w3resource

SQL Exercise: List employees id, salary, and commission

SQL employee Database: Exercise-9 with Solution

[An editor is available at the bottom of the page to write and execute the scripts.]

9. From the following table, write a SQL query to find the employee ID, salary, and commission of all the employees.

Sample table: employees


Pictorial Presentation:

SQL exercises on employee Database: List the emp_id, salary, and commission of all the employees

Sample Solution:

SELECT emp_id,
       salary,
       commission
FROM employees;

Sample Output:

 emp_id | salary  | commission
--------+---------+------------
  68319 | 6000.00 |
  66928 | 2750.00 |
  67832 | 2550.00 |
  65646 | 2957.00 |
  67858 | 3100.00 |
  69062 | 3100.00 |
  63679 |  900.00 |
  64989 | 1700.00 |     400.00
  65271 | 1350.00 |     600.00
  66564 | 1350.00 |    1500.00
  68454 | 1600.00 |       0.00
  68736 | 1200.00 |
  69000 | 1050.00 |
  69324 | 1400.00 |
(14 rows)

Explanation:

The given query in SQL that selects the emp_id, salary, and commission columns from the 'employees' table.

Relational Algebra Expression:

Relational Algebra Expression: List the emp_id, salary, and commission of all the employees.

Relational Algebra Tree:

Relational Algebra Tree: List the emp_id, salary, and commission of all the employees.

Practice Online


Sample Database: employee

employee database structure

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous SQL Exercise: Without the spaces, count each employee name characters.
Next SQL Exercise: Display the unique department with jobs.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.



Follow us on Facebook and Twitter for latest update.

SQL: Tips of the Day

Grouped LIMIT in PostgreSQL: Show the first N rows for each group?

db=# SELECT * FROM xxx;
 id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  3 |          1 | C
  4 |          1 | D
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
  8 |          2 | H
(8 rows)

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
(5 rows)

PostgreSQL v9.3 you can do a lateral join

select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
    select * from t t_inner
    where t_inner.section_id = t_outer.section_id
    order by t_inner.name
    limit 2
) t_top on true
order by t_outer.section_id;

Database: PostgreSQL

Ref: https://bit.ly/3AfYwZI

 





We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook