SQL Exercise: List the employees who joined before 1 April 1991

Last update on April 18 2023 06:46:15 (UTC/GMT +8 hours)

SQL employee Database: Exercise-25 with Solution

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25. From the following table, write a SQL query to find those employees who joined before 1st April 1991. Return employee ID, employee name, hire date and salary.

Sample table: employees

Pictorial Presentation:

SQL exercises on employee Database: List the name, id, hire_date, and salary of all the employees joined before 1 apr 91

Sample Solution:

SELECT e.emp_id,
       e.emp_name,
       e.hire_date,
       e.salary
FROM employees e
WHERE hire_date <'1991-04-01';

Sample Output:

 emp_id | emp_name | hire_date  | salary
--------+----------+------------+---------
  63679 | SANDRINE | 1990-12-18 |  900.00
  64989 | ADELYN   | 1991-02-20 | 1700.00
  65271 | WADE     | 1991-02-22 | 1350.00
(3 rows)

Explanation:

The said query in SQL that returns a list of all employees who were hired before April 1st, 1991, along with their ID, name, hire date, and salary from the 'employees' table.

The WHERE clause filters the results to include only those employees who were hired before April 1st, 1991.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Sample Database: employee

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Previous SQL Exercise: List the employees who have joined in the year 1991.
Next SQL Exercise: List the employee who are not working under a manager.

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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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