SQL12 获取每个部门中当前员工薪水最高的相关信息

描述

有一个员工表dept_emp简况如下:

有一个薪水表salaries简况如下:

获取每个部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary，按照部门编号dept_no升序排列，以上例子输出如下:

（注意: Mysql与Sqlite select 非聚合列的结果可能不一样)

示例1

输入：

drop table if exists  `dept_emp` ; 
drop table if exists  `salaries` ; 
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01');
INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01');
INSERT INTO dept_emp VALUES(10003,'d002','1996-08-03','9999-01-01');

INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');
INSERT INTO salaries VALUES(10003,92527,'2001-08-02','9999-01-01');

复制输出：

d001|10001|88958
d002|10003|92527

select dept_no,s.emp_no,salary maxSalary
from dept_emp de join salaries s on de.emp_no=s.emp_no 
where (dept_no,salary) in (select dept_no,max(salary)
                              from dept_emp de join salaries s 
               on de.emp_no=s.emp_no 
               where s.to_date='9999-01-01' and de.to_date='9999-01-01' 
               group by dept_no)
order by dept_no