Product Final Price after Many Discount given - sql

Question

I have two tables.
One table of Ids and their prices, and second table of discounts per Id.
In the table of discounts an Id can has many Discounts, and I need to know the final price of an Id.
What is the Best way to query it (in one query) ?
The query should be generic for many discounts per id (not only 2 as mentioned below in the example)
For example
Table one
id price
1 2.00
2 2.00
3 2.00
Table two
id Discount
1 0.20
1 0.30
2 0.40
3 0.50
3 0.60
Final result:
id OrigPrice PriceAfterDiscount
1 2.00 1.12
2 2.00 1.20
3 2.00 0.40

Answer 1

Here's another way to do it:
SELECT T1.ID, T1.Price, T1.Price * EXP(SUM(LOG(1 - T2.Discount)))
FROM T1 INNER JOIN T2 ON T1.ID = T2.ID
GROUP BY T1.ID, T1.Price
The EXP/LOG trick is just another way to do multiplication.
If you have entries in T1 without discounts in T2, you could change the INNER JOIN to a LEFT JOIN. You would end up with the following:
ID Price Discount
4 2.00 NULL
Your logic can either account for the null in the discounted price column and take the original price instead, or just add a 0 discount record for those.

Answer 2

Generally it can be done with a trick with LOG/EXP functions but it is complex.
Here is a basic example:
declare #p table(id int, price money)
declare #d table(id int, discount money)
insert into #p values
(1, 2),
(2, 2),
(3, 2)
insert into #d values
(1, 0.2),
(1, 0.3),
(2, 0.4),
(3, 0.5),
(3, 0.6)
select p.id,
p.price,
p.price * ca.discount as PriceAfterDiscount
from #p p
cross apply (select EXP(SUM(LOG(1 - discount))) as discount FROM #d where id = p.id) ca
For simpler(cursor based approach) you will need a recursive CTE, but in this case you need some unique ordering column in Discounts table to run it correctly. This is shown in #Tanner`s answer.
And finally you can approach this with a regular cursor

Answer 3

I believe this produces the desired results using a CTE to iterate through the discounts. The solution below is re-runnable in isolation.
Edited: to include data that might not have any discounts applied in the output with a left join in the first part of the CTE.
CREATE TABLE #price
(
id INT,
price DECIMAL(5, 2)
);
CREATE TABLE #discount
(
id INT,
discount DECIMAL(5, 2)
);
INSERT INTO #price
(
id,
price
)
VALUES
(1, 2.00),
(2, 2.00),
(3, 2.00),
(4, 3.50); -- no discount on this item
INSERT INTO #discount
(
id,
discount
)
VALUES
(1, 0.20),
(1, 0.30),
(2, 0.40),
(3, 0.50),
(3, 0.60);
-- new temporary table to add a row number to discounts so we can iterate through them
SELECT d.id,
d.discount,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY d.discount) rn
INTO #GroupedDiscount
FROM #discount AS d;
-- note left join in first part of cte to get prices that aren't discounted included
WITH cte
AS (SELECT p.id,
p.price,
CASE
WHEN gd.discount IS NULL THEN
p.price
ELSE
CAST(p.price * (1.0 - gd.discount) AS DECIMAL(5, 2))
END AS discountedPrice,
gd.rn
FROM #price AS p
LEFT JOIN #GroupedDiscount AS gd
ON gd.id = p.id
AND gd.rn = 1
UNION ALL
SELECT cte.id,
cte.price,
CAST(cte.discountedPrice * (1.0 - gd.discount) AS DECIMAL(5, 2)) AS discountedPrice,
cte.rn + 1 AS rn
FROM cte
INNER JOIN #GroupedDiscount AS gd
ON gd.id = cte.id
AND gd.rn = cte.rn + 1
)
SELECT cte.id,
cte.price,
MIN(cte.discountedPrice) AS discountedPrice
FROM cte
GROUP BY id,
cte.price;
DROP TABLE #price;
DROP TABLE #discount;
DROP TABLE #GroupedDiscount;
Results:
id price discountedPrice
1 2.00 1.12
2 2.00 1.20
3 2.00 0.40
4 3.50 3.50 -- no discount

Answer 4

As others have said, EXP(SUM(LOG())) is the way to do the calculation. Here is basically same approach from yet another angle:
WITH CTE_Discount AS
(
SELECT Id, EXP(SUM(LOG(1-Discount))) as TotalDiscount
FROM TableTwo
GROUP BY id
)
SELECT t1.id, CAST(Price * COALESCE(TotalDiscount,1) AS Decimal(18,2)) as FinalPRice
FROM TableOne t1
LEFT JOIN CTE_Discount d ON t1.id = d.id
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