SQL Exercise: Employees joining on given days, ascending in seniority

Last update on April 24 2023 12:28:44 (UTC/GMT +8 hours)

SQL employee Database: Exercise-72 with Solution

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72. From the following table, write a SQL query to find those employees who joined on 1-MAY-91, or 3-DEC-91, or 19-JAN-90. Sort the result-set in ascending order by hire date. Return complete information about the employees.

Pictorial Presentation:

SQL exercises on employee Database: List the employees in ascending order of seniority who joined on 1-MAY-91,or 3-DEC-91, or 19-JAN-90

Sample table: employees

Sample Solution:

SELECT *
FROM employees
WHERE hire_date IN ('1991-5-01',
                    '1991-12-03',
                    '1990-01-19')
ORDER BY hire_date ASC;

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
  69062 | FRANK    | ANALYST  |      65646 | 1991-12-03 | 3100.00 |            |   2001
  69000 | JULIUS   | CLERK    |      66928 | 1991-12-03 | 1050.00 |            |   3001
(3 rows)

Explanation:

The said query in SQL that returns all the rows and columns from the 'employees' table where the "hire_date" column matches either '1991-5-01', '1991-12-03', or '1990-01-19', and sorts the result set in ascending order on "hire_date" column.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Sample Database: employee

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Previous SQL Exercise: Display the location of CLARE.
Next SQL Exercise: Sort the employees by salary less than 1000.

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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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