SQL Grouping and dense rank concept - sql

Question

I have a data set that looks like:
cust city hotel_id amount
-------------------------------
A 1 252 3160
B 1 256 1893
C 2 105 2188
D 2 105 3054
E 3 370 6107
F 2 110 3160
G 2 150 1893
H 3 310 2188
I 1 252 3160
J 1 250 4000
K 3 370 5000
L 3 311 1095
Query to display the top 3 hotels by revenue (Sum of amount) for each city?
Since same hotel can be booked by other customer in same city so we need to sum the amount to find total amount.
Expected output:
city hotel_id amount
---------------------------
1 252 6320
1 250 4000
1 256 1893
2 105 5242
2 110 3160
2 150 1893
3 370 11107
3 310 2188
3 311 1095

Answer 1

SELECT
t.city, t.hotel_id, t.amount
FROM
(
SELECT city, hotel_id, SUM(amount) AS amount,
ROW_NUMBER() OVER (PARTITION BY city ORDER BY SUM(amount) DESC) AS rn
FROM yourTable
GROUP BY city, hotel_id
) t
WHERE t.rn <= 3
ORDER BY t.city, t.amount DESC;
Demo here:
Rextester

Answer 2

To get the total sum for each hotel_id you need to group by that column first, then group by the city for syntax purposes. The #tmp table here should have all of the data you need, so then you just have to select the top 3 entries for each city from there.
SELECT city, hotel_id, SUM(amount) AS 'total' INTO #tmp
FROM [table]
GROUP BY hotel_id, city
(SELECT TOP 3 *
FROM #tmp
WHERE city = 1)
UNION
(SELECT TOP 3 *
FROM #tmp
WHERE city = 2)
UNION
(SELECT TOP 3 *
FROM #tmp
WHERE city = 3)