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SQL Exercise: List employees id, salary, and commission

SQL employee Database: Exercise-9 with Solution

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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

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

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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

 

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