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SQL Exercise: Employees work as SALESMEN, sorted by their salary

SQL employee Database: Exercise-79 with Solution

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79. From the following table, write a SQL query to list the employees who works as a SALESMAN. Sort the result set in ascending order of annual salary. Return employee id, name, annual salary, daily salary of all the employees.

Pictorial Presentation:

SQL exercises on employee Database: List the employee id, name, annual salary, daily salary of all the employees in the ascending order of annual salary who works as a SALESMAN

Sample table: employees


Sample Solution:

SELECT e.emp_id,
       e.emp_name,
       12*salary "Annual Salary",
       (12*salary)/365 "Daily Salary"
FROM employees e
WHERE e.job_name = 'SALESMAN'
ORDER BY "Annual Salary" ASC;

Sample Output:

 emp_id | emp_name | Annual Salary |    Daily Salary
--------+----------+---------------+---------------------
  65271 | WADE     |      16200.00 | 44.3835616438356164
  66564 | MADDEN   |      16200.00 | 44.3835616438356164
  68454 | TUCKER   |      19200.00 | 52.6027397260273973
  64989 | ADELYN   |      20400.00 | 55.8904109589041096
(4 rows)

Explanation:

The said query in SQL that selects the employee ID, employee name, annual salary, and daily salary of all employees from 'employees' table whose job name is 'SALESMAN', and orders the results in ascending order by annual salary.

The calculation 12*salary is used to calculate the annual salary, and the "Annual Salary" alias is assigned to the result.

The WHERE clause filters the results to include only those employees whose job name is 'SALESMAN'.

The ORDER BY clause ordered the result set by annual salary in ascending order.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: Sort employees in ascending order by annual salary.
Next SQL Exercise: List employees in ascending order on their experiences.

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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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