﻿ SQL: Employees in the ascending order on annual salary

# SQL Exercise: Employees in the ascending order on annual salary

## SQL employee Database: Exercise-69 with Solution

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69. From the following table, write a SQL query to rank the employees according to their annual salary in ascending order. Return employee ID, employee name, monthly salary, salary/30 as Daily_Salary, and 12*salary as Anual_Salary.

Pictorial Presentation:

Sample table: employees

Sample Solution:

``````SELECT emp_id,
emp_name,
salary Monthly_Salary,
salary/30 Daily_Salary,
12*salary Anual_Salary
FROM employees
ORDER BY Anual_Salary ASC;
``````

Sample Output:

``` emp_id | emp_name | monthly_salary |     daily_salary     | anual_salary
--------+----------+----------------+----------------------+--------------
63679 | SANDRINE |         900.00 |  30.0000000000000000 |     10800.00
69000 | JULIUS   |        1050.00 |  35.0000000000000000 |     12600.00
68736 | ADNRES   |        1200.00 |  40.0000000000000000 |     14400.00
65271 | WADE     |        1350.00 |  45.0000000000000000 |     16200.00
66564 | MADDEN   |        1350.00 |  45.0000000000000000 |     16200.00
69324 | MARKER   |        1400.00 |  46.6666666666666667 |     16800.00
68454 | TUCKER   |        1600.00 |  53.3333333333333333 |     19200.00
64989 | ADELYN   |        1700.00 |  56.6666666666666667 |     20400.00
67832 | CLARE    |        2550.00 |  85.0000000000000000 |     30600.00
66928 | BLAZE    |        2750.00 |  91.6666666666666667 |     33000.00
65646 | JONAS    |        2957.00 |  98.5666666666666667 |     35484.00
69062 | FRANK    |        3100.00 | 103.3333333333333333 |     37200.00
67858 | SCARLET  |        3100.00 | 103.3333333333333333 |     37200.00
68319 | KAYLING  |        6000.00 | 200.0000000000000000 |     72000.00
(14 rows)
```

Explanation:

The said query in SQL that retrieves the employee ID, name, monthly salary, daily salary, and annual salary from the 'employees' table and order the records based on the annual salary in ascending order.

The salary column is aliased as "Monthly_Salary", the salary column is divided by 30 is aliased as "Daily_Salary", and the salary column is multiplied by 12 is aliased as "Anual_Salary".

The results then arranged in ascending order on "Anual_Salary" column. This means that the employee with the lowest annual salary will come first.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Display all the unique job in descending order.
Next SQL Exercise: Employees who are CLERK, ANALYST in descending order.

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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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