Re: COUNT(DISTINCT field) OVER (PARTITION BY another_field)

Auto-answer:

WITH uniques AS (
SELECT group_key, COUNT(DISTINCT reference) AS unique_references
FROM record_data
GROUP BY group_key
)
SELECT
group_key,
record_data.*,
uniques.unique_references
FROM record_data
JOIN uniques
USING (group_key)
ORDER BY
group_key;

Super-PostgreSQL is my hero...

2009/7/21 Daniel Cristian Cruz <danielcristian(at)gmail(dot)com>

> Hi list,
>
> Does someone knows why this is not a supported feature?
>
> I was happily SQL querying when I woke up from my dreams with this message:
>
> ERROR: DISTINCT is not implemented for window functions
> LINE 6: COUNT(DISTINCT handle) OVER (PARTITION BY pk_pessoas) AS ha...
>
> It could be very useful to find unique references within groups of clone
> records. Or there is another way to write this kind of query?
>
> Example (reference is a column of record_data):
>
> SELECT
> group_key,
> record_data.*,
> COUNT(DISTINCT reference) OVER (PARTITION BY group_key) AS
> unique_references
> FROM record_data
> ORDER BY
> group_key;
>
> Regards,
> --
> Daniel Cristian Cruz
> クルズ クリスチアン ダニエル
>

--
Daniel Cristian Cruz
クルズ クリスチアン ダニエル