﻿ SQL: Employees with more than 27 years of experience

# SQL Exercise: Employees with more than 27 years of experience

## SQL employee Database: Exercise-21 with Solution

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21. From the following table, write a SQL query to identify employees with more than 27 years of experience. Return complete information about the employees.

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT *
FROM employees
WHERE EXTRACT(YEAR
FROM age(CURRENT_DATE, hire_date)) > 27;
``````

Sample Output:

``` emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------
64989 | ADELYN   | SALESMAN |      66928 | 1991-02-20 | 1700.00 |     400.00 |   3001
65271 | WADE     | SALESMAN |      66928 | 1991-02-22 | 1350.00 |     600.00 |   3001
65679 | SANDRINE | CLERK    |      69062 | 1990-12-18 | 900.00  |            |   2001
(3 rows)
```

Explanation:

The said query in SQL that retrieves all the rows from the 'employees' table where the number of years between the "hire_date" column and the current date is greater than 27 years.

The "age" function calculates the time interval between the current date and the "hire_date" column and returns it as an interval value. The interval value is then passed to the "EXTRACT" function to extract the year value.

The result will include all columns of the selected rows.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: List all the employees whose designation is CLERK.
Next SQL Exercise: List the employees whose salaries are less than 3500.

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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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