SQL Exercise: Employees with more than 27 years of experience
SQL employee Database: Exercise-21 with Solution
[An editor is available at the bottom of the page to write and execute the scripts.]
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

Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous SQL Exercise: List all the employees whose designation is CLERK.
Next SQL Exercise: List the employees whose salaries are less than 3500.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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
- Weekly Trends
- Python Interview Questions and Answers: Comprehensive Guide
- Scala Exercises, Practice, Solution
- Kotlin Exercises practice with solution
- MongoDB Exercises, Practice, Solution
- SQL Exercises, Practice, Solution - JOINS
- Java Basic Programming Exercises
- SQL Subqueries
- Adventureworks Database Exercises
- C# Sharp Basic Exercises
- SQL COUNT() with distinct
- JavaScript String Exercises
- JavaScript HTML Form Validation
- Java Collection Exercises
- SQL COUNT() function
- SQL Inner Join
We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook