【Oracle】oracle取最大值和最小值的几个方法汇总

(1)oracle使用keep分析函数取最值记录

-- 取工资sal最大的雇员姓名及其工资,以及工资sal最少的雇员姓名及其工资
select
deptno,
empno,
ename,
sal,
max(ename) keep(dense_rank FIRST order by sal) over (partition by deptno) as min_sal_man,
max(sal) keep(dense_rank FIRST order by sal) over (partition by deptno) as min_sal,
max(ename) keep(dense_rank LAST order by sal) over (partition by deptno) as max_sal_man,
max(sal) keep(dense_rank LAST order by sal) over (partition by deptno) as max_sal
from emp
where deptno=10

结果如下:

【Oracle】oracle取最大值和最小值的几个方法汇总

从语句中可以看到,ename和sal都是用的max(),这样做的目的是为了去除由于keep()函数得到的有重复值的数据结果集。这样用有一个弊端,加入部门20有两个相同的最大SAL的人,部门30有两个相同的最小SAL的人,如果按照这种方法取出来的数据,就不一定准确了,重复的人会被去除掉。

我们用下面的语句来修改一下:

select
deptno,
empno,
ename,
sal,
max(ename) keep(dense_rank FIRST order by sal) over (partition by deptno) as min_sal_man,
max(sal) keep(dense_rank FIRST order by sal) over (partition by deptno) as min_sal,
max(ename) keep(dense_rank LAST order by sal) over (partition by deptno) as max_sal_man,
max(sal) keep(dense_rank LAST order by sal) over (partition by deptno) as max_sal,
wmsys.wm_concat(ename) keep(dense_rank LAST order by sal) over (partition by deptno) as 工资最高的人,
wmsys.wm_concat(ename) keep(dense_rank FIRST order by sal) over (partition by deptno) as 工资最低的人
from emp
where deptno=20
order by 1, 2 ;

我们新增了两个列:工资最高的人,工资最低的人。执行看一下结果:

【Oracle】oracle取最大值和最小值的几个方法汇总

可以看到,deptno=20时,SCOTT和FORD两个人的工资SAL都是3000,如果用MAX()就只能取出其中一个人的姓名,显然是不对的。

然后,我们再来看一下deptno=30时的情况:

select
deptno,
empno,
ename,
sal,
max(ename) keep(dense_rank FIRST order by sal) over (partition by deptno) as min_sal_man,
max(sal) keep(dense_rank FIRST order by sal) over (partition by deptno) as min_sal,
max(ename) keep(dense_rank LAST order by sal) over (partition by deptno) as max_sal_man,
max(sal) keep(dense_rank LAST order by sal) over (partition by deptno) as max_sal,
wmsys.wm_concat(ename) keep(dense_rank LAST order by sal) over (partition by deptno) as 工资最高的人,
wmsys.wm_concat(ename) keep(dense_rank FIRST order by sal) over (partition by deptno) as 工资最低的人
from emp
where deptno=30
order by 1, 2 ;

deptno=30时的结果如下:

【Oracle】oracle取最大值和最小值的几个方法汇总

可以看到,deptno=30时,WARD和MARTIN两人的工资最小且均为1250,如果用MAX()的方式,就只能取出其中一个人的名称。

这就是因为keep()取出来的数据集是包含多个数据结果的,所以,在语句中使用了wmsys.wm_concat()函数,该函数的作用是以逗号分隔连接列的值。

注:wm_concat()的功能有点儿类似分析函数listagg() within group() 。

(2)使用SQL子查询和聚合函数,查询出最大值和最小值

-- 使用子查询查询出最大值和最小值
select * from
(
select
deptno,
listagg(ename,',') within group (order by deptno) as dept_max_ename,
max(sal) as dept_max_sal
from emp
where (deptno,sal) in (select deptno, max(sal) as max_sal from emp group by deptno)
group by deptno
) A
inner join
(
select
deptno,
listagg(ename,',') within group (order by deptno) as dept_min_ename,
min(sal) as dept_min_sal
from emp
where (deptno,sal) in (select deptno, min(sal) as min_sal from emp group by deptno)
group by deptno
) B
on A.deptno = B.deptno

结果如下:

【Oracle】oracle取最大值和最小值的几个方法汇总

在这个方案里面,还使用了listagg()分析函数将最值有重复姓名的人合并在一起,用wm_concat()函数替代listagg()也可以

wm_concat(ename) as dept_max_ename,
wm_concat(ename) as dept_min_ename,

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