﻿ SQL: List those employees whose salary contain only 3 digits

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

## SQL employee Database: Exercise-32 with Solution

[An editor is available at the bottom of the page to write and execute the scripts.]

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:

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

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

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.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.

﻿

## 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

We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook