﻿ SQL: Employees whose salary is greater than their managers

# SQL Exercise: Employees whose salary is greater than their managers

## SQL employee Database: Exercise-54 with Solution

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54. From the following table, write a SQL query to identify employees whose salaries are higher than their managers' salaries. Return employee name, job name, manager ID, salary, manager name, manager's salary.

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT w.emp_name,
w.job_name,
w.manager_id,
w.salary,
m.emp_name "Manager",
m.emp_id,
m.salary "Manager_Salary"
FROM employees w,
employees m
WHERE w.manager_id = m.emp_id
AND w.salary > m.salary;
``````

Sample Output:

``` emp_name | job_name | manager_id | salary  | Manager | emp_id | Manager_Salary
----------+----------+------------+---------+---------+--------+----------------
SCARLET  | ANALYST  |      65646 | 3100.00 | JONAS   |  65646 |        2957.00
FRANK    | ANALYST  |      65646 | 3100.00 | JONAS   |  65646 |        2957.00
(2 rows)
```

Explanation:

The given query in SQL that selects the employee name, job name, manager ID, salary, manager name aliased as "Manager", manager ID, and manager salary aliased as "Manager_Salary" for all employees from the employees table.

The query is using a self-join to combine data from the employees table with itself based on the manager_id column.

The WHERE clause filters those employees whose salary is greater than their manager's salary.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Employees earning 60000 or not working as analysts.
Next SQL Exercise: Employees whose salary is in a range and location PERTH.

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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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