SQL12 获取每个部门中当前员工薪水最高的相关信息

有一个员工表dept_emp简况如下:

emp_no	dept_no	from_date	to_date
10001	d001	1986-06-26	9999-01-01
10002	d001	1996-08-03	9999-01-01
10003	d002	1996-08-03	9999-01-01

有一个薪水表salaries简况如下:

emp_no	salary	from_date	to_date
10001	88958	2002-06-22	9999-01-01
10002	72527	2001-08-02	9999-01-01
10003	92527	2001-08-02	9999-01-01

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:

dept_no	emp_no	maxSalary
d001	10001	88958
d002	10003	92527

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');

输出：


d001|10001|88958
d002|10003|92527

SQL:

select d1.dept_no,d1.emp_no,s.salary
from dept_emp d1 join salaries s on d1.emp_no = s.emp_no
where 1 > (select count(*) from dept_emp d2 join salaries s2 on d2.emp_no = s2.emp_no where s2.salary > s.salary and d1.dept_no = d2.dept_no)
ORDER BY d1.dept_no