Hello.
I have already solved the issue by using subquery:
SELECT
t.id AS t_id,
t.rownum AS t_rownum
FROM (
SELECT
FROM
foo.id AS id,
row_number() OVER (ORDER BY foo.id) AS rownum
) AS t
WHERE rownum % 50 = 1
I have just tried your suggestion about using HAVING instead of WHERE, but that
fails with the same error. Thus a label cannot be used inside a query.
However, I am still curious whether the original WindowedRangeQuery recipe at
http://www.sqlalchemy.org/trac/wiki/UsageRecipes/WindowedRangeQuery
works or also has this error.
Thank you,
Ladislav Lenart
On 6.6.2013 18:26, Charlie Clark wrote:
> Am 06.06.2013, 16:30 Uhr, schrieb Ladislav Lenart <[email protected]>:
>
>> The returned query q produces the following SQL:
>> SELECT
>> foo.id AS foo_id,
>> row_number() OVER (ORDER BY foo.id) AS rownum
>> FROM foo
>> WHERE rownum % 2 = 1
>> When executed, it generates the following error:
>> sqlalchemy.exc.ProgrammingError:
>> (ProgrammingError) column "rownum" does not exist
>
> Just a guess but the error is coming from Postgres. You'll need to
> doublecheck the docs but I don't think "rownum" is available for the WHERE
> clause because OVER works like the aggregate functions. You can either try
> wrapping the SELECT in correlated subquery as the example shows or simply
> use HAVING rownum % 2 = 1 (I think this is best approach but I'm far from
> an expert). Try the SQL directly in the DB to see which works.
>
> Charlie
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