标签:experience 员工 project Employee LeeCode1077 employee years III id
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
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