Oracle SQL查询使用两个where子句获取差异值

Question

我正在尝试查询 avg(score1+score2/2) 与 'current' 和最新的 'archived' 之间的差异。在 Oracle Apex 上做一个图表

表名：我的表

id | score1 | score2 | status | date
------------------------------------------
1  |   10   |   20   |  current|  07/09/19
2  |   20   |   30   |archived| 04/09/19
3  |   15   |   35   |archived| 02/09/19

想要结果：
(avg(score1 + score2/2) where status = 'current') - (avg(score1 + score2/2) where status = 'archived' only the most recent)

我试过了

Answer 1

一种选择是使用

min/max(score1) keep (dense_rank first order by "date" desc) over (partition by status)

计算存档案例，以及当前案例的普通算术平均计算（根据样本数据，当前案例仅存在一行）

with myTable( id, score1, score2, status, "date" )as
(
 select 1, 10, 20, 'current' , date'2019-09-07' from dual union all
 select 2, 20, 30, 'archived', date'2019-09-04' from dual union all
 select 3, 15, 35, 'archived', date'2019-09-02' from dual 
), t as
(
 select
   case when status = 'current' then ( score1 + score2 ) / 2 end as curr,
   case when status = 'archived' then
   (
    (
     min(score1) keep (dense_rank first order by "date" desc) over (partition by status)+
     min(score2) keep (dense_rank first order by "date" desc) over (partition by status)
    )/2
   )
   end as arch
   from myTable
)
select max(curr)-max(arch) as "Avg.Result"
  from t;

Demo

Answer 2

你想要这个吗？

select status , avg(score1 + score2/2) from you_table
group by status

或

select (select  avg(score1 + score2/2) from you_table
where  status='current')-(select  avg(score1 + score2/2) from you_table
where  status='archived') diff from dual

Answer 3

嗯。 . .一种方法是条件聚合：

select max(case when status = 'current' then score_avg end), as current_score,
       max(case when status = 'archive' then score_avg end), as last_archive_score,
       (max(case when status = 'current' then score_avg end) -
        max(case when status = 'archive' then score_avg end)
       ) as diff
from (select t.*,
             row_number() over (partition by status order by date desc) as seqnum,
             (score1 + score2) / 2 as score_avg
      from t
     ) t
where seqnum = 1;

我猜你真的想要(score1 + score2) / 2)。但是，如果您想要 score1 + score2 / 2，请改用该表达式。