not able to come up with a way to express the following
postgresql
query with sqlalchemy:
SELECT users.username, roles.role
FROM ( SELECT * FROM users ORDER BY username LIMIT 2 ) AS users
LEFT JOIN roles ON users.username = roles.username;
i hope someone can help me out here :)
what i want to achieve (and what the postgresql query does):
i have to tables one with users one with all the roles of an user.
and i want to get 2 users but with all the roles they have.
thanks in advance!
sifu
ps: sqlalchemy is awesome! the only limiting factor seems to be my
stupidity :)
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