﻿ SQL: Display the unique department with jobs

# SQL Exercise: Display the unique department with jobs

## SQL employee Database: Exercise-10 with Solution

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10. From the following table, write a SQL query to find the unique department with jobs. Return department ID, Job name.

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT DISTINCT dep_id,
job_name
FROM employees ;
``````

Sample Output:

``` dep_id | job_name
--------+-----------
3001 | MANAGER
2001 | ANALYST
3001 | SALESMAN
1001 | MANAGER
1001 | PRESIDENT
2001 | MANAGER
2001 | CLERK
1001 | CLERK
3001 | CLERK
(9 rows)
```

Explanation:

The said query in SQL that selects distinct combinations of the dep_id and job_name columns from the 'employees' table. The "DISTINCT" keyword ensures that each combination of dep_id and job_name appears only once in the result set, even if there are multiple employees with the same job and department.

A query like this might be used to get all unique job titles and departments in the company based on the "employees" table.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: List employees id, salary, and commission.
Next SQL Exercise: List the employees not in department 2001.

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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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