Question

我遇到了返回前N对产品的问题，客户共同购买。假设我有以下查询：

SELECT c.CustomerKey, ProductAlternateKey
FROM 
    [AdventureWorksDW2008R2].[dbo].[FactInternetSales] FIS JOIN
    dbo.DimCustomer C ON C.CustomerKey = FIS.CustomerKey JOIN
    dbo.DimProduct P ON P.ProductKey = FIS.ProductKey

GROUP BY 
    C.CustomerKey, P.ProductAlternateKey

它返回以下结果：

CustomerKey  | ProductAlternateKey| 
-------------+-----------+
0001         | A                  |
0001         | B                  |
0002         | A                  | 
0002         | B                  |
0002         | C                  |
0003         | B                  |
0003         | C                  |
0004         | A                  |
0004         | B                  |
0005         | A                  |
0005         | C                  |

从这个查询我想找到客户购买的前2对产品，所以在我的例子中 - ＆gt; （A，B）3次出现和（B，C） - > 2次出现

Answer 1

使用RANK或DENSERANK功能。排名函数的MSDN：http://msdn.microsoft.com/en-us/library/ms176102.aspx

使用自解压SQL的示例：

declare @Person Table ( personID int identity, person varchar(8));

insert into @Person values ('Brett'),('Sean'),('Chad'),('Michael'),('Ray'),('Erik'),('Queyn');

declare @Orders table ( OrderID int identity, PersonID int, Desciption varchar(32), Amount int);

insert into @Orders values (1, 'Shirt', 20),(1, 'Shoes', 50),(2, 'Shirt', 22),(2, 'Shoes', 20),(3, 'Shirt', 20),(3, 'Shoes', 50),(3, 'Hat', 20),(4, 'Shirt', 20),(5, 'Shirt', 20),(5, 'Pants', 30),
(6, 'Shirt', 20),(6, 'RunningShoes', 70),(7, 'Shirt', 22),(7, 'Shoes', 40),(7, 'Coat', 80);

with a as 
    (
    Select
        person
    ,   o.Desciption
    ,   o.Amount
    ,   rank() over(partition by p.personId order by Amount) as Ranking
    ,   Dense_rank() over(partition by p.personId order by Amount) as DenseRanking
    from @Person p
        join @Orders o on p.personID = o.PersonID
    )
select *
from a 
where Ranking <= 2  -- determine top 2, 3, etc.... whatever you want.
order by person, amount

Answer 2

给这个测试：

declare @Orders table ( OrderID int identity, PersonID int,  Description varchar(32), ProductId int, Amount int);

insert into @Orders values (1, 'Shirt', 1, 20),(1, 'Shoes', 2, 50),(2, 'Shirt', 1, 22),(2, 'Shoes', 2, 20),(3, 'Shirt', 1, 20),(3, 'Shoes', 2, 50),(3, 'Hat',   3, 20),(4, 'Shirt', 1, 20),(5, 'Shirt', 1, 20),(5, 'Pants', 4, 30),(6, 'Shirt', 1, 20),(6, 'RunningShoes', 5, 70),(6, 'Hat', 3, 20),(7, 'Shirt', 1, 22),(7, 'Shoes', 2, 40),(7, 'Coat', 6, 80);

select
     top 2       
     I.Description,
     M.Description
from @Orders as I
inner join @Orders as C on C.ProductId = I.ProductId   and C.PersonId <> I.PersonId
inner join @Orders as M on C.PersonId = M.PersonId and C.ProductId <> M.ProductId
where 
    I.productid = 3
    and I.ProductId<>M.ProductId
group by 
    I.Description
    ,M.Description
order
    by count(2) desc

获得TOP N对产品，客户通常购买

2 个答案: