| comments | difficulty | edit_url | tags | |
|---|---|---|---|---|
true |
Medium |
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Table: Project
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| project_id | int |
| employee_id | int |
+-------------+---------+
(project_id, employee_id) is the primary key (combination of columns with unique values) of this table.
employee_id is a foreign key (reference column) 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 (column with unique values) of this table. Each row of this table contains information about one employee.
Write a solution to report the most experienced employees in each project. In case of a tie, report all employees with the maximum number of experience years.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: 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 | 3 | | 4 | Doe | 2 | +-------------+--------+------------------+ Output: +-------------+---------------+ | project_id | employee_id | +-------------+---------------+ | 1 | 1 | | 1 | 3 | | 2 | 1 | +-------------+---------------+ Explanation: Both employees with id 1 and 3 have the most experience among the employees of the first project. For the second project, the employee with id 1 has the most experience.
We can first perform an inner join between the Project table and the Employee table, and then use the window function rank() to group the Project table, sort it in descending order by experience_years, and finally select the most experienced employee for each project.
# Write your MySQL query statement below
WITH
T AS (
SELECT
*,
RANK() OVER (
PARTITION BY project_id
ORDER BY experience_years DESC
) AS rk
FROM
Project
JOIN Employee USING (employee_id)
)
SELECT project_id, employee_id
FROM T
WHERE rk = 1;