LeetCode:Database 44.项目员工 III
Posted Xiao Miao
tags:
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要求: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。
项目表 Project的结构:
+-------------+---------+
| 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的结构:
+------------------+---------+
| Column Name | Type |
+------------------+---------+
| employee_id | int |
| name | varchar |
| experience_years | int |
+------------------+---------+
employee_id 是这个表的主键
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 Table:
+-------------+---------------+
| 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。
SQL语句:
select s1 as project_id ,s2 as employee_id
from(
select s1,s2,
dense_rank() over(partition by s1
order by s3 desc ) as r
from
(select a.project_id as s1,a.employee_id as s2,
b.experience_years as s3
from project a
join employee b
on a.employee_id=b.employee_id)c
)d
where r=1;
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