﻿ SQL: Employees whose salary is in a range and location PERTH

# SQL Exercise: Employees whose salary is in a range and location PERTH

## SQL employee Database: Exercise-55 with Solution

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55. From the following table, write a SQL query to find those employees whose salary is between 2000 and 5000 (Begin and end values are included.) and location is PERTH. Return employee name, department ID, salary, and commission.

Sample table: employees

Sample table: department

Pictorial Presentation:

Sample Solution:

``````SELECT e.emp_name,
e.dep_id,
e.salary,
e.commission
FROM employees e,
department d
WHERE e.dep_id = d.dep_id
AND d.dep_location = 'PERTH'
AND e.salary BETWEEN 2000 AND 5000;
``````

Sample Output:

``` emp_name | dep_id | salary  | commission
----------+--------+---------+------------
BLAZE    |   3001 | 2750.00 |
(1 row)
```

Explanation:

This is a SQL query that selects the employee name, department ID, salary, and commission for all employees from the employees and department tables.

The query is using an inner join in the WHERE clause to combine data from the employees and department tables based on their common column dep_id. It then includes the employees who work in the department located in Perth and whose salary is between 2000 and 5000.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Employees whose salary is greater than their managers.
Next SQL Exercise: Employees joined before a date but not graded 4.

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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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