LeetCode 資料庫 部門工資最高的員工

【題目描述】

employee表包含所有員工資訊，每個員工有其對應的 id, salary 和 department id。

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

| id | name  | salary | departmentid |
+----+-------+--------+--------------+
| 1  | joe   | 70000  | 1            |
| 2  | henry | 80000  | 2            |
| 3  | sam   | 60000  | 2            |
| 4  | max   | 90000  | 1            |
+----+-------+--------+--------------+

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

| id | name     |
+----+----------+
| 1  | it       |
| 2  | sales    |
+----+----------+

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

| department | employee | salary |
+------------+----------+--------+
| it         | max      | 90000  |
| sales      | henry    | 80000  |
+------------+----------+--------+

首先建表、生成資料

drop table if exists employee;

drop table if exists department;
create table `employee` (
`id` int(11) not null auto_increment,
`name` char(20) not null,
`salary` int(20) not null,
`departmentid` int(11) not null,
primary key(`id`)
)engine=innodb charset=utf8;
create table `department` (
`id` int(11) not null auto_increment,
`name` char(20) not null,
primary key(`id`)
)engine=innodb charset=utf8;
insert into employee(name, salary, departmentid) value("joe", 70000, 1),("8enry", 60000, 2),("sam", 60000, 2),("max", 90000, 1);
insert into department(name) value("it"),("sales");

mysql> select d. name as department, e. name as employee, e.salary from departme

nt d, employee e where e.departmentid = d.id and e.salary = ( select max(salary)
from employee where departmentid = d.id );
+------------+----------+--------+
| department | employee | salary |
+------------+----------+--------+
| sales      | henry    |  80000 |
| it         | max      |  90000 |
+------------+----------+--------+
2 rows in set (0.00 sec)

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