﻿ SQL: List the employee who are not working under a manager

# SQL Exercise: List the employee who are not working under a manager

## SQL employee Database: Exercise-26 with Solution

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26. From the following table, write a SQL query identify the employees who do not report to a manager. Return employee name, job name.

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT e.emp_name,
e.job_name
FROM employees e
WHERE manager_id IS NULL;
``````

Sample Output:

``` emp_name     | job_name
--------------+--------------
KAYLING     | PRESIDENT
```

Explanation:

The said query in SQL that retrieves a list of all employees who do not have a manager, along with their name and job position from the 'employees' table.

The WHERE clause that filters the results so that, it only includes the employees who do not have a value in the "manager_id" column, which indicates whether or not the "manager_id" column is null.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: List the employees who joined before 1 April 1991.
Next SQL Exercise: List all the employees joined on 1st may 91.

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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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