add 1075

Author: fishercoder1534

README.md

@@ -855,6 +855,7 @@ _If you like this project, please leave me a star._ ★
 |1084|[Sales Analysis III](https://leetcode.com/problems/sales-analysis-iii/)|[Solution](../master/database/_1084.sql) |  | Easy |
 |1083|[Sales Analysis II](https://leetcode.com/problems/sales-analysis-ii/)|[Solution](../master/database/_1083.sql) |  | Easy |
 |1082|[Sales Analysis I](https://leetcode.com/problems/sales-analysis-i/)|[Solution](../master/database/_1082.sql) |  | Easy |
+|1075|[Project Employees I](https://leetcode.com/problems/project-employees-i/)|[Solution](../master/database/_1075.sql) |  | Easy |
 |1069|[Product Sales Analysis II](https://leetcode.com/problems/product-sales-analysis-ii/)|[Solution](../master/database/_1069.sql) |  | Easy |
 |1068|[Product Sales Analysis I](https://leetcode.com/problems/product-sales-analysis-i/)|[Solution](../master/database/_1068.sql) | | Easy |
 |627|[Swap Salary](https://leetcode.com/problems/swap-salary/)|[Solution](../master/database/_627.sql) | | Easy |

database/_1075.sql

@@ -0,0 +1,65 @@
+--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
+
+--# Write your MySQL query statement below
+select project_id, round(avg(experience_years), 2) as average_years
+from Project
+join Employee
+using (employee_id)
+group by project_id
+order by project_id