1. 题目
员工表：Employees
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| employee_id | int |
| employee_name | varchar |
| manager_id | int |
+---------------+---------+
employee_id 是这个表的主键。
这个表中每一行中，employee_id 表示职工的 ID，
employee_name 表示职工的名字，
manager_id 表示该职工汇报工作的直线经理。
这个公司 CEO 是 employee_id = 1 的人。
用 SQL 查询出所有直接或间接向公司 CEO 汇报工作的职工的 employee_id 。
由于公司规模较小，经理之间的间接关系不超过 3 个经理。
可以以任何顺序返回的结果，不需要去重。
查询结果示例如下：
Employees table:
+-------------+---------------+------------+
| employee_id | employee_name | manager_id |
+-------------+---------------+------------+
| 1 | Boss | 1 |
| 3 | Alice | 3 |
| 2 | Bob | 1 |
| 4 | Daniel | 2 |
| 7 | Luis | 4 |
| 8 | Jhon | 3 |
| 9 | Angela | 8 |
| 77 | Robert | 1 |
+-------------+---------------+------------+
Result table:
+-------------+
| employee_id |
+-------------+
| 2 |
| 77 |
| 4 |
| 7 |
+-------------+
公司 CEO 的 employee_id 是 1.
employee_id 是 2 和 77 的职员直接汇报给公司 CEO。
employee_id 是 4 的职员间接汇报给公司 CEO 4 --> 2 --> 1 。
employee_id 是 7 的职员间接汇报给公司 CEO 7 --> 4 --> 2 --> 1 。
employee_id 是 3, 8 ，9 的职员不会直接或间接的汇报给公司 CEO。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/all-people-report-to-the-given-manager
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2. 解题
- 一次连接
select *
from Employees e1 left join Employees e2
on e1.manager_id = e2.employee_id
{"headers": ["employee_id", "employee_name", "manager_id", "employee_id", "employee_name", "manager_id"],
"values": [
[1, "Boss", 1, 1, "Boss", 1],
[3, "Alice", 3, 3, "Alice", 3],
[2, "Bob", 1, 1, "Boss", 1],
[4, "Daniel", 2, 2, "Bob", 1],
[7, "Luis", 4, 4, "Daniel", 2],
[8, "John", 3, 3, "Alice", 3],
[9, "Angela", 8, 8, "John", 3],
[77, "Robert", 1, 1, "Boss", 1]]}
- 二次连接
select *
from Employees e1 left join Employees e2
on e1.manager_id = e2.employee_id
left join Employees e3
on e2.manager_id = e3.employee_id
{"headers": ["employee_id", "employee_name", "manager_id", "employee_id", "employee_name", "manager_id", "employee_id", "employee_name", "manager_id"],
"values": [
[1, "Boss", 1, 1, "Boss", 1, 1, "Boss", 1],
[3, "Alice", 3, 3, "Alice", 3, 3, "Alice", 3],
[2, "Bob", 1, 1, "Boss", 1, 1, "Boss", 1],
[4, "Daniel", 2, 2, "Bob", 1, 1, "Boss", 1],
[7, "Luis", 4, 4, "Daniel", 2, 2, "Bob", 1],
[8, "John", 3, 3, "Alice", 3, 3, "Alice", 3],
[9, "Angela", 8, 8, "John", 3, 3, "Alice", 3],
[77, "Robert", 1, 1, "Boss", 1, 1, "Boss", 1]]}
- 答案
# Write your MySQL query statement below
select e1.employee_id
from Employees e1 left join Employees e2
on e1.manager_id = e2.employee_id
left join Employees e3
on e2.manager_id = e3.employee_id
where e3.manager_id = 1 and e1.employee_id != 1
173 ms
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