用listagg怎么替代这个写法

Posted 2023-05-01

参考技术A （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_salfrom emp
where deptno=10
结果如下：
技术分享
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：
技术分享
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：
技术分享
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：
技术分享
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,本回答被提问者采纳

Object.assign替代写法

export function ensureAppTimeouts(timeouts = {}) {
    return {
        ...DEFAULT_TIMEOUTS,
        ...timeouts
    }
    return Object.assign({},DEFAULT_TIMEOUTS,timeouts)
}