Question

我有2张桌子，一张带有销售，一张带有公司：

销售表

Transaction_Id   Shop_id     Sale_date    Client_ID

92356            24234       11.09.2018    12356
92345            32121       11.09.2018    32121
94323            24321       11.09.2018    21231
94278            45321       11.09.2018    42123

公司表

Client_ID  Company_name

12345      ABC 
13322      ABC  
32321      BCD
22221      BCD

我想要实现的是每对商店中每个公司的客户数（在两个商店中至少有一笔交易的客户）：

Shop_Id_1     Shop_id_2   Company_name    Count(distinct Client_id)

12356         12345       ABC             31
12345         14278       ABC             23
14323         12345       BCD             32
14278         12345       BCD             43

我认为我必须使用自我联接，但是即使使用过滤器一个星期，我的查询也会杀死DB，对此有什么想法吗？我正在使用Microsoft SQL Server 2012。

谢谢

Answer 1

我认为您的问题有些问题。我将其解释为公司表包含商店ID，而不是ClienId。

首先，您可以创建一个解决方案，以将商店作为每个公司的行。在这里，我每个公司最多选择5家商店。不要忘记在CTE之前的上一个语句中的分号。

WITH CTE_Comp AS 
(
    SELECT *, ROW_NUMBER() OVER (PARTITION BY CompanyName ORDER BY ShopID) AS RowNumb 
    FROM Company AS C
)

SELECT C1.ShopID, 
      C2.ShopID AS ShopID_2, 
      C3.ShopID AS ShopID_3, 
      C4.ShopID AS ShopID_4, 
      C5.ShopID AS ShopID_5, 
      C1.CompanyName
INTO ShopsByCompany
FROM CTE_Comp AS C1
    LEFT JOIN CTE_Comp AS C2 ON C1.CompanyName= C2.CompanyName AND RowNumb = 2 
    LEFT JOIN CTE_Comp AS C2 ON C1.CompanyName= C3.CompanyName AND RowNumb = 3 
    LEFT JOIN CTE_Comp AS C2 ON C1.CompanyName= C4.CompanyName AND RowNumb = 4 
    LEFT JOIN CTE_Comp AS C2 ON C1.CompanyName= C5.CompanyName AND RowNumb = 5 
WHERE C1.RowNumb = 1

在那之后，只需几个步骤，我认为您可以获得预期的结果：

WITH ClientsPerShop AS 

(
    SELECT ShopID,
         COUNT (DISTINCT ClientID) AS TotalClients
    FROM Sales
    GROUP BY ShopID
)

, ClienstsPerCompany AS 

(
    SELECT CompanyName,
         SUM (TotalClients) AS ClientsPerComp
    FROM Company AS C
    INNER JOIN ClientsPerShop AS CPS ON C.ShopID = CPS.ShopID
    GROUP BY CompanyName
)

SELECT * 
FROM ClienstsPerCompany AS CPA
INNER JOIN ShopsByCompany AS SBC ON SBC.CompanyName = CPA.CompanyName

希望这将使您更接近您的解决方案，祝您好运！

Answer 2

我认为这是一种自我结合和聚合，但又有所不同。不同之处在于，您希望将公司包含在每个sales记录中，以便可以在自联接中使用它：

with sc as (
      select s.*, c.company_name
      from sales s join
           companies c
           on s.client_id = c.client_id
     )
select sc1.shop_id, sc2.shop_id, sc1.company_name, count(distinct sc1.client_id)
from sc sc1 join
     sc sc2
     on sc1.client_id = sc2.client_id and
        sc1.company_name = sc2.company_name
group by sc1.shop_id, sc2.shop_id, sc1.company_name;

每个公司的最受欢迎的商店对工人

2 个答案: