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SQL Exercise: List those employees whose salary contain only 3 digits

SQL employee Database: Exercise-32 with Solution

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32. From the following table, write a SQL query to identify employees whose salaries contain only three digits. Return complete information about the employees.

Sample table: employees


Pictorial Presentation:

SQL exercises on employee Database: List those employees whose salary contain only 3 digits

Sample Solution:

SELECT *
FROM employees
WHERE length(TRIM(TO_CHAR(salary,'9999'))) = 3;

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

Explanation:

The said query in SQL that selects all columns from the table 'employees' where the length of the salary value, after converting it to a 4-character string with leading zeros, is equal to 3.

The TO_CHAR converts the "salary" column to a character string using the format mask "9999", which means that the salary value will be displayed with leading zeros if necessary to make it 4 digits long. The TRIM function removes any leading or trailing spaces from the resulting string.

The WHERE clause checks if the length of the resulting string after trimming any spaces is exactly 3 characters long.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: List those employees whose salary is an odd value.
Next SQL Exercise: List the employees who joined in the month of APRIL.

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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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