﻿ SQL: List the employees who are senior to their own manager

# SQL Exercise: List the employees who are senior to their own manager

## SQL employee Database: Exercise-63 with Solution

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63. From the following table, write a SQL query to find those employees who are senior to their manager. Return complete information about the employees.

Pictorial Presentation:

Sample table: employees

Sample Solution:

``````SELECT *
FROM employees w,
employees m
WHERE w.manager_id = m.emp_id
AND w.hire_date < m.hire_date;
``````

Sample Output:

``` emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id | emp_id | emp_name | job_name  | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------+--------+----------+-----------+------------+------------+---------+------------+--------
66928 | BLAZE    | MANAGER  |      68319 | 1991-05-01 | 2750.00 |            |   3001 |  68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001
67832 | CLARE    | MANAGER  |      68319 | 1991-06-09 | 2550.00 |            |   1001 |  68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001
65646 | JONAS    | MANAGER  |      68319 | 1991-04-02 | 2957.00 |            |   2001 |  68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001
63679 | SANDRINE | CLERK    |      69062 | 1990-12-18 |  900.00 |            |   2001 |  69062 | FRANK    | ANALYST   |      65646 | 1991-12-03 | 3100.00 |            |   2001
64989 | ADELYN   | SALESMAN |      66928 | 1991-02-20 | 1700.00 |     400.00 |   3001 |  66928 | BLAZE    | MANAGER   |      68319 | 1991-05-01 | 2750.00 |            |   3001
65271 | WADE     | SALESMAN |      66928 | 1991-02-22 | 1350.00 |     600.00 |   3001 |  66928 | BLAZE    | MANAGER   |      68319 | 1991-05-01 | 2750.00 |            |   3001
(6 rows)
```

Explanation:

The said query in SQL that selects all columns from the 'employees' table twice, with different aliases "w" and "m" to find all employees who were hired before their manager.

The query joins the 'employees' table to itself, to create a relationship between the employee and their manager based on the emp_id and manager_id columns, and the hire date comparison then filters the results to only show employees who were hired earlier than their manager using the WHERE clause.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Employees location, grade, and experience over 25 years.
Next SQL Exercise: List the employee with their grade for the grade 4.

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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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