﻿ SQL: Compute the total salary for each job in 1991

# SQL Exercise: Compute the total salary for each job in 1991

## SQL employee Database: Exercise-92 with Solution

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92. From the following table, write a SQL query to calculate the total annual salary distributed across each job in 1991. Return job name, total annual salary.

Sample table: employees

Sample Solution:

SELECT job_name,
sum(12*salary)
FROM employees
WHERE to_char(hire_date,'YYYY') = '1991'
GROUP BY job_name;

Sample Output:

job_name  |   sum
-----------+----------
CLERK     | 12600.00
PRESIDENT | 72000.00
SALESMAN  | 72000.00
ANALYST   | 37200.00
MANAGER   | 99084.00
(5 rows)

Explanation:

The said query in SQL that selects job_name and the total salary for every job type from the 'employees' table for those employees who were hired in 1991.

The WHERE clause filters only those employees who were hired in the year 1991.

Each row of the results table represents a different job type, and the total salary is calculated based on all employees in the 'employees' table who have that job type and were hired in 1991.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Average salary, total remuneration for each type of job.
Next SQL Exercise: List id, name, department id, location of all employees.

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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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