How to make a query showing purchases of a client on the same day, but only if those were made in diffrent stores (oracle)? - sql

Question

I want to show cases of clients with at least 2 purchases on the same day. But I only want to count those purchases that were made in different stores.
So far I have:
Select Purchase.PurClientId, Purchase.PurDate, Purchase.PurId
from Purchase
join
(
Select count(Purchase.PurId),
Purchase.PurClientId,
to_date(Purchase.PurDate)
from Purchases
group by Purchase.PurClientId,
to_date(Purchase.PurDate)
having count (Purchase.PurId) >=2
) k
on k.PurClientId=Purchase.PurClientId
But I have no clue how to make it count purchases only if those were made in different stores. The column which would allow to identify shop is Purchase.PurShopId.
Thanks for help!

Answer 1

You can use:
SELECT PurId,
PurDate,
PurClientId,
PurShopId
FROM (
SELECT p.*,
COUNT(DISTINCT PurShopId) OVER (
PARTITION BY PurClientId, TRUNC(PurDate)
) AS num_stores
FROM Purchase p
)
WHERE num_stores >= 2;
Or
SELECT *
FROM Purchase p
WHERE EXISTS(
SELECT 1
FROM Purchase x
WHERE p.purclientid = x.purclientid
AND p.purshopid != x.purshopid
AND TRUNC(p.purdate) = TRUNC(x.purdate)
);
Which, for the sample data:
CREATE TABLE purchase (
purid PRIMARY KEY,
purdate,
purclientid,
PurShopId
) AS
SELECT 1, DATE '2021-01-01', 1, 1 FROM DUAL UNION ALL
SELECT 2, DATE '2021-01-02', 1, 1 FROM DUAL UNION ALL
SELECT 3, DATE '2021-01-02', 1, 2 FROM DUAL UNION ALL
SELECT 4, DATE '2021-01-03', 1, 1 FROM DUAL UNION ALL
SELECT 5, DATE '2021-01-03', 1, 1 FROM DUAL UNION ALL
SELECT 6, DATE '2021-01-04', 1, 2 FROM DUAL;
Both output:
PURID
PURDATE
PURCLIENTID
PURSHOPID
2
2021-01-02 00:00:00
1
1
3
2021-01-02 00:00:00
1
2
db<>fiddle here