SQL Exercise: Sort unique jobs by department in descending order

Last update on April 24 2023 12:27:34 (UTC/GMT +8 hours)

SQL employee Database: Exercise-76 with Solution

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76. From the following table, write a SQL query to list the unique jobs of department 2001 and 3001 in descending order. Return job name.

Pictorial Presentation:

SQL exercises on employee Database: List the unique jobs of department 2001 and 3001 in descending order

Sample table: employees

Sample Solution:

SELECT DISTINCT job_name
FROM employees
WHERE dep_id IN (2001,
                 3001)
ORDER BY job_name DESC;

Sample Output:

 job_name
----------
 SALESMAN
 MANAGER
 CLERK
 ANALYST
(4 rows)

Explanation:

The said query in SQL that selects the distinct job names of employees from the 'employees' table who work in departments 2001 or 3001, and orders the results in descending order by job name.

The DISTINCT keyword eliminates duplicates in the result set, ensuring that each job name appears only once.

The WHERE clause filters the results to only include employees who work in departments 2001 or 3001.

The ORDER BY clause orders the result set by job name in descending order.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Sample Database: employee

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Previous SQL Exercise: Ascending by job name and descending by employee ID.
Next SQL Exercise: Employees except given job in ASC order of salaries.

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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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