SQL，从，分隔行到列

输出

Create table test123
(
SeqId int,
string1 varchar(100),
string2 varchar(100)
)

insert into test123(SeqId,string1,string2)values
(10,'a,b,c','x,y,z'),
(20,'d,e,f','p,q,r')

---

这将允许字符串中的项目数量可变/不均匀

几乎任何拆分/解析函数都可以。我提供了一个，它还提供了项目的序列号

选项1：不带解析函数

SeqId RowNum String1 String2
10  1   a   x
10  2   b   y
10  3   c   z
20  1   d   p
20  2   e   q
20  3   f   r

Declare @YourTable table (SeqId int,string1 varchar(100),string2 varchar(100))
insert into @YourTable values
(10,'a,b,c','x,y,z'),
(20,'d,e,f','p,q,r')

Select A.SeqId
      ,B.*
 From  @YourTable A
 Cross Apply (
                Select RowNum  = RetSeq
                      ,String1 = max(String1)
                      ,String2 = max(String2)
                 From (
                        Select RetSeq  = Row_Number() over (Order By (Select null))
                              ,String1 = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
                              ,String2 = ''
                        From  (Select x = Cast('<x>' + replace((Select replace(A.String1,',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A 
                        Cross Apply x.nodes('x') AS B(i)
                        Union All
                        Select RetSeq  = Row_Number() over (Order By (Select null))
                              ,String1 = ''
                              ,String2 = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
                        From  (Select x = Cast('<x>' + replace((Select replace(A.String2,',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A 
                        Cross Apply x.nodes('x') AS B(i)
                      ) B1
                 Group By RetSeq
             ) B

两者都返回

SeqId RowNum String1 String2
10  1   a   x
10  2   b   y
10  3   c   z
20  1   d   p
20  2   e   q
20  3   f   r

Declare @YourTable table (SeqId int,string1 varchar(100),string2 varchar(100))
insert into @YourTable values
(10,'a,b,c','x,y,z'),
(20,'d,e,f','p,q,r')

Select A.SeqId
      ,B.*
 From  @YourTable A
 Cross Apply (
                Select RowNum  = RetSeq
                      ,String1 = max(String1)
                      ,String2 = max(String2)
                 From (
                        Select RetSeq  = Row_Number() over (Order By (Select null))
                              ,String1 = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
                              ,String2 = ''
                        From  (Select x = Cast('<x>' + replace((Select replace(A.String1,',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A 
                        Cross Apply x.nodes('x') AS B(i)
                        Union All
                        Select RetSeq  = Row_Number() over (Order By (Select null))
                              ,String1 = ''
                              ,String2 = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
                        From  (Select x = Cast('<x>' + replace((Select replace(A.String2,',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A 
                        Cross Apply x.nodes('x') AS B(i)
                      ) B1
                 Group By RetSeq
             ) B

有兴趣的UDF

Declare @YourTable table (SeqId int,string1 varchar(100),string2 varchar(100))
insert into @YourTable values
(10,'a,b,c','x,y,z'),
(20,'d,e,f','p,q,r')

Select A.SeqId
      ,B.*
 From  @YourTable A
 Cross Apply (
                Select RowNum = RetSeq
                      ,String1= max(String1)
                      ,String2= max(String2)
                 From (
                        Select RetSeq,String1=RetVal,String2='' from [dbo].[udf-Str-Parse](A.string1,',')
                        Union All
                        Select RetSeq,String1='',String2=RetVal from [dbo].[udf-Str-Parse](A.string2,',')
                      ) B1
                 Group By RetSeq

             ) B

---

使用动态SQL检查以下解决方案

declare@t表（SeqId int，string1-varchar（100），string2-varchar（100））
声明@result表（SeqId int、RowNum int、String1-varchar（10）、String2-varchar（10））
声明@SeqId int
声明@string1 varchar（100）
声明@string2 varchar（100）
声明@char1 varchar（5）
声明@char2 varchar（5）
声明@count1 int
声明@count2 int
声明@rownum int
从test123插入@t select*
存在时（从@t中选择*
开始
从@t中选择top 1@SeqId=SeqId、@string1=string1、@string2=string2
设置@rownum=1
while@rownum您的问题是什么？可能重复的方法大约有一百万种和一种分割逗号分隔字符串的方法。随便找一个。e、 g.此处缺少大量可能的信息：逗号分隔的字符串在string1和string2中是否始终具有相同数量的元素？行之间的元素数是否相同？无论如何，为什么要以这种方式存储数据？好奇和乐于助人的人想知道。也许使用字符串分割，像宝石一样工作，真的帮了我很多。我知道当它只是一列时该怎么做，对于不止一列，你的解决方案非常有效。@SrinivasaRao很高兴它有帮助