【题目】
项目表 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。
员工表 Employee:
+------------------+---------+
| Column Name | Type |
+------------------+---------+
| employee_id | int |
| name | varchar |
| experience_years | int |
+------------------+---------+
主键是 employee_id。
请写一个 SQL 语句,查询每一个项目中员工的 平均 工作年限,精确到小数点后两位。
查询结果的格式如下:
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 | 1 |
| 4 | Doe | 2 |
+-------------+--------+------------------+
Result 表:
+-------------+---------------+
| project_id | average_years |
+-------------+---------------+
| 1 | 2.00 |
| 2 | 2.50 |
+-------------+---------------+
第一个项目中,员工的平均工作年限是 (3 + 2 + 1) / 3 = 2.00;第二个项目中,员工的平均工作年限是 (3 + 2) / 2 = 2.50
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/project-employees-i
【思路】
先左连接,然后按project_id进行分组,然后累加后除
【代码】
SELECT project_id,ROUND(SUM(experience_years)/COUNT(project_id),2) as average_years FROM Project Left Join Employee On Project.employee_id = Employee.employee_id GROUP BY project_id;
标签:项目,1075,员工,project,Employee,employee,years,id From: https://www.cnblogs.com/End1ess/p/17381640.html