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

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

## SQL employee Database: Exercise-91 with Solution

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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 Tree:

## Practice Online

Sample Database: employee

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

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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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