﻿ SQL: Find employees and departments with a given budget

# SQL Exercise: Find employees and departments with a given budget

## SQL JOINS: Exercise-28 with Solution

From the following tables write a SQL query to find the departments with budgets more than Rs. 50000 and display the first name and last name of employees.

Sample table:emp_department

Sample table: emp_details

Sample Solution:

``````SELECT emp_details.emp_fname AS "First Name", emp_lname AS "Last Name"
FROM emp_details
INNER JOIN emp_department
ON emp_details.emp_dept = emp_department.dpt_code
AND emp_department.dpt_allotment > 50000;
``````

Output of the Query:

```First Name	Last Name
Alan		Snappy
Maria		Foster
Michale		Robbin
Enric		Dosio
Joseph		Dosni
Zanifer		Emily
Kuleswar	Sitaraman
Henrey		Gabriel
Alex		Manuel
George		Mardy
```

Explanation:

The said SQL query is selecting specific columns from the emp_details table and renaming them as 'First Name' and 'Last Name'. It then joins the emp_details table with the emp_department table based on a matching value in the 'emp_dept' column of the emp_details table and the 'dpt_code' column of the emp_department table. The query also includes a filter that only returns the rows where the 'dpt_allotment' column in the emp_department table is greater than 50000.
The INNER JOIN clause is used which is used to only return the rows where there is a match in both tables, it discards the unmatched rows from both tables.

## Query Visualization:

Duration:

Rows:

Cost:

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Previous SQL Exercise: Employee and sanction amount for their department.
Next SQL Exercise: Departments where more than two employees are working.

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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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