一哥们出条sql题给我玩,将下面sql改成不使用keep分析函数的写法。
select deptno,
ename,
sal,
hiredate,
min(sal) keep(dense_rank first order by hiredate) over(partition by deptno) min_sal,
max(sal) keep(dense_rank last order by hiredate) over(partition by deptno) max_sal
from emp;
我一开始改错了,被这哥们喷菜鸡,我草。
-- 错误等价改写,逻辑不等价
with x as (
select e1.deptno,
e1.ename,
e1.sal,
e1.hiredate,
row_number() over (partition by DEPTNO order by HIREDATE) rn_first,
row_number() over (partition by DEPTNO order by HIREDATE DESC) rn_last
from EMP e1)
select
e.deptno,
e.ename,
e.sal,
e.hiredate,
x1.SAL,
x2.SAL
from emp e
inner join x x1 on e.DEPTNO = x1.DEPTNO and x1.rn_first = 1
inner join x x2 on e.DEPTNO = x2.DEPTNO and x2.rn_last = 1;
我换了张表测试下,发现上面改写是逻辑有问题,如果同一个组内有相同日期的 ,分组字段内有NULL值的,确实会导致SQL结果集不一致。
-- 将EMP表替换成EMPLOYEES,如果使用上面等价改写就错误了。
select DEPARTMENT_ID,
FIRST_NAME,
SALARY,
HIRE_DATE,
min(SALARY) keep(dense_rank first order by HIRE_DATE) over(partition by DEPARTMENT_ID) min_sal,
max(SALARY) keep(dense_rank last order by HIRE_DATE) over(partition by DEPARTMENT_ID) max_sal
from EMPLOYEES;
最终等价改写的SQL,增加了分组字段内有NULL值的逻辑 ,和处理一个组内有相同日期的逻辑。
select e.DEPARTMENT_ID,
e.FIRST_NAME,
e.SALARY,
e.HIRE_DATE,
(select MIN_SALARY
from (select DEPARTMENT_ID, MIN(SALARY) MIN_SALARY
from (select DEPARTMENT_ID,
SALARY,
HIRE_DATE,
dense_rank() over (PARTITION BY DEPARTMENT_ID ORDER BY HIRE_DATE) RN
from EMPLOYEES)
WHERE RN = 1
GROUP BY DEPARTMENT_ID) e1
where case when e1.DEPARTMENT_ID is null then 99999 else e1.DEPARTMENT_ID end = case when e.DEPARTMENT_ID is null then 99999 else e.DEPARTMENT_ID end) a_min,
(select MAX_SALARY
from (select DEPARTMENT_ID, MAX(SALARY) MAX_SALARY
from (select DEPARTMENT_ID,
SALARY,
HIRE_DATE,
dense_rank() over (PARTITION BY DEPARTMENT_ID ORDER BY HIRE_DATE DESC) RN
from EMPLOYEES)
WHERE RN = 1
GROUP BY DEPARTMENT_ID) e1
where case when e1.DEPARTMENT_ID is null then 99999 else e1.DEPARTMENT_ID end = case when e.DEPARTMENT_ID is null then 99999 else e.DEPARTMENT_ID end ) a_max
FROM EMPLOYEES e;
差集比较后是等价的: