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질문 & 답변
카테고리
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세부 분야
데이터베이스
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해결 여부
미해결
구문 정리
22.11.24 15:29 작성 조회수 106
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with a as (select emp.*, row_number () over (partition by deptno order by sal desc) as high_sal
, row_number () over (partition by deptno order by sal) as low_sal
from emp),
b as (select *
from a where high_sal = 1 or low_sal = 1),
c as (select distinct a.*, case when a.high_sal=1 then '최고'
when a.low_sal =1 then '최저'
end as 급여수준
from a join b on a.deptno = b.deptno
where a.high_sal = 1 or a.low_sal =1),
d as (select deptno , max(sal) as max_sal, min(sal) as min_sal
from c group by deptno)
select c.*, d.max_sal - d.min_sal
from c join d on c.deptno = d.deptno
order by deptno
이 구문을 좀 더 깔끔하게 정리하고 싶은데 혹시 필요없는 절이라던가 생략 가능한 부분이 있을까요?
답변을 작성해보세요.
0
권 철민
지식공유자2022.11.24
아래 SQL은 어떤신지요?
with a as
(
select emp.*, row_number () over (partition by deptno order by sal desc) as high_sal
, row_number () over (partition by deptno order by sal) as low_sal
from emp
),
b as (
select deptno, max(sal) as max_sal_dept, min(sal) as min_sal_dept
from emp group by deptno
)
select a.*
, case when a.high_sal=1 then '최고'
when a.low_sal =1 then '최저' end as 급여수준
, (max_sal_dept - min_sal_dept) as sal_diff
from a join b on a.deptno = b.deptno
where a.high_sal = 1 or a.low_sal=1;
답변 1