Sql中的不同行

Question

我有一个记录事件的表，因此可以多次记录相同的ID。我想为每个唯一的RefID返回单行，因为'Y'代替相关列中的'N'。下面是一个模拟

    DECLARE @Ref TABLE 
    (   
      RefID        INT
     , InvoiceNo    INT 
     , InvoicedDate Date
     , CustID       INT 
     , PaidOnTime   CHAR(1)
     , Paidlate     CHAR(1)
     , PaidByCash   CHAR(1) 
     , PaidByCard   CHAR(1)
    )
   INSERT INTO @Ref VALUES
   (23,50,'22-jun-2015', 11,'Y','N','Y','N')
 , (23,50,'22-jun-2015', 11,'Y','N','N','Y')
 , (27,11,'12-Aug-2015', 11,'Y','N','N','Y')
 , (27,11,'22-Aug-2015', 11,'N','Y','N','Y')
 , (45,67,'28-jun-2015', 11,'N','Y','Y','N')
 , (45,67,'28-jun-2015', 11,'N','N','N','Y')
 , (48,51,'18-jun-2015', 11,'Y','N','Y','N') 

 SELECT * FROM @Ref --would return values like so:

例如，23的RefID应为“23,50,22 / 06 / 2015,11，Y，N，Y，Y”

Answer 1

按您想要唯一的列进行分组，并使用max()获取每个组的最高值（因为Y按字母顺序高于N）

SELECT RefID, InvoiceNo, InvoicedDate, CustID,
       max(PaidOnTime), max(Paidlate), max(PaidByCash), max(PaidByCard)
FROM @Ref 
GROUP BY RefID, InvoiceNo, InvoicedDate, CustID