问题很简单，但问题很热门：）
所以我为string\u split（）创建了一些 Package 器，它以更一般的方式产生pivot结果。它是一个返回值的表函数（nn，value1，value2，值50）-对于大多数csv行足够了。如果有更多的值，它们将换行到下一行-nn表示行号。将第三个参数@columncnt=[yournumber]设置为在特定位置换行：

alter FUNCTION fn_Split50
(   
    @str varchar(max),
    @delim char(1), 
    @columnCnt int = 50
)
RETURNS TABLE 
AS
RETURN 
(
    SELECT * 
    FROM (SELECT 
            nn = (nn - 1) / @columnCnt + 1, 
            nnn = 'value' + cast(((nn - 1) % @columnCnt) + 1 as varchar(10)), 
            value
        FROM (SELECT 
            nn = ROW_NUMBER() over (order by (select null)),
            value
            FROM string_split(@str, @delim) aa
            ) aa
        where nn > 0
    ) bb
    PIVOT  
    (  
    max(value) 
    FOR nnn IN (    
        value1, value2, value3, value4, value5, value6, value7, value8, value9, value10,
        value11, value12, value13, value14, value15, value16, value17, value18, value19, value20,
        value21, value22, value23, value24, value25, value26, value27, value28, value29, value30,
        value31, value32, value33, value34, value35, value36, value37, value38, value39, value40,
        value41, value42, value43, value44, value45, value46, value47, value48, value49, value50        
     )  
    ) AS PivotTable 
)

使用示例：

select * from dbo.fn_split50('zz1,aa2,ss3,dd4,ff5', ',', DEFAULT)

select * from dbo.fn_split50('zz1,aa2,ss3,dd4,ff5,gg6,hh7,jj8,ww9,qq10', ',', 3)

select * from dbo.fn_split50('zz1,11,aa2,22,ss3,33,dd4,44,ff5,55,gg6,66,hh7,77,jj8,88,ww9,99,qq10,1010', ',',2)

希望能有所帮助：）

我在上面重新写了一个答案，让它变得更好：

CREATE FUNCTION [dbo].[CSVParser]
(
  @s        VARCHAR(255),
  @idx      NUMERIC
)
RETURNS VARCHAR(12)
BEGIN
    DECLARE @comma int
    SET @comma = CHARINDEX(',', @s)
    WHILE 1=1
    BEGIN
        IF @comma=0
            IF @idx=1
                RETURN @s
            ELSE
                RETURN ''
        IF @idx=1
        BEGIN
            DECLARE @word VARCHAR(12)
            SET @word=LEFT(@s, @comma - 1)
            RETURN @word
        END
        SET @s = RIGHT(@s,LEN(@s)-@comma)
        SET @comma = CHARINDEX(',', @s)
        SET @idx = @idx - 1
    END
    RETURN 'not used'
END

用法示例：

SELECT dbo.CSVParser(COLUMN, 1),
       dbo.CSVParser(COLUMN, 2),
       dbo.CSVParser(COLUMN, 3)
FROM   TABLE

Select distinct PROJ_UID,PROJ_NAME,RES_UID from E2E_ProjectWiseTimesheetActuals
where   CHARINDEX(','+cast(PROJ_UID as varchar(8000))+',', @params) > 0 and  CHARINDEX(','+cast(RES_UID as varchar(8000))+',', @res) > 0

select distinct modelFileId,F4.*
from contract
cross apply (select XmlList=convert(xml, '<x>'+replace(modelFileId,';','</x><x>')+'</x>').query('.')) F2
cross apply (select mfid1=XmlNode.value('/x[1]','varchar(512)')
,mfid2=XmlNode.value('/x[2]','varchar(512)')
,mfid3=XmlNode.value('/x[3]','varchar(512)')
,mfid4=XmlNode.value('/x[4]','varchar(512)') from XmlList.nodes('x') F3(XmlNode)) F4
where modelFileId like '%;%'
order by modelFileId

我发现如上所述使用parsename会导致任何带有句点的名称为空。
因此，如果名称中有一个首字母或标题后跟一个点，则返回null。
我发现这对我有用：

SELECT 
REPLACE(SUBSTRING(FullName, 1,CHARINDEX(',', FullName)), ',','') as Name,
REPLACE(SUBSTRING(FullName, CHARINDEX(',', FullName), LEN(FullName)), ',', '') as Surname
FROM Table1

ALTER function get_occurance_index(@delimiter varchar(1),@occurence int,@String varchar(100))
returns int
AS Begin
--Declare @delimiter varchar(1)=',',@occurence int=2,@String varchar(100)='a,b,c'
Declare @result int
 ;with T as (
    select 1 Rno,0 as row, charindex(@delimiter, @String) pos,@String st
    union all
    select Rno+1,pos + 1, charindex(@delimiter, @String, pos + 1), @String
    from T
    where pos > 0
)
select  @result=pos 
from T 
where pos > 0   and rno = @occurence 
return isnull(@result,0)
ENd
declare @data as table (data varchar(100))
insert into @data values('1,2,3') 
insert into @data values('aaa,bbbbb,cccc') 
select top  3 Substring (data,0,dbo.get_occurance_index( ',',1,data)) ,--First Record always starts with 0
Substring (data,dbo.get_occurance_index( ',',1,data)+1,dbo.get_occurance_index( ',',2,data)-dbo.get_occurance_index( ',',1,data)-1) ,
Substring (data,dbo.get_occurance_index( ',',2,data)+1,len(data)) , -- Last record cant be more than len of actual data
data 
From @data

您可能会发现在sql用户定义函数中解析分隔字符串的解决方案很有帮助（来自代码项目）。
这是本页的代码部分：

CREATE FUNCTION [fn_ParseText2Table]
  (@p_SourceText VARCHAR(MAX)
  ,@p_Delimeter VARCHAR(100)=',' --default to comma delimited.
  )
 RETURNS @retTable
  TABLE([Position] INT IDENTITY(1,1)
   ,[Int_Value] INT
   ,[Num_Value] NUMERIC(18,3)
   ,[Txt_Value] VARCHAR(MAX)
   ,[Date_value] DATETIME
   )
AS
/*
********************************************************************************
Purpose: Parse values from a delimited string
  & return the result as an indexed table
Copyright 1996, 1997, 2000, 2003 Clayton Groom (<A href="mailto:Clayton_Groom@hotmail.com">Clayton_Groom@hotmail.com</A>)
Posted to the public domain Aug, 2004
2003-06-17 Rewritten as SQL 2000 function.
 Reworked to allow for delimiters > 1 character in length
 and to convert Text values to numbers
2016-04-05 Added logic for date values based on "new" ISDATE() function, Updated to use XML approach, which is more efficient.
********************************************************************************
* /
BEGIN
 DECLARE @w_xml xml;
 SET @w_xml = N'<root><i>' + replace(@p_SourceText, @p_Delimeter,'</i><i>') + '</i></root>';
 INSERT INTO @retTable
     ([Int_Value]
    , [Num_Value]
    , [Txt_Value]
    , [Date_value]
     )
     SELECT CASE
       WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST(CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC) AS INT)
      END AS [Int_Value]
    , CASE
       WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC(18, 3))
      END AS [Num_Value]
    , [i].value('.', 'VARCHAR(MAX)') AS [txt_Value]
    , CASE
       WHEN ISDATE([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST([i].value('.', 'VARCHAR(MAX)') AS DATETIME)
      END AS [Num_Value]
     FROM @w_xml.nodes('//root/i') AS [Items]([i]);
 RETURN;
END;
GO

DECLARE @INPUT VARCHAR (MAX)='N,A,R,E,N,D,R,A'
DECLARE @ELIMINATE_CHAR CHAR (1)=','
DECLARE @L_START INT=1
DECLARE @L_END INT=(SELECT LEN (@INPUT))
DECLARE @OUTPUT CHAR (1)
WHILE @L_START <=@L_END
BEGIN
    SET @OUTPUT=(SUBSTRING (@INPUT,@L_START,1))
    IF @OUTPUT!=@ELIMINATE_CHAR
    BEGIN
        PRINT @OUTPUT
    END
    SET @L_START=@L_START+1
END

很简单，你可以通过下面的查询：

DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3'
SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))

我的表格：

Value  ColOne
--------------------
1      Cleo, Smith

如果没有太多的列，下面的操作应该可以

ALTER TABLE mytable ADD ColTwo nvarchar(256);
UPDATE mytable SET ColTwo = LEFT(ColOne, Charindex(',', ColOne) - 1);
--'Cleo' = LEFT('Cleo, Smith', Charindex(',', 'Cleo, Smith') - 1)
UPDATE mytable SET ColTwo = REPLACE(ColOne, ColTwo + ',', '');
--' Smith' = REPLACE('Cleo, Smith', 'Cleo' + ',')
UPDATE mytable SET ColOne = REPLACE(ColOne, ',' + ColTwo, ''), ColTwo = LTRIM(ColTwo);
--'Cleo' = REPLACE('Cleo, Smith', ',' + ' Smith', '')

结果：

Value  ColOne ColTwo
--------------------
1      Cleo   Smith

试试这个：

declare @csv varchar(100) ='aaa,bb,csda,daass';
set @csv = @csv+',';
with cte as
(
    select SUBSTRING(@csv,1,charindex(',',@csv,1)-1) as val, SUBSTRING(@csv,charindex(',',@csv,1)+1,len(@csv)) as rem 
    UNION ALL
    select SUBSTRING(a.rem,1,charindex(',',a.rem,1)-1)as val, SUBSTRING(a.rem,charindex(',',a.rem,1)+1,len(A.rem)) 
    from cte a where LEN(a.rem)>=1
    ) select val from cte

这对我有用

CREATE FUNCTION [dbo].[SplitString](
    @delimited NVARCHAR(MAX),
    @delimiter NVARCHAR(100)
) RETURNS @t TABLE ( val NVARCHAR(MAX))
AS
BEGIN
    DECLARE @xml XML
    SET @xml = N'<t>' + REPLACE(@delimited,@delimiter,'</t><t>') + '</t>'
    INSERT INTO @t(val)
    SELECT  r.value('.','varchar(MAX)') as item
    FROM  @xml.nodes('/t') as records(r)
    RETURN
END

使用instring函数：）

select Value, 
       substring(String,1,instr(String," ") -1) Fname,  
       substring(String,instr(String,",") +1) Sname 
from tablename;

使用了两种功能，

1. substring(string, position, length) ==>返回从位置到长度的字符串
2. instr(string,pattern) ==>返回图案的位置。
   如果我们在子字符串中不提供长度参数，它将返回到字符串的末尾

我遇到了一个类似的问题，但一个复杂的一个，因为这是第一个线程，我发现关于这个问题，我决定张贴我的发现。我知道这是一个简单问题的复杂解决方案，但我希望我可以帮助其他人谁去寻找一个更复杂的解决方案这个线程。我不得不拆分一个包含5个数字（列名：levelsfeed）的字符串，并在单独的列中显示每个数字。例如：8,1,2,2,2应显示为：

1  2  3  4  5
-------------
8  1  2  2  2

解决方案1：使用xml函数：这是迄今为止最慢的解决方案

SELECT Distinct FeedbackID, 
, S.a.value('(/H/r)[1]', 'INT') AS level1
, S.a.value('(/H/r)[2]', 'INT') AS level2
, S.a.value('(/H/r)[3]', 'INT') AS level3
, S.a.value('(/H/r)[4]', 'INT') AS level4
, S.a.value('(/H/r)[5]', 'INT') AS level5
FROM (            
    SELECT *,CAST (N'<H><r>' + REPLACE(levelsFeed, ',', '</r><r>')  + '</r> </H>' AS XML) AS [vals]
    FROM Feedbacks 
)  as d
CROSS APPLY d.[vals].nodes('/H/r') S(a)

解决方案2：使用split函数和pivot(split函数将字符串拆分为具有列名（data）的行

SELECT FeedbackID, [1],[2],[3],[4],[5]
FROM (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY feedbackID ORDER BY (SELECT  null)) as rn 
FROM (
    SELECT FeedbackID, levelsFeed
    FROM Feedbacks 
) as a
CROSS APPLY dbo.Split(levelsFeed, ',')
) as SourceTable
PIVOT
(
    MAX(data)
    FOR rn IN ([1],[2],[3],[4],[5])
)as pivotTable

解决方案3：使用字符串操作函数-比解决方案2更快

SELECT FeedbackID,
SUBSTRING(levelsFeed,0,CHARINDEX(',',levelsFeed)) AS level1,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),4) AS level2,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),3) AS level3,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),2) AS level4,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),1) AS level5
FROM Feedbacks

因为levelsfeed包含5个字符串值，所以我需要对第一个字符串使用substring函数。
我希望我的解决方案能帮助其他人找到更复杂的拆分为列的方法

此功能最快：

CREATE FUNCTION dbo.F_ExtractSubString
(
  @String VARCHAR(MAX),
  @NroSubString INT,
  @Separator VARCHAR(5)
)
RETURNS VARCHAR(MAX) AS
BEGIN
    DECLARE @St INT = 0, @End INT = 0, @Ret VARCHAR(MAX)
    SET @String = @String + @Separator
    WHILE CHARINDEX(@Separator, @String, @End + 1) > 0 AND @NroSubString > 0
    BEGIN
        SET @St = @End + 1
        SET @End = CHARINDEX(@Separator, @String, @End + 1)
        SET @NroSubString = @NroSubString - 1
    END
    IF @NroSubString > 0
        SET @Ret = ''
    ELSE
        SET @Ret = SUBSTRING(@String, @St, @End - @St)
    RETURN @Ret
END
GO

用法示例：

SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
       dbo.F_ExtractSubString(COLUMN, 2, ', '),
       dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM   TABLE