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
Was it helpful?

Solution

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.

OTHER TIPS

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

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

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

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

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

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
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