w3resource

SQL Exercise: Average salary, total remuneration for each type of job

SQL employee Database: Exercise-91 with Solution

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

91. From the following table, write a SQL query to find the average salary and average total remuneration (salary and commission) for each type of job. Return name, average salary and average total remuneration.

Sample table: employees


Sample Solution:

SELECT job_name,
       avg(salary),
       avg(salary+commission)
FROM employees
GROUP BY job_name;

Sample Output:

 job_name  |          avg          |          avg
-----------+-----------------------+-----------------------
 CLERK     | 1137.5000000000000000 |
 SALESMAN  | 1500.0000000000000000 | 2125.0000000000000000
 MANAGER   | 2752.3333333333333333 |
 PRESIDENT | 6000.0000000000000000 |
 ANALYST   | 3100.0000000000000000 |
(5 rows)

Explanation:

The said query in SQL that selects job_name, the average salary, and the average total remuneration from the 'employees' table. The query then applies a grouping using the GROUP BY clause on the job_name column.

Relational Algebra Expression:

Relational Algebra Expression: Find the average salary and average total remuneration for each type of job.

Relational Algebra Tree:

Relational Algebra Tree: Find the average salary and average total remuneration for each type of job.

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: Find the highest salary from all the employees.
Next SQL Exercise: Compute the total salary for each job in 1991.

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