solved

Author: akscoder082003

MySQL/1075. Project Employees.sql

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+-- 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 Project.project_id, 
+round(avg(Employee.experience_years), 2) as average_years 
+from Project left join Employee 
+on Project.employee_id=Employee.employee_id 
+group by Project.project_id;