SQL Exercise: List all the employees joined on 1st may 91
SQL employee Database: Exercise-27 with Solution
[An editor is available at the bottom of the page to write and execute the scripts.]
27. From the following table, write a SQL query to find the employees who joined on the 1st of May 1991. Return complete information about the employees.
Sample table: employees
Pictorial Presentation:

Sample Solution:
SELECT *
FROM employees
WHERE hire_date = '1991-05-01';
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+----------+------------+------------+---------+------------+-------- 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 (1 row)
Explanation:
The said query in SQL that returns a list of all employees who were hired on May 1st, 1991, along with all of their information from the 'employees' table.
The WHERE clause filters the results to only include employees who were hired on May 1st, 1991.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous SQL Exercise: List the employee who are not working under a manager.
Next SQL Exercise: Experiences of all employees working with Manger 68319.
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