The following SQL schema will produce the table below.
Create table If Not Exists Employees (employee_id int, name varchar(30), salary int);
Truncate table Employees;
insert into Employees (employee_id, name, salary) values ('2', 'Meir', '3000');
insert into Employees (employee_id, name, salary) values ('3', 'Michael', '3800');
insert into Employees (employee_id, name, salary) values ('7', 'Addilyn', '7400');
insert into Employees (employee_id, name, salary) values ('8', 'Juan', '6100');
insert into Employees (employee_id, name, salary) values ('9', 'Kannon', '7700');
employee_id | name | salary |
---|---|---|
2 | Meir | 3000 |
3 | Michael | 3800 |
7 | Addilyn | 7400 |
8 | Juan | 6100 |
9 | Kannon | 7700 |
Let's say we want to create a new table that shows us which employees need a raise. To do this, we will have two columns: employee_id and needs_raise. If the employees salary is less than 5000, they should have a "yes" value for needs_raise. Otherwise, they should have "no".
We can achieve this with the following query that uses a conditional statement.
SELECT employee_id,
CASE
WHEN salary < 5000 THEN "yes"
ELSE "no"
END AS needs_raise
FROM Employees;
employee_id | needs_raise |
---|---|
2 | yes |
3 | yes |
7 | no |
8 | no |
9 | no |