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SQL Exercise: Display all the unique job in descending order

SQL employee Database: Exercise-68 with Solution

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68. From the following table, write a SQL query to sort the unique jobs in descending order. Return job name.

Pictorial Presentation:

SQL exercises on employee Database: Display all the unique job in descending order

Sample table: employees


Sample Solution:

SELECT DISTINCT job_name
FROM employees
ORDER BY job_name DESC;

Sample Output:

 job_name
-----------
 SALESMAN
 PRESIDENT
 MANAGER
 CLERK
 ANALYST
(5 rows)

Explanation:

The given query in SQL that retrieves all unique values of the "job_name" column from the 'employees' table and order them in descending order.

The query returns all unique job names that exist in the 'employees' table, with each job name appearing only once in the result set. The job names will be ordered in descending order.

The "DISTINCT" keyword eliminates any duplicate values in the result set.

Relational Algebra Expression:

Relational Algebra Expression: Display all the unique job in descending order.

Relational Algebra Tree:

Relational Algebra Tree: Display all the unique job in descending order.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: Employees in ascending and descending order.
Next SQL Exercise: Employees in the ascending order on annual salary.

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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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