我只是很好奇,在Postgres中有办法做到这一点吗?
SUM(DISTINCT column_a) OVER(PARTITION BY column_b, column_c)
使用DISTINCT会导致错误:窗口功能未实现DISTINCT
答案 0 :(得分:0)
这应该可以解决问题:
SELECT column_a,
column_b,
column_c,
sum(column_a) FILTER (WHERE is_new) OVER w
FROM (SELECT column_a,
column_b,
column_c,
column_a IS DISTINCT FROM lag(column_a) OVER w AS is_new
FROM atable
WINDOW w AS (PARTITION BY column_b, column_c ORDER BY column_a)
) AS q
WINDOW w AS (PARTITION BY column_b, column_c ORDER BY column_a);
在内部查询中,column_a
的所有重复项都将得到is_new = FALSE
,因此这些重复项不在外部查询中计算。