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SQL Exercise: List those employees whose salary is an odd value

SQL employee Database: Exercise-31 with Solution

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31. From the following table, write a SQL query to identify the employees whose salaries are odd. Return complete information about the employees.

Sample table: employees


Pictorial Presentation:

SQL exercises on employee Database: List those employees whose salary is an odd value

Sample Solution:

SELECT *
FROM employees
WHERE mod(salary,2) = 1;

Sample Output:

 emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------
  65646 | JONAS    | MANAGER  |      68319 | 1991-04-02 | 2957.00 |            |   2001
(1 row)

Explanation:

The said query in SQL that selects all fields from the employees table where the salary field is an odd number.

The mod() function returns the remainder when the salary field is divided by 2. If the remainder is 1, it means that the salary is an odd number.

The WHERE clause filters the results to only include rows where the salary is odd.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: Employees retiring after 31-Dec-99 after 8 years.
Next SQL Exercise: List those employees whose salary contain only 3 digits.

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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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