Andrey wrote:
> select id, first_value(id) over(order by id), last_value(id) over(order by
> id) from t;
>
> RESULT:
> id | first_value | last_value
> ----+-------------+------------
> 1 | 1 | 1
> 2 | 1 | 2
> 3 | 1 | 3
> (3 rows)
>
> fist_value - good, last_value - bad
Looks ok to me. What did you expect?
--
Heikki Linnakangas
EnterpriseDB http://www.enterprisedb.com