Question

我有两张桌子：

tbl A           tbl B
ID | Value      ID |  tblA_ID  | Qty
1  | CGK        1  |  2,3      | 100
2  | BNK        2  |  2        | 85
3  | DPK        3  |  1,3,4    | 200
4  | CMG        4  |  1,2      | 120

加入时我想要这个输出：

 tblBID|  tblA_Values  | Qty
 1     |  BNK,DPK      | 100
 2     |  BNK          | 85
 3     |  CGK,DPK,CMG  | 200
 4     |  CGK,BNK      | 120

我不知道搜索问题和答案的关键字，这就是我发布此问题的原因。

Answer 1

不仅仅是简单的加入。实际上有三个主要部分需要关注。

1）解析或拆分数据

2）使用tblA

加入已解析的结果

3）通过XML / Stuff（）

重建字符串

如果您无法使用UDF，则可以轻松将其修改为内联流程。

SQL or dbFiddle

Select tblBID      = A.ID
      ,tblA_Values = B.S
      ,A.Qty
 From TblB A
 Cross Apply (
               Select S=Stuff((Select ',' +Value 
                                From  (
                                      Select B1.RetSeq,Value            
                                       From [dbo].[udf-Str-Parse](A.[tblA_ID],',') B1
                                       Join TblA B2 on B1.RetVal = B2.ID
                                      ) X1
                                Order By RetSeq
                                For XML Path (''))
                              ,1,1,'')
             )B

<强>返回

感兴趣的UDF

CREATE FUNCTION [dbo].[udf-Str-Parse] (@String varchar(max),@Delimiter varchar(10))
Returns Table 
As
Return (  
    Select RetSeq = Row_Number() over (Order By (Select null))
          ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
    From  (Select x = Cast('<x>' + replace((Select replace(@String,@Delimiter,'§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A 
    Cross Apply x.nodes('x') AS B(i)
);
--Thanks Shnugo for making this XML safe
--Select * from [dbo].[udf-Str-Parse]('Dog,Cat,House,Car',',')
--Select * from [dbo].[udf-Str-Parse]('John Cappelletti was here',' ')

编辑 - 内联选项（无UDF）

Select tblBID      = A.ID
      ,tblA_Values = B.S
      ,A.Qty
 From TblB A
 Cross Apply (
               Select S=Stuff((Select ',' +Value 
                                From  (
                                      Select B1.RetSeq,Value            
                                       From (
                                                Select RetSeq = Row_Number() over (Order By (Select null))
                                                      ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
                                                From  (Select x = Cast('<x>' + replace((Select replace(A.[tblA_ID],',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A 
                                                Cross Apply x.nodes('x') AS B(i)
                                            ) B1
                                       Join TblA B2 on B1.RetVal = B2.ID
                                      ) X1
                                Order By RetSeq
                                For XML Path (''))
                              ,1,1,'')
             )B

如何以逗号

1 个答案: