Optimize subquery in SELECT - mysql

Question

My table schema is as follow:
Indexes:
products.id PRIMARY KEY
products.description UNIQUE
expenses.id PRIMARY KEY
expenses.product_id FOREIGN KEY to product.id
My goal is to load
Cost of each product of current month (AS costs_november)
Cost of each product of last month (AS costs_october)
Change in costs of current month compared to last (current month costs - last month costs) (AS costs)
Percentage change of current month costs compared to last (last month costs * 100 / current month costs) (AS percent_diff)
I've managed to code SQL that does exactly that:
SELECT description, (SUM(cost) - IFNULL(
(
SELECT SUM(cost)
FROM expenses
WHERE month = 9 AND year = 2019 AND product_id = e.product_id
GROUP BY product_id
), 0)) AS costs,
SUM(cost) * 100 /
(
SELECT SUM(cost)
FROM expenses
WHERE month = 9 AND year = 2019 AND product_id = e.product_id
GROUP BY product_id
) AS percent_diff,
SUM(cost) AS costs_october,
IFNULL(
(
SELECT SUM(cost)
FROM expenses
WHERE month = 9 AND year = 2019 AND product_id = e.product_id
GROUP BY product_id
), 0) AS costs_september
FROM expenses e
JOIN products p ON (e.product_id = p.id)
WHERE month = 10 AND year = 2019
GROUP BY product_id
ORDER BY product_id;
But is copy-pasting the same subquery three times really the solution? In theory it requires to run four queries per product. Is there a more elegant way?
Appreciate for any help!

Answer 1

I would address this with conditional aggregation:
select
p.description,
sum(case when e.month = 11 then e.cost else 0 end) costs_november,
sum(case when e.month = 10 then e.cost else 0 end) costs_october,
sum(case when e.month = 11 then e.cost else -1 * e.cost end) costs,
sum(case when e.month = 10 then e.cost else 0 end)
* 100
/ nullif(
sum(case when e.month = 11 then e.cost else 0 end),
0
) percent_diff
from expenses e
inner join products p on p.id = e.product_id
where e.year = 2019 and e.month in (10, 11)
goup by e.product_id
You can avoid repeating the same conditional sums by using a subquery (your RDBMS would probably optimize it anyway, but this tends to make the query more readable):
select
description,
costs_november,
costs_october,
costs_november - costs_october costs,
costs_october * 100 / nullif(costs_november, 0) percent_diff
from (
select
p.description,
sum(case when e.month = 11 then e.cost else 0 end) costs_november,
sum(case when e.month = 10 then e.cost else 0 end) costs_october
from expenses e
inner join products p on p.id = e.product_id
where e.year = 2019 and e.month in (10, 11)
goup by e.product_id
) t

Answer 2

You can calculate for all months and all products at one time:
SELECT year, month,
SUM(costs) as curr_month_costs,
LAG(SUM(costs)) OVER (ORDER BY year, month) as prev_month_costs,
(SUM(costs) -
LAG(SUM(costs)) OVER (ORDER BY year, month)
) as diff,
LAG(SUM(costs)) OVER (ORDER BY year, month) * 100 / SUM(costs)
FROM expenses e JOIN
products p
ON e.product_id = p.id
GROUP BY product_id, year, month
ORDER BY year, month, product_id;
You can use a subquery if you want to select only the current month.