首先给出本文所需的表格：

emp表：

mysql> select * from emp;
+-------+--------+-----------+------+------------+---------+---------+--------+
| EMPNO | ENAME  | JOB       | MGR  | HIREDATE   | SAL     | COMM    | DEPTNO |
+-------+--------+-----------+------+------------+---------+---------+--------+
|  7369 | SMITH  | CLERK     | 7902 | 1980-12-17 |  880.00 |    NULL |     20 |
|  7499 | ALLEN  | SALESMAN  | 7698 | 1981-02-20 | 1760.00 |  300.00 |     30 |
|  7521 | WARD   | SALESMAN  | 7698 | 1981-02-22 | 1375.00 |  500.00 |     30 |
|  7566 | JONES  | MANAGER   | 7839 | 1981-04-02 | 2975.00 |    NULL |     20 |
|  7654 | MARTIN | SALESMAN  | 7698 | 1981-09-28 | 1250.00 | 1400.00 |     30 |
|  7698 | BLAKE  | MANAGER   | 7839 | 1981-05-01 | 2850.00 |    NULL |     30 |
|  7782 | CLARK  | MANAGER   | 7839 | 1981-06-09 | 2450.00 |    NULL |     10 |
|  7788 | SCOTT  | ANALYST   | 7566 | 1987-04-19 | 3000.00 |    NULL |     20 |
|  7839 | KING   | PRESIDENT | NULL | 1981-11-17 | 5000.00 |    NULL |     10 |
|  7844 | TURNER | SALESMAN  | 7698 | 1981-09-08 | 1500.00 |    0.00 |     30 |
|  7876 | ADAMS  | CLERK     | 7788 | 1987-05-23 | 1100.00 |    NULL |     20 |
|  7900 | JAMES  | CLERK     | 7698 | 1981-12-03 |  950.00 |    NULL |     30 |
|  7902 | FORD   | ANALYST   | 7566 | 1981-12-03 | 3000.00 |    NULL |     20 |
|  7934 | MILLER | CLERK     | 7782 | 1982-01-23 | 1300.00 |    NULL |     10 |
+-------+--------+-----------+------+------------+---------+---------+--------+
14 rows in set (0.00 sec)

dept表：

mysql> select * from dept;
+--------+------------+----------+
| DEPTNO | DNAME      | LOC      |
+--------+------------+----------+
|     10 | ACCOUNTING | NEW YORK |
|     20 | RESEARCH   | DALLAS   |
|     30 | SALES      | CHICAGO  |
|     40 | OPERATIONS | BOSTON   |
+--------+------------+----------+
4 rows in set (0.00 sec)

方法一：建立两个左连接

#第一步：获得每个部门的最高薪水

mysql> 
select 
    deptno,max(sal) maxsal 
from 
    emp 
group by 
    deptno;


+--------+---------+

| deptno | maxsal |

+--------+---------+

| 20     | 3000.00 |

| 30     | 2850.00 |

| 10     | 5000.00 |

+--------+---------+


#第二步建立两个左连接表

mysql> 
select 
    d.dname,t.deptno,e.ename,t.maxsal 
from 
    emp e 
left join 
    (select deptno,max(sal) maxsal from emp group by deptno) t 
on 
    t.deptno = e.deptno and t.maxsal=e.sal 
left join 
    dept as d 
on 
    e.deptno = d.deptno;

+------------+--------+-------+---------+
| dname      | deptno | ename | maxsal  |
+------------+--------+-------+---------+
| RESEARCH   |     20 | SCOTT | 3000.00 |
| RESEARCH   |     20 | FORD  | 3000.00 |
| SALES      |     30 | BLAKE | 2850.00 |
| ACCOUNTING |     10 | KING  | 5000.00 |
+------------+--------+-------+---------+
4 rows in set (0.00 sec)

法二:建立一个左连接

#第一步找出每个部门最高的薪水值
mysql> 
select d
    eptno ,max(sal) 
from 
    emp 
group by 
    deptno;

+--------+----------+
| deptno | max(sal) |
+--------+----------+
|     20 |  3000.00 |
|     30 |  2850.00 |
|     10 |  5000.00 |
+--------+----------+
3 rows in set (0.00 sec)

#第二步：上面的表为临时表，将emp表与dept表之间建立一个左连接，条件是emp表中的部门和薪水等于临时表中部门和薪水
mysql>
select 
    d.dname,e.deptno,e.ename,e.sal 
from 
    emp as e join dept as d 
on 
    e.deptno = d.deptno 
where 
    (e.deptno,e.sal) IN (select deptno ,max(sal) from emp group by deptno);

+------------+--------+-------+---------+
| dname      | deptno | ename | sal     |
+------------+--------+-------+---------+
| SALES      |     30 | BLAKE | 2850.00 |
| RESEARCH   |     20 | SCOTT | 3000.00 |
| ACCOUNTING |     10 | KING  | 5000.00 |
| RESEARCH   |     20 | FORD  | 3000.00 |
+------------+--------+-------+---------+
4 rows in set (0.00 sec)