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SQL Exercise: Find the salaries of all employees

SQL employee Database: Exercise-2 with Solution

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2. From the following table, write a SQL query to find the salaries of all employees. Return salary.

Sample table: employees


Pictorial Presentation:

SQL exercises on employee Database: Find the salaries of all employees

Sample Solution:

SELECT salary
FROM employees;

Sample Output:

 salary
---------
 6000.00
 2750.00
 2550.00
 2957.00
 3100.00
 3100.00
  900.00
 1700.00
 1350.00
 1350.00
 1600.00
 1200.00
 1050.00
 1400.00
(14 rows)

Explanation:

The said query in SQL that retrieves only the "salary" column from the 'employees' table.

The query returns a result set that includes all the values stored in the "salary" column for all rows of the 'employees' table.

Relational Algebra Expression:

Relational Algebra Expression: Find the salaries of all employees.

Relational Algebra Tree:

Relational Algebra Tree: Find the salaries of all employees.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: Display all the information of the employees.
Next SQL Exercise: Display the unique designations for the 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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