SQL Server - Top Saleperson Per Region - sql-server

Question

SELECT region, person, sum(dollars) as thousands
FROM sales
GROUP BY region, person
ORDER BY region, sum(dollars) desc
The SQL above produces a complete list of sales people per region like this
region person thousands
canada mike smith $114
canada joe blog $76
canada pete dodd $45
usa john doe $253
usa jane smyth $120
europe pieter tsu $546
europ mike lee $520
If I'm only interested in showing the top salesperson per region (as below), how can I best do that?
region person thousands
canada mike smith $114
usa john doe $253
europe pieter tsu $546

Answer 1

I've done something like burnall suggested. I wasn't getting much love with the "top 1 with ties" part, so I made the whole thing a subquery and chose rows where ranking = 1.
select *
from
(
select region,
person,
rank() over(partition by region order by sum(dollars) desc) as ranking
from sales
group by region,
person
) temp
where ranking = 1
Note that this also works for ties since rank() seems to place the same ranking on sums that are equal.

Answer 2

Using Sql Server 2005+ you could do this using a ROW_NUMBER()
Have a look at this full example.
DECLARE #sales TABLE(
region VARCHAR(50),
person VARCHAR(50),
Sales FLOAT
)
INSERT INTO #sales SELECT 'canada','mike smith',1
INSERT INTO #sales SELECT 'canada','mike smith',1
INSERT INTO #sales SELECT 'canada','mike smith',1
INSERT INTO #sales SELECT 'canada','mike smith',1
INSERT INTO #sales SELECT 'canada','joe blog',1
INSERT INTO #sales SELECT 'canada','joe blog',1
INSERT INTO #sales SELECT 'canada','pete dodd',1
INSERT INTO #sales SELECT 'usa','john doe',1
INSERT INTO #sales SELECT 'usa','john doe',1
INSERT INTO #sales SELECT 'usa','jane smyth',1
INSERT INTO #sales SELECT 'europe','pieter tsu',1
INSERT INTO #sales SELECT 'europe','pieter tsu',1
INSERT INTO #sales SELECT 'europe','mike lee',1
;WITH Counts AS(
SELECT region,
person,
count(*) as thousands
FROM #sales
GROUP BY region,
person
), CountVals AS(
SELECT *,
ROW_NUMBER() OVER(PARTITION BY region ORDER BY thousands DESC) ROWID
FROM Counts
)
SELECT *
FROM CountVals
WHERE ROWID = 1

Answer 3

In SQL Server 2005 and above use ROW_NUMBER with PARTITION BY. Following should work (not tested, and probably can be shortened):
WITH total_sales
AS (SELECT region, person, count(*) as thousands
FROM sales
GROUP BY region, person
ORDER BY region, count(*) desc
)
, ranked_sales
AS (SELECT region, person, thousands,
ROW_NUMBER() OVER (PARTITION BY region ORDER BY thousands DESC, person) AS region_rank
FROM total_sales
)
SELECT region, person, thousands
FROM ranked_sales
WHERE region_rank = 1

Answer 4

First of all I do not understand why count(*) is in $.
My solution is similar to existing, but shorter and I believe faster
select top 1 with ties region, person, rank() over(partition by region order by count(*) desc)
from sales
group by region, person
order by 3

Answer 5

This isn't too difficult. This query will do exactly what you want.
select distinct region,
(select top 1 person
from Sales s2 where s2.region = s1.region
group by person
order by SUM(dollars) desc) as person,
(select top 1 SUM(dollars)
from Sales s2 where s2.region = s1.region
group by person
order by SUM(dollars) desc) as thousands
from sales s1

Answer 6

You can use the max() aggregate. It's probably less efficient than the other alternatives because you'll be doing group by twice
SELECT region,person,max(thousands) FROM
(SELECT region, person, count(*) as thousands
FROM sales
GROUP BY region, person) tmp
GROUP BY region, person
ORDER BY region, max(thousands) desc

Answer 7

to find top 5 salesperson by sales for each region
select *
from
(
select region,
[Customer Name],
rank() over(partition by region order by sum(sales) desc) as ranking
from Orders
group by region, [Customer Name]
) temp
where ranking between 1 and 5