On Mar 28, 2008, at 11:53 AM, Julien wrote:
>
> It doesn't work because more than one row are returned by the subquery
> used in the expression ...
thats not what "as_scalar()" means (not to be confused with
"scalar()"). it just means the select() object will be treated as
though it were a column, and will not export its FROM to the from
clause of the enclosing select (i.e., its "scalar" within the row).
like so:
from sqlalchemy import *
from sqlalchemy.sql import table, column
t = table('t', column('a'), column('b'))
s = select([t.c.a])
>>> print select([func.count(s)])
SELECT count((SELECT t.a
FROM t)) AS count_1
FROM (SELECT t.a AS a
FROM t)
>>> print select([func.count(s.as_scalar())])
SELECT count((SELECT t.a
FROM t)) AS count_1
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