SQL Exercise: List the employees whose names contain AR together

Last update on April 25 2023 12:57:10 (UTC/GMT +8 hours)

SQL employee Database: Exercise-109 with Solution

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109. From the following table, write a SQL query to find those employees whose names contain the character set 'AR' together. Return complete information about the employees.

Sample table: employees

Sample Solution:

SELECT *
FROM employees
WHERE emp_name LIKE '%AR%';

Sample Output:

 emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------
  67832 | CLARE    | MANAGER  |      68319 | 1991-06-09 | 2550.00 |            |   1001
  67858 | SCARLET  | ANALYST  |      65646 | 1997-04-19 | 3100.00 |            |   2001
  69324 | MARKER   | CLERK    |      67832 | 1992-01-23 | 1400.00 |            |   1001
(3 rows)

Explanation:

The said query in SQL that selects all columns from the 'employees' table where the "emp_name" column contains the characters "AR" anywhere within the string. The "%" symbol is a wildcard that matches any sequence of characters.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Sample Database: employee

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Previous SQL Exercise: Employees joined with a as the 2nd character of a month.
Next SQL Exercise: List the employees those who joined in 90's.

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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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