﻿ SQL: Employees who are CLERK, ANALYST in descending order

# SQL Exercise: Employees who are CLERK, ANALYST in descending order

## SQL employee Database: Exercise-70 with Solution

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70. From the following table, write a SQL query to find those employees who are either 'CLERK' or 'ANALYST’. Sort the result set in descending order on job_name. Return complete information about the employees.

Pictorial Presentation:

Sample table: employees

Sample Solution:

``````SELECT *
FROM employees
WHERE job_name='CLERK'
OR job_name='ANALYST'
ORDER BY job_name DESC;
``````

Sample Output:

``` emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------
63679 | SANDRINE | CLERK    |      69062 | 1990-12-18 |  900.00 |            |   2001
68736 | ADNRES   | CLERK    |      67858 | 1997-05-23 | 1200.00 |            |   2001
69000 | JULIUS   | CLERK    |      66928 | 1991-12-03 | 1050.00 |            |   3001
69324 | MARKER   | CLERK    |      67832 | 1992-01-23 | 1400.00 |            |   1001
67858 | SCARLET  | ANALYST  |      65646 | 1997-04-19 | 3100.00 |            |   2001
69062 | FRANK    | ANALYST  |      65646 | 1991-12-03 | 3100.00 |            |   2001
(6 rows)
```

Explanation:

The said query in SQL that retrieves all the columns and rows from the 'employees' table where the job name is either 'CLERK' or 'ANALYST'. The result set is then arranged in descending order on the job name, that means the the job name beginning with letter 'Z' will come before the letter 'A'.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Employees in the ascending order on annual salary.
Next SQL Exercise: Display the location of CLARE.

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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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