How not to calculate a value twice in select? - postgresql

Question

psql (9.6.1, server 9.5.5)
employees
Column | Type | Modifiers | Storage | Stats target | Description
----------------+-----------------------------+-----------------------------------------------------------------+----------+--------------+---- ---------
employee_id | integer | not null default nextval('employees_employee_id_seq'::regclass) | plain | |
first_name | character varying(20) | | extended | |
last_name | character varying(25) | not null | extended | |
email | character varying(25) | not null | extended | |
phone_number | character varying(20) | | extended | |
hire_date | timestamp without time zone | not null | plain | |
job_id | character varying(10) | not null | extended | |
salary | numeric(8,2) | | main | |
commission_pct | numeric(2,2) | | main | |
manager_id | integer | | plain | |
department_id | integer
I need to extract employee number, last name, salary, salary increased by 15.5 % (expressed as a whole number), and the difference between the new and old salary.
I have done like this:
select employee_id,
last_name,
salary,
round(salary * 1.155, 0) as "New Salary",
round(salary * 1.155, 0) - salary as "Increase"
from employees;
What troubles me is that I have calculated the new salary twice.
I tried to use alias in the same select. Experimented like this:
select 2 as val_a, val_a - 4; --not working
Well, my solution outputs acceptable result. But isn't there a better solution?

Answer 1

That calculation is really nothing if you are worried about performance. Some optimizers may even reuse the calculations internally.
If you must do it yourself, you can use subquery like this:
select t.*,
New_Salary - salary as Increase
from (
select employee_id,
last_name,
salary,
round(salary * 1.155, 0) as New_Salary,
from employees
) t;

Answer 2

You can write it with a subquery, if you are "picky" about not computing twice the same values:
SELECT
*, "New Salary" - salary as "Increase"
FROM
(
SELECT
employee_id,
last_name,
salary,
round(salary * 1.155, 0) as "New Salary"
FROM
employees
) AS s0 ;
In practice, the difference when you run it a few times is neglectable:
dbfiddle here