﻿ SQL: Find the salaries of all employees

# 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:

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 Tree:

## Practice Online

Sample Database: employee

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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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