Question

我使用窗口函数创建了一个名为custom_rank的列。有没有办法获得每个用户ID的特定列的最小值和最大值？我基本上试图在他们的注册期内获得给定用户的最小和最大权重。

select top 1000 
  c.weight, c.units, 
  rank() over (partition by uuid, enrolled_on order by enrolled_on desc, input_date asc) as "weight_rank"
  from tableA a
join tableB b
  on (b.member_no = a.member_no)
join tableC c
  on (c.userId = b.uuid)
where input_date >= enrolled_on and input_date < cancel_date

Answer 1

使用CTE，然后获得每个用户的最大/最小值。

with mm as
(
    select top 1000 
      c.weight, c.units, c.userId
      rank() over (partition by uuid, enrolled_on order by enrolled_on desc, input_date asc) as "weight_rank"
      from tableA a
    join tableB b
      on (b.member_no = a.member_no)
    join tableC c
      on (c.userId = b.uuid)
    where input_date >= enrolled_on and input_date < cancel_date
)
select   userId, min(weight_rank), max(weight_rank)
from     mm
group by userId;

Answer 2

只需使用max()和min()作为窗口函数：

with cte as (
      select c.weight, c.units, uuid,
             rank() over (partition by uuid, enrolled_on
                          order by enrolled_on desc, input_date asc) as weight_rank
      from tableA a join
           tableB b
           on b.member_no = a.member_no join
           tableC c
           on c.userId = b.uuid
      where input_date >= enrolled_on and input_date < cancel_date
     )
select top 1000 cte.*,
       max(weighted_rank) over (partition by uuid) as max_weightedrank_user,
       min(weighted_rank) over (partition by uuid) as min_weightedrank_user,
from cte;

使用窗口函数创建的列上的MIN（）MAX（）

2 个答案: