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SQL Exercise: List all the employees joined on 1st may 91

SQL employee Database: Exercise-27 with Solution

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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:

SQL exercises on employee Database: List all the employees joined on 1st may 91

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 Expression: List all the employees joined on 1st may 91.

Relational Algebra Tree:

Relational Algebra Tree: List all the employees joined on 1st may 91.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: List the employee who are not working under a manager.
Next SQL Exercise: Experiences of all employees working with Manger 68319.

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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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