Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1075.html
1075. Project Employees I
Level
Easy
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.
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
Solution
Join tables Project and Employee using the field employee_id and select project_id and avg(experience_years). Use round to round the result to 2 digits. Use ifnull to avoid dividing by zero.
# Write your MySQL query statement below
select project_id, ifnull(round(avg(experience_years), 2), 0) as average_years
from Project left join Employee
on Project.employee_id = Employee.employee_id
group by project_id;