LeeCode1077. 项目员工 III

SQL架构

项目表 Project：

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) 是这个表的主键
employee_id 是员工表 Employee 的外键

员工表 Employee：

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id 是这个表的主键

写 一个 SQL 查询语句，报告在每一个项目中经验最丰富的雇员是谁。如果出现经验年数相同的情况，请报告所有具有最大经验年数的员工。

查询结果格式在以下示例中：

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

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

Result 表：
+-------------+---------------+
| project_id  | employee_id   |
+-------------+---------------+
| 1           | 1             |
| 1           | 3             |
| 2           | 1             |
+-------------+---------------+
employee_id 为 1 和 3 的员工在 project_id 为 1 的项目中拥有最丰富的经验。在 project_id 为 2 的项目中，employee_id 为 1 的员工拥有最丰富的经验。

题解一

select 
    project_id,employee_id
from    
    (select a.project_id,a.employee_id,
        dense_rank() over(partition by a.project_id order by b.experience_years desc )rk
    from project a join employee b on a.employee_id=b.employee_id) t
where 
    rk = 1
;

题解二

select 
    a.project_id,a.employee_id 
from 
    Project a 
join 
    Employee b
on 
    a.employee_id = b.employee_id
where 
    (a.project_id,b.experience_years) 
in
    (select a.project_id,max(b.experience_years) from Project a join Employee b
    on a.employee_id = b.employee_id group by project_id)
;

标签：experience,员工,project,Employee,LeeCode1077,employee,years,III,id
来源： https://blog.csdn.net/haohaoxuexiyai/article/details/115426757