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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:

SQL exercises on employee Database: List the emp_name and salary is increased by 15% and expressed as no.of Dollars

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
 ADELYN   | $  1,955
 WADE     | $  1,553
 MADDEN   | $  1,553
 TUCKER   | $  1,840
 ADNRES   | $  1,380
 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

employee database structure

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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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