SQL Exercise: List the employees who joined before 1991

Last update on April 18 2023 06:45:06 (UTC/GMT +8 hours)

SQL employee Database: Exercise-12 with Solution

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12. From the following table, write a SQL query to find those employees who joined before 1991. Return complete information about the employees.

Sample table: employees

Pictorial Presentation:

SQL exercises on employee Database: List the employees who joined before 1991

Sample Solution:

SELECT *
FROM employees
WHERE hire_date<('1991-1-1');

Sample Output:

 emp_id | emp_name | job_name | manager_id | hire_date  | salary | commission | dep_id
--------+----------+----------+------------+------------+--------+------------+--------
  63679 | SANDRINE | CLERK    |      69062 | 1990-12-18 | 900.00 |            |   2001
(1 row)

Explanation:

The said query in SQL that selects all columns from the 'employees' table where the hire_date is before January 1, 1991.

The "WHERE" clause filters the results for all employees whose hire date is earlier than January 1, 1991.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Sample Database: employee

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Previous SQL Exercise: List the employees not in department 2001.
Next SQL Exercise: Average salaries of the employees works as ANALYST.

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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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