task: #1075

Author: Ivanbyone

README.md

@@ -49,6 +49,7 @@ Have a good contributing!
     - [619. Biggest Single Number](./leetcode/easy/619.%20Biggest%20Single%20Number.sql)
     - [620. Not Boring Movies](./leetcode/easy/620.%20Not%20Boring%20Movies.sql)
     - [1068. Product Sales Analysis I](./leetcode/easy/1068.%20Product%20Sales%20Analysis%20I.sql)
+    - [1075. Project Employees I](./leetcode/easy/1075.%20Project%20Employees%20I.sql)
     - [1327. List the Products Ordered in a Period](./leetcode/easy/1327.%20List%20the%20Products%20Ordered%20in%20a%20Period.sql)
     - [1378. Replace Employee ID With The Unique Identifier](./leetcode/easy/1378.%20Replace%20Employee%20ID%20With%20The%20Unique%20Identifier.sql)
     - [1407. Top Travellers](./leetcode/easy/1407.%20Top%20Travellers.sql)

leetcode/easy/1075. Project Employees I.sql

@@ -0,0 +1,43 @@
+/*
+Question 1075. Project Employees I
+Link: https://leetcode.com/problems/project-employees-i/description/
+
+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. It's guaranteed that experience_years is not NULL.
+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.
+*/
+
+SELECT
+    p.project_id,
+    ROUND(AVG(e.experience_years), 2) AS average_years
+FROM Project AS p
+LEFT JOIN 
+    Employee AS e
+    ON p.employee_id = e.employee_id
+GROUP BY p.project_id