﻿ SQL: Employee names and salaries are increased by 15% in \$

# SQL Exercise: Employee names and salaries are increased by 15% in \$

## SQL employee Database: Exercise-4 with Solution

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4. From the following table, write a SQL query to list the employees’ name, increased their salary by 15%, and expressed as number of Dollars.

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT emp_name,
to_char(1.15*salary,'\$99,999') AS "Salary"
FROM employees;
``````

Sample Output:

``` emp_name |  Salary
----------+----------
KAYLING  | \$  6,900
BLAZE    | \$  3,163
CLARE    | \$  2,933
JONAS    | \$  3,401
SCARLET  | \$  3,565
FRANK    | \$  3,565
SANDRINE | \$  1,035
TUCKER   | \$  1,840
JULIUS   | \$  1,208
MARKER   | \$  1,610
(14 rows)
```

Explanation:

The provided query in SQL that retrieves names of all the employees, and their corresponding salaries, adjusted by a 15% increase aliased as "Salary". The AS keyword is used to assign the name "Salary" to the newly created column.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Display the unique designations for the employees.
Next SQL Exercise: Produce the output of employees name and job name.

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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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