我们需要在 sql server 中为两列拆分数据答案

【问题标题】：We need to split data for two columns in sql server我们需要在 sql server 中为两列拆分数据
【发布时间】：2016-09-30 15:45:17
【问题描述】：

我们需要在 sql server 中为两列拆分数据。

CREATE TABLE #t (Id int,Number varchar(32),Name varchar(32));    

Insert Into #t(Id, Number,Name)
Values(123, '1,2,3','a~b~c') ,
(456, '1,2,3,4','a~b~c~d'),
(789, '1,2','a~b')

SELECT *FROM #t

输入：

Id  Number  Name
123 1,2,3   a~b~c
456 1,2,3,4 a~b~c~d
789 1,2 a~b

我们需要以下格式的输出：

Id  Number  Name
123 1   a
123 2   b
123 3   c
456 1   a
456 2   b
456 3   c
456 4   d
789 1   a
789 2   b

【问题讨论】：

为什么要投反对票？ :-|
到目前为止你尝试了什么？
看看这个类似的答案：stackoverflow.com/questions/5493510/…
同时检查stackoverflow.com/questions/5123585/…
我尝试使用 xml 和光标，但它只适用于一列。我无法从 2 列中拆分相同时间的数据。

标签： sql sql-server sql-server-2008

【解决方案1】：

这是使用cte的方法..

;WITH cte_1
AS
    (SELECT Id,ROW_NUMBER()OVER(PARTITION BY ID ORDER BY ID,LTRIM(RTRIM(m.n.value('.[1]','varchar(8000)'))))RNO
    ,LTRIM(RTRIM(m.n.value('.[1]','varchar(8000)'))) AS Number

    FROM
    (
    SELECT Id
           ,CAST('<XMLRoot><RowData>' + REPLACE(Number,',','</RowData><RowData>') + '</RowData></XMLRoot>' AS XML) AS x
    FROM   #t 
    )t
    CROSS APPLY x.nodes('/XMLRoot/RowData')m(n))
,cte_2
AS
    (SELECT Id,ROW_NUMBER()OVER(PARTITION BY ID ORDER BY ID,LTRIM(RTRIM(m.n.value('.[1]','varchar(8000)'))))RNO
    ,LTRIM(RTRIM(m.n.value('.[1]','varchar(8000)'))) AS Name

    FROM
    (
    SELECT Id,CAST('<XMLRoot><RowData>' + REPLACE(Name,'~','</RowData><RowData>') + '</RowData></XMLRoot>' AS XML) AS x
    FROM   #t 
    )t
    CROSS APPLY x.nodes('/XMLRoot/RowData')m(n))
SELECT a.id,a.Number,b.name
FROM cte_1 a JOIN cte_2 b
    ON a.id=b.id and a.RNo=b.RNo

【讨论】：

【解决方案2】：

借助解析器和交叉应用

Select A.ID
      ,B.*
 From  #t A
 Cross Apply (  Select Number=C.RetVal
                      ,Name  =D.RetVal
                  From (Select * from [dbo].[udf-Str-Parse](A.Number,',')) C
                  Join (Select * from [dbo].[udf-Str-Parse](A.Name,'~')) D
                    on (C.RetSeq=D.RetSeq)
             ) B

退货

ID  Number  Name
123 1       a
123 2       b
123 3       c
456 1       a
456 2       b
456 3       c
456 4       d
789 1       a
789 2       b

UDF（如果需要）

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(@String,@Delimiter,'</x><x>')+'</x>' as xml).query('.')) as A 
    Cross Apply x.nodes('x') AS B(i)
);

如果您不能使用 UDF，请编辑 - 我们可以将 UDF 代码移动到两个子查询中以产生相同的结果

Select A.ID
      ,B.*
 From  #t A
 Cross Apply (  Select Number=C.RetVal
                      ,Name  =D.RetVal
                  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(A.Number,',','</x><x>')+'</x>' as xml).query('.')) as A 
                            Cross Apply x.nodes('x') AS B(i)
                        ) C
                  Join (
                        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(A.Name,'~','</x><x>')+'</x>' as xml).query('.')) as A 
                            Cross Apply x.nodes('x') AS B(i)
                        ) D
                    on (C.RetSeq=D.RetSeq)
             ) B

【讨论】：

【解决方案3】：

XML 和 CTE 的另一种方式：

DECLARE @x xml

SELECT @x = (
    SELECT  Id as '@id',
            CAST('<n number="'+REPLACE(Number,',','"/><n number="')+'"/>' as xml),
            CAST('<w name="'+REPLACE([Name],'~','"/><w name="')+'" />' as xml)
    FROM #t 
    FOR XML PATH('i')
)

;WITH numbers AS (
SELECT  t.v.value('../@id','int') as id,
        t.v.value('@number','nvarchar(10)') as number,
        ROW_NUMBER() OVER (PARTITION BY t.v.value('../@id','int') ORDER BY (SELECT NULL)) as RN
FROM @x.nodes('/i/n') as t(v)
), names AS (
SELECT  t.v.value('../@id','int') as id,
        t.v.value('@name','nvarchar(10)') as [name],
        ROW_NUMBER() OVER (PARTITION BY t.v.value('../@id','int') ORDER BY (SELECT NULL)) as RN
FROM @x.nodes('/i/w') as t(v)
)

SELECT  n.id,
        n.number,
        nn.[name]
FROM numbers n
INNER JOIN names nn
    ON n.id = nn.id and n.rn = nn.rn

输出：

id  number  name
123 1       a
123 2       b
123 3       c
456 1       a
456 2       b
456 3       c
456 4       d
789 1       a
789 2       b

我在 ROW_NUMBER() 部分使用SELECT NULL 来按原样订购。但它可能会导致一些问题，例如不正确的排序。

【讨论】：