1075-ProjectEmployeesI.sql

-- 1075. Project Employees I
-- Table: Project
-- +-------------+---------+
-- | Column Name | Type    |
-- +-------------+---------+
-- | project_id  | int     |
-- | employee_id | int     |
-- +-------------+---------+
-- (project_id, employee_id) is the primary key of this table.
-- employee_id is a foreign key to Employee table.
-- Each row of this table indicates that the employee with employee_id is working on the project with project_id.
--  
-- Table: Employee
-- +------------------+---------+
-- | Column Name      | Type    |
-- +------------------+---------+
-- | employee_id      | int     |
-- | name             | varchar |
-- | experience_years | int     |
-- +------------------+---------+
-- employee_id is the primary key of this table.
-- Each row of this table contains information about one employee.
--  
-- Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.
-- Return the result table in any order.
-- The query result format is in the following example.
--
-- Example 1:
-- Input:
-- 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                |
-- +-------------+--------+------------------+
-- Output:
-- +-------------+---------------+
-- | project_id  | average_years |
-- +-------------+---------------+
-- | 1           | 2.00          |
-- | 2           | 2.50          |
-- +-------------+---------------+
-- Explanation: The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00 and for the second project is (3 + 2) / 2 = 2.50

-- Create table If Not Exists Project (project_id int, employee_id int)
-- Create table If Not Exists Employee (employee_id int, name varchar(10), experience_years int)
-- Truncate table Project
-- insert into Project (project_id, employee_id) values ('1', '1')
-- insert into Project (project_id, employee_id) values ('1', '2')
-- insert into Project (project_id, employee_id) values ('1', '3')
-- insert into Project (project_id, employee_id) values ('2', '1')
-- insert into Project (project_id, employee_id) values ('2', '4')
-- Truncate table Employee
-- insert into Employee (employee_id, name, experience_years) values ('1', 'Khaled', '3')
-- insert into Employee (employee_id, name, experience_years) values ('2', 'Ali', '2')
-- insert into Employee (employee_id, name, experience_years) values ('3', 'John', '1')
-- insert into Employee (employee_id, name, experience_years) values ('4', 'Doe', '2')

-- Write your MySQL query statement below
SELECT
    p.project_id,
    ROUND(SUM(e.experience_years) / COUNT(*),2) AS average_years
FROM
    Project  AS p,
    Employee AS e
WHERE
    p.employee_id = e.employee_id
    AND e.experience_years IS NOT NULL -- 处理 experience_years 为 null 的情况
GROUP BY
    p.project_id