LeetCode MySQL 1076. 项目员工II

1. 题目

Table: Project

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
主键为 (project_id, employee_id)。
employee_id 是员工表 Employee 表的外键。

Table: Employee

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
主键是 employee_id。

编写一个SQL查询，报告所有雇员最多的项目。

查询结果格式如下所示：

Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result table:
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
第一个项目有3名员工，第二个项目有2名员工。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/project-employees-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

# Write your MySQL query statement below
select project_id 
from Project p
group by project_id
having count(p.employee_id) =
                (
                    select count(*) maxnumofEmp
                    from Project
                    group by project_id
                    order by maxnumofEmp desc
                    limit 1
                )

或者用 all 函数

# Write your MySQL query statement below
select project_id 
from Project p
group by project_id
having count(p.employee_id) >=
             all(
                    select count(*) num
                    from Project
                    group by project_id
                )

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