How to get biggest differences in column compared to last month? - mysql

Question

I have two tables - Products and Prices
Every month table Prices is populated with new prices for each products. How can I get 5 products whose prices have the biggest incremental difference from last month prices?
table Products
id | name
1 | apples
2 | pears
3 | bananas
table Prices
id | price | product_id | created_at
1 | 10 | 1 | 2017-02-07 07:00:00
2 | 10 | 2 | 2017-02-07 07:00:00
3 | 15 | 3 | 2017-02-07 07:00:00
5 | 15 | 1 | 2017-03-07 07:00:00
6 | 20 | 2 | 2017-03-07 07:00:00
7 | 25 | 3 | 2017-03-07 07:00:00
The result would be to find out that
1. Bananas has prices by 15 higher (lastMonth: 15, now: 25)
2. Pears 2 has prices by 10 higher (lastMonth: 10, now: 20)
3. Apples has prices by 5 higher (lastMonth: 10, now: 15)
I was thinking something like this (uff I know this is terrible)
SELECT products.id, products.name, prices.beforePrice, prices.afterPrice, prices.difference
FROM products
INNER JOIN prices ON products.id = prices.product_id
WHERE
(
SELECT *biggest-difference*
FROM prices
WHERE *difference_between_last_2_months*
GROUP BY product_id
LIMIT 5
)

Answer 1

Create table/insert data
CREATE TABLE Products
(`id` INT, `name` VARCHAR(7))
;
INSERT INTO Products
(`id`, `name`)
VALUES
(1, 'apples'),
(2, 'pears'),
(3, 'bananas')
;
CREATE TABLE Prices
(`id` INT, `price` INT, `product_id` INT, `created_at` DATETIME)
;
INSERT INTO Prices
(`id`, `price`, `product_id`, `created_at`)
VALUES
(1, 10, 1, '2017-02-07 07:00:00'),
(2, 10, 2, '2017-02-07 07:00:00'),
(3, 15, 3, '2017-02-07 07:00:00'),
(5, 15, 1, '2017-03-07 07:00:00'),
(6, 20, 2, '2017-03-07 07:00:00'),
(7, 25, 3, '2017-03-07 07:00:00')
;
Query
SELECT
Products.id
, Products.name
, (current_month.price - last_month.price) AS difference
, (
CASE
WHEN last_month.price > current_month.price
THEN 'lower'
WHEN last_month.price < current_month.price
THEN 'higher'
END
) AS incremental
, last_month.price 'lastMonth'
, current_month.price 'now'
FROM (
SELECT
*
FROM
Prices
WHERE
MONTH(created_at) = MONTH((CURDATE() - INTERVAL 1 MONTH))
)
AS
last_month
INNER JOIN (
SELECT
*
FROM
Prices
WHERE
MONTH(created_at) = MONTH((CURDATE()))
)
AS
current_month
ON
last_month.product_id = current_month.product_id
INNER JOIN
Products
ON
last_month.product_id = Products.id
WHERE
last_month.price < current_month.price #incremental should be higher
ORDER BY
difference DESC
LIMIT 5
Result
id name difference incremental lastMonth now
------ ------- ---------- ----------- --------- --------
2 pears 10 higher 10 20
3 bananas 10 higher 15 25
1 apples 5 higher 10 15

Answer 2

You can use a proper joins based on fliterd select by month.
This should return the value you need ( you can add the literal string you need )
select p.name, m1.price as this_month, m2.price as prev_month, m2.price-m1.price as diff
from product
left join (
select price, product_id
from Prices
where month(created_at) = month(NOW())
and year(created_at) = year(NOW())
) m1 on m1.product_id = p.id
left join (
select price, product_id
from Prices
where month(created_at) = MONT(DATE_SUB(NOW(), INTERVAL 1 MONTH)
and year(created_at) = year(DATE_SUB(NOW(), INTERVAL 1 MONTH)
) m2 on m2.product_id = p.id
order by diff desc
limit 5