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SQL Exercise: Sort the employees by salary less than 1000

SQL employee Database: Exercise-73 with Solution

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73. From the following table, write a SQL query to find those employees who earn less than 1000. Sort the result-set in ascending order by salary. Return complete information about the employees.

Pictorial Presentation:

SQL exercises on employee Database: List the employees who are drawing the salary less than 1000 and sort the output in ascending order on salary

Sample table: employees


Sample Solution:

SELECT *
FROM employees
WHERE salary < 1000
ORDER BY salary;

Sample Output:

 emp_id | emp_name | job_name | manager_id | hire_date  | salary | commission | dep_id
--------+----------+----------+------------+------------+--------+------------+--------
  63679 | SANDRINE | CLERK    |      69062 | 1990-12-18 | 900.00 |            |   2001
(1 row)

Explanation:

The said query in SQL that returns all the rows and columns from the 'employees' table where the "salary" column is less than 1000, and sorts the result set in ascending order on "salary" column.

Relational Algebra Expression:

Relational Algebra Expression: List the employees who are drawing the salary less than 1000 and sort the output in ascending order on salary.

Relational Algebra Tree:

Relational Algebra Tree: List the employees who are drawing the salary less than 1000 and sort the output in ascending order on salary.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: Employees joining on given days, ascending in seniority.
Next SQL Exercise: Sort the employees in ascending order on the salary.

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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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