20_employee_salaries.sql

/*
Employee Salaries

Write a query that prints a list of employee names (i.e.: the name attribute) for employees in Employee 
having a salary greater than $2000 per month who have been employees for less than months. 
Sort your result by ascending employee_id.

- Input Format
The Employee table containing employee data for a company is described as follows:
    +-------------+----------+
    | Column      |  Type    |
    +-------------+----------+
    | employee_id | Integer  |
    | name        | String   |
    | months      | Integer  |
    | salary      | Integer  |
    +-------------+----------+

where employee_id is an employee's ID number, name is their name, months is the total number 
of months they've been working for the company, and salary is their monthly salary.
    +-------------+----------+--------+--------+
    | employee_id | name     | months | salary |
    +-------------+----------+--------+--------+
    | 12228       | Rose     | 15     | 1968   |    
    | 33645       | Angela   | 1      | 33443  |
    | 45692       | Frank    | 17     | 1608   |
    | 56118       | Patrick  | 7      | 1345   |
    | 59725       | Lisa     | 11     | 2330   |   
    | 74197       | Kimberly | 16     | 4372   |   
    | 78454       | Bonnie   | 8      | 1771   |   
    | 83565       | Michael  | 6      | 2017   |   
    | 98607       | Todd     | 5      | 3396   |   
    | 99989       | Joe      | 9      | 3573   |   
    +-------------+----------+--------+--------+

- Sample Output
    Angela
    Michael
    Todd
    Joe

- Explanation
Angela has been an employee for 1 month and earns 3443 per month.
Michael has been an employee for 6 months and earns 2017 per month.
Todd has been an employee for 5 months and earns 3396 per month.
Joe has been an employee for 9 months and earns 3573 per month.
We order our output by ascending employee_id.
*/

SELECT 
    name 
FROM 
    employee
WHERE 
    salary > 2000 AND 
    months < 10
ORDER BY 
    employee_id
;