PARTITION based on a column and GROUP BY another column [duplicate] - sql

Question

This question already has answers here:
Get top 1 row of each group
(19 answers)
Select top 10 records for each category
(14 answers)
Closed 6 months ago.
Please consider the below script:
declare #tbl Table
(
CustomerId INT,
CountryID int,
Amount int
);
insert into #tbl values
(1,1,100),
(1,2,200),
(1,3,300),
(1,4,400),
(2,1,800),
(2,1,1000),
(3,1,500),
(2,4,200),
(2,3,900),
(3,1,3000),
(5,1,100),
(5,2,200),
(5,4,5000),
(6,1,1000),
(6,3,900),
(7,2,400),
(8,3,4000),
(2,1,100),
(1,1,100)
Declare #Result Table
(
CountryID int,
CustomerID int,
SumAmount int
);
Declare #CountryID int;
DECLARE db_cursor CURSOR FOR
SELECT distinct CountryID
FROM #tbl
OPEN db_cursor
FETCH NEXT FROM db_cursor INTO #CountryID
WHILE ##FETCH_STATUS = 0
BEGIN
insert into #Result
select top 2 #CountryID, CustomerID, SUM(Amount)
from #tbl
where CountryID = #CountryID
group by CustomerId
order by 3 desc
FETCH NEXT FROM db_cursor INTO #CountryID
END
CLOSE db_cursor
DEALLOCATE db_cursor
select *
from #Result
It returns this result :
CountryID CustomerID SumAmount
----------------------------------
1 3 3500
1 2 1900
2 7 400
2 5 200
3 8 4000
3 6 900
4 5 5000
4 1 400
In fact I want to get Top 2 customers that have maximum Amount in each Country.
How can I get that result without CURSOR and single query?
Thanks

Answer 1

The solution is :
WITH T AS
(
SELECT CountryID, CustomerId, SUM(Amount) AS SumAmount,
RANK() OVER(PARTITION BY CountryID ORDER BY SUM(Amount) DESC) AS R
FROM #tbl
GROUP BY CountryID, CustomerId
)
SELECT *
FROM T
WHERE R <= 2
But remember that when you want a top n rank , you will not systematically have exactly n rows returning because of ex aequo... You can have more, you can have less, depending of which ranking function you use and how many equal mesure you are ranking...