﻿ SQL: Employees under a given department ORDER BY dep_id ASC

# SQL Exercise: Employees under a given department ORDER BY dep_id ASC

## SQL employee Database: Exercise-82 with Solution

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82. From the following table, write a SQL query to find the location of all the employees working in the FINANCE or AUDIT department. Sort the result-set in ascending order by department ID. Return complete information about the employees.

Pictorial Presentation:

Sample table: employees

Sample table: department

Sample Solution:

``````SELECT *
FROM employees e,
department d
WHERE (dep_name = 'FINANCE'
OR dep_name ='AUDIT')
AND e.dep_id = d.dep_id
ORDER BY e.dep_id ASC;
``````

OR

``````SELECT *
FROM employees e,
department d
WHERE d.dep_name IN ('FINANCE',
'AUDIT')
AND e.dep_id = d.dep_id
ORDER BY e.dep_id ASC;
``````

Sample Output:

``` emp_id | emp_name | job_name  | manager_id | hire_date  | salary  | commission | dep_id | dep_id | dep_name | dep_location
--------+----------+-----------+------------+------------+---------+------------+--------+--------+----------+--------------
68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001 |   1001 | FINANCE  | SYDNEY
67832 | CLARE    | MANAGER   |      68319 | 1991-06-09 | 2550.00 |            |   1001 |   1001 | FINANCE  | SYDNEY
69324 | MARKER   | CLERK     |      67832 | 1992-01-23 | 1400.00 |            |   1001 |   1001 | FINANCE  | SYDNEY
67858 | SCARLET  | ANALYST   |      65646 | 1997-04-19 | 3100.00 |            |   2001 |   2001 | AUDIT    | MELBOURNE
69062 | FRANK    | ANALYST   |      65646 | 1991-12-03 | 3100.00 |            |   2001 |   2001 | AUDIT    | MELBOURNE
63679 | SANDRINE | CLERK     |      69062 | 1990-12-18 |  900.00 |            |   2001 |   2001 | AUDIT    | MELBOURNE
68736 | ADNRES   | CLERK     |      67858 | 1997-05-23 | 1200.00 |            |   2001 |   2001 | AUDIT    | MELBOURNE
65646 | JONAS    | MANAGER   |      68319 | 1991-04-02 | 2957.00 |            |   2001 |   2001 | AUDIT    | MELBOURNE
(8 rows)
```

Explanation:

The said query in SQL that selects all columns (*) from the employees table and the department table where the department name is either 'FINANCE' or 'AUDIT', and the dep_id column in the employees table matches the dep_id column in the department table.

The WHERE clause includes only those employees who belong to the 'FINANCE' or 'AUDIT' and joins the employees and department tables on the dep_id column.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Sort employees by designation, joining after 1991.
Next SQL Exercise: List the employees along with grades in ascending order.

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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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