Question

我需要找到订单相同的客户。（使用T-SQL）

订单

OrderID Customerer  
1   2   
2   5   
3   6   
4   2   
5   4   
6   6   
7   8

OrderLine

OrderLineID OrderID OrderDate   OrderType   Quantity    Reference  
1   1   01/01/2011  1   1   Coca Cola  
2   1   01/01/2011  1   3   Tea  
3   2   02/02/2011  2   1   Coffee  
4   2   02/02/2011  2   2   Solo  
5   2   03/02/2011  1   1   Soda  
6   3   03/02/2011  1   3   Tea  
7   3   03/02/2011  1   1   Coca Cola  
8   4   05/06/2011  1   1   Beer  
9   5   06/06/2011  2   1   Tea  
10  5   06/06/2011  2   1   Coca Cola  
11  6   07/07/2011  1   1   Coffee  
12  6   07/07/2011  1   2   Solo  
13  6   07/07/2011  1   1   Soda  
14  6   07/07/2011  1   1   Beer  
15  7   08/08/2011  1   1   Beer

此处订单ID为1和3的订单被认为是相同的，因为订单行的数量，“数量”和“参考”在两个订单上都是相同的。意味着客户2和6已经下了相同的订单。

订单5与订单1和3不同，因为数量不同。

订单2与订单6不同，因为订单行不同。

第4和第7顺序也相同。

我正在寻找这样的结果：

IdenticalOrders

OrderID CustomeerID  
1   2  
3   6  
4   2  
7   8

这似乎是一项简单的任务，但我无法理解从哪里开始。（我还是t-sql的新手:-)）

Answer 1

这是一种方式。

SELECT  O1.OrderID ,
        O1.Customer ,
        O2.OrderID ,
        O2.Customer
FROM    [Order] O1
        JOIN [Order] O2 ON O1.OrderID < O2.OrderID
                           AND O1.Customer <> O2.Customer
WHERE   NOT EXISTS ( SELECT Quantity ,
                            Reference
                     FROM   OrderLine
                     WHERE  O1.OrderID = OrderLine.OrderID
                     EXCEPT
                     SELECT Quantity ,
                            Reference
                     FROM   OrderLine
                     WHERE  O2.OrderID = OrderLine.OrderID )
        AND NOT EXISTS ( SELECT Quantity ,
                                Reference
                         FROM   OrderLine
                         WHERE  O2.OrderID = OrderLine.OrderID
                         EXCEPT
                         SELECT Quantity ,
                                Reference
                         FROM   OrderLine
                         WHERE  O1.OrderID = OrderLine.OrderID )

您还可以使用XML PATH来模拟GROUP_CONCAT，然后JOIN两个结果集

DECLARE @T TABLE
    (
      OrderId INT PRIMARY KEY,
      Customer INT ,
      complete_order VARCHAR(MAX)
    )  

    INSERT  INTO @T
            SELECT  *
            FROM    [Order] O
                    CROSS APPLY ( SELECT    CAST(Quantity AS VARCHAR(30))
                                            + '~' + Reference + '~~'
                                  FROM      OrderLine OL
                                  WHERE     OL.OrderID = O.OrderID
                                  ORDER BY  Reference ,
                                            Quantity
                                FOR
                                  XML PATH('')
                                ) T ( complete_order )

SELECT  T1.OrderId, 
        T1.Customer
FROM    @T T1
WHERE   EXISTS ( SELECT *
                 FROM   @T T2
                 WHERE  T1.Customer <> T2.Customer
                        AND T1.OrderId <> T2.OrderId
                        AND T1.complete_order = T2.complete_order )

Answer 2

这是马丁第二个建议的延伸。这将显示所有匹配的组合而不重复。

;With FmtOL(customer, orderid, complete_order) as
(            
SELECT  customer, orderid, complete_order
     FROM    Order O
                    cross apply ( SELECT    CAST(Quantity AS VARCHAR(30))
                                            + '~' + Reference + '~~'
                                  FROM      OrderLine OL
                                  WHERE     OL.OrderID = O.OrderID
                                  ORDER BY  Reference ,
                                            Quantity
                                FOR
                                  XML PATH('')
                                ) T ( complete_order )
)
SELECT  T1.OrderId, 
        T1.Customer,
        STUFF(C1.a, 1, 2, '') as [SameAs]
FROM    FmtOL T1
Cross apply ( SELECT '; ' + 'Customer ' + Cast(T2.Customer as varchar(30)) 
                  + '''s order ' + Cast(T2.OrderID as varchar(30))
                 FROM   FmtOL T2
                 WHERE  T1.Customer < T2.Customer
                   AND T1.OrderId < T2.OrderId
                   AND T1.complete_order = T2.complete_order
                 order by ';' + Cast(T2.Customer as varchar(30)) 
                    + '''s order ' + Cast(T2.OrderID as varchar(30))
                    , t2.orderid
                 for xml path('')
             ) C1 (a)
  where C1.a is not null

结果应如下所示：

OrderId Customer SameAs 
1       2        Customer 6's order 3 
4       2        Customer 8's order 7

Answer 3

这是最简单的方法。

-- sample table
create table x
    (
         LineId int identity(1, 1)
        ,InvoiceFk int
        ,ProductFk int
        ,Quantity int
    )

-- sample data
insert into x 
(InvoiceFk, ProductFk, Quantity) values
     (11, 1, 1)
    ,(11, 2, 1)
    ,(11, 3, 1)
    ,(12, 1, 2)
    ,(12, 2, 2)
    ,(12, 3, 2)
    ,(13, 1, 3)
    ,(13, 2, 3)
    ,(13, 3, 3)

-- your order, probably from a parameter
declare @order table
    (
         InvoiceFk int
        ,ProductFk int
        ,Quantity int
    )
insert into @order
(InvoiceFk, ProductFk, Quantity) values
     (14, 1, 1) -- duplicate invoice 11
    ,(14, 2, 1)
    ,(14, 3, 1)

-- your order unique checksum
declare @orderCheck int
select @orderCheck = checksum_agg(checksum(ProductFk, Quantity))
from @order

-- test your order in existing data
declare @match int
select @match = 
    (
        select TOP 1 InvoiceFk from
        (
            select
                 InvoiceFk
                ,checksum_agg(Col1) as Col2
            from
                (
                select 
                     InvoiceFk
                    ,checksum(productfk, quantity) as Col1
                from x
                ) as T1
            group by
                InvoiceFk
        ) as T2
        where 
            T2.Col2 = @orderCheck
    )


-- evaluate if your order is unique or not
if (@match is not null)
begin
    print 'Identical to invoice: ' + Str(@match);
end
else
begin
    print 'Order is unique';
end

-- clean up sample table
drop table x

祝你好运！

找到订单相同的客户

3 个答案: