当 else with partition by 在 redshift 查询中不起作用时

Question

我想从 TAG_SALES_by_month 中排除标签的类别 sub_tag1、sub_tag2 和 sub_tag3，但其余的无论我在 where 条件中提到的是什么计入计数。我无法达到预期的效果 result.can 谁能帮助我达到同样的效果，我将不胜感激。

select o.tag, 

 o.SOME, o.THING, o.ILIKE, o.date, c.THE, c.MOST,
date_part(month, o.date) as Month,
date_part(day, o.date) as day,
count(o.id) over (partition by day, CUST_Id) as SALE_NO,
count(o.id) over (partition by Month, CUST_Id) as SALE_NO_by_month,
count(case when (tag <> 'sub_tag1' AND tag <> 'sub_tag2' AND tag <> 'sub_tag3')  then o.id else 0 END) over (partition by Month, CUST_Id) as TAG_SALES_by_month,
c.id as CUST_Id

from order_info o
left join config c on o.SOME = c.SOME

where date >= '05/01/2021' AND tag in ('sub_tag1', 'sub_tag2', 'sub_tag3', 'sub_tag4', 'sub_tag5',
 'sub_tag6') AND ILIKE = 'JACK'

 group by o.tag, o.SOME, o.THING, o.ILIKE, o.date, c.THE, c.MOST, CUST_Id, o.id
 order by date

Answer 1

根据评论，这里的问题是 COUNT 将 return 1 用于任何值，它计算 存在与不存在 value/row.

因此 COUNT(CASE WHEN... ELSE 0...) 仍会在 ELSE 条件下计数 1，因为 0 是一个存在的值。

解决方案是使用ELSE NULL或省略ELSE子句，默认为NULL，因为NULL不会被计算在内。

当 else with partition by 在 redshift 查询中不起作用时

When else with partition by isn't working in redshift queries

sql

amazon-redshift

partition