Question

Oracle Beginner在这里尝试从每家商店返回收入最高的员工。是否可以通过执行INNER JOIN ON MAX / SAL+COMM表达式和第二列（store_id）来返回此结果，其格式类似于下面的第一个查询（两个 - 列连接）？替代解决方案返回我想要的结果，但仅基于总薪资，允许在更大的数据集中产生无关的结果。

ENAME   SAL     COMM    STORE_ID
KING    5000    -       3
BLAKE   2850    -       3
CLARK   2450    -       3
JONES   2975    -       1
SCOTT   3000    400     4
FORD    3000    0       4
SMITH   800     700     4

SELECT store_id, ename, (salary+NVL(commission,0)) total_pay 
FROM employees E1
INNER JOIN
(SELECT store_id, MAX(sal+NVL(comm,0)) highest_paid
FROM employees
GROUP BY store_id) E2
ON E1.store_id = E2.store_id
AND total_pay = highest_paid

替代解决方案

SELECT * 
FROM EMPLOYEES
WHERE (SAL+NVL(COMM,0)) IN (SELECT MAX(SAL+NVL(COMM,0))
                            FROM EMPLOYEES
                            GROUP BY store_id)

Answer 1

我不确定你在寻找什么，但表达这种逻辑的一种非常自然的方式是使用窗口函数：

SELECT storeid, empname, agg
FROM (SELECT e.*, (salary+ COALESCE(commission, 0)) AS agg,
             MAX(salary + COALESCE(commission, 0)) OVER (PARTITION BY storeid) as max_agg 
      FROM table.employees e
     ) e
WHERE agg = max_agg;

使用列和聚合结果集来连接两个查询

1 个答案: