﻿ SQL: Employees in ascending and descending order

# SQL Exercise: Employees in ascending and descending order

## SQL employee Database: Exercise-67 with Solution

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67. From the following table, write a SQL query to list employees in ascending order on department ID and descending order on jobs. Return complete information about the employees.

Pictorial Presentation:

Sample table: employees

Sample Solution:

``````SELECT *
FROM employees
ORDER BY dep_id ASC,
job_name DESC;
``````

Sample Output:

``` emp_id | emp_name | job_name  | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+-----------+------------+------------+---------+------------+--------
68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001
67832 | CLARE    | MANAGER   |      68319 | 1991-06-09 | 2550.00 |            |   1001
69324 | MARKER   | CLERK     |      67832 | 1992-01-23 | 1400.00 |            |   1001
65646 | JONAS    | MANAGER   |      68319 | 1991-04-02 | 2957.00 |            |   2001
63679 | SANDRINE | CLERK     |      69062 | 1990-12-18 |  900.00 |            |   2001
68736 | ADNRES   | CLERK     |      67858 | 1997-05-23 | 1200.00 |            |   2001
69062 | FRANK    | ANALYST   |      65646 | 1991-12-03 | 3100.00 |            |   2001
67858 | SCARLET  | ANALYST   |      65646 | 1997-04-19 | 3100.00 |            |   2001
65271 | WADE     | SALESMAN  |      66928 | 1991-02-22 | 1350.00 |     600.00 |   3001
64989 | ADELYN   | SALESMAN  |      66928 | 1991-02-20 | 1700.00 |     400.00 |   3001
66564 | MADDEN   | SALESMAN  |      66928 | 1991-09-28 | 1350.00 |    1500.00 |   3001
68454 | TUCKER   | SALESMAN  |      66928 | 1991-09-08 | 1600.00 |       0.00 |   3001
66928 | BLAZE    | MANAGER   |      68319 | 1991-05-01 | 2750.00 |            |   3001
69000 | JULIUS   | CLERK     |      66928 | 1991-12-03 | 1050.00 |            |   3001
(14 rows)
```

Explanation:

The said query in SQL that retrieves all the rows from the 'employees' table and order them first by the "dep_id" column in ascending order, and then by the "job_name" column in descending order.

This means that the employee records will be grouped first by department ID in ascending order, and within each department, the records will be ordered by job name in descending order.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: List the employees in ascending order of their salaries.
Next SQL Exercise: Display all the unique job in descending 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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