text: /*
  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.
  Table: Employee
  
  +------------------+---------+
  | Column Name      | Type    |
  +------------------+---------+
  | employee_id      | int     |
  | name             | varchar |
  | experience_years | int     |
  +------------------+---------+
  employee_id is the primary key of this table.
   
  
  Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.
  
  The query result format is in the following example:
  
  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                |
  +-------------+--------+------------------+
  
  Result table:
  +-------------+---------------+
  | project_id  | average_years |
  +-------------+---------------+
  | 1           | 2.00          |
  | 2           | 2.50          |
  +-------------+---------------+
  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
  */
  SELECT project_id, Round(Avg(experience_years), 2) AS average_years 
  FROM   project AS p 
  INNER JOIN employee AS e 
  ON p.employee_id = e.employee_id 
  GROUP  BY project_id 
  ORDER  BY NULL 
  
  