我有一个包含字符串的SQL Server 2008 R2列，我需要用逗号分隔。我在StackOverflow上看到了很多答案，但它们都不适用于R2。我确保我对任何拆分函数示例都有选择权限。非常感谢任何帮助。
答案 0 :(得分：187)
我之前使用过这个SQL可能对你有用： -
CREATE FUNCTION dbo.splitstring ( @stringToSplit VARCHAR(MAX) )
RETURNS
@returnList TABLE ([Name] [nvarchar] (500))
AS
BEGIN
DECLARE @name NVARCHAR(255)
DECLARE @pos INT
WHILE CHARINDEX(',', @stringToSplit) > 0
BEGIN
SELECT @pos = CHARINDEX(',', @stringToSplit)
SELECT @name = SUBSTRING(@stringToSplit, 1, @pos-1)
INSERT INTO @returnList
SELECT @name
SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, LEN(@stringToSplit)-@pos)
END
INSERT INTO @returnList
SELECT @stringToSplit
RETURN
END
并使用它： -
SELECT * FROM dbo.splitstring('91,12,65,78,56,789')
答案 1 :(得分：54)
有没有人考虑过基于集合的方法，而不是递归CTE和while循环？
CREATE FUNCTION [dbo].[SplitString]
(
@List nvarchar(MAX),
@Delim nvarchar(255)
)
RETURNS TABLE
AS
RETURN ( SELECT [Value] FROM
(
SELECT
[Value] = LTRIM(RTRIM(SUBSTRING(@List, [Number],
CHARINDEX(@Delim, @List + @Delim, [Number]) - [Number])))
FROM (SELECT Number = ROW_NUMBER() OVER (ORDER BY name)
FROM sys.all_columns) AS x
WHERE Number <= LEN(@List)
AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim)) = @Delim
) AS y
);
如果你想避免字符串长度的限制＆lt; = sys.all_columns中的行数（SQL Server 2017中model中的9,980};在你自己的用户数据库中要高得多），您可以使用其他方法来获取数字，例如构建自己的table of numbers。如果您不能使用系统表或创建自己的系统表，也可以使用递归CTE：
CREATE FUNCTION [dbo].[SplitString]
(
@List nvarchar(MAX),
@Delim nvarchar(255)
)
RETURNS TABLE WITH SCHEMABINDING
AS
RETURN ( WITH n(n) AS (SELECT 1 UNION ALL SELECT n+1 FROM n
WHERE n < LEN(@List))
SELECT [Value] FROM
(
SELECT
[Value] = LTRIM(RTRIM(SUBSTRING(@List, n,
CHARINDEX(@Delim, @List + @Delim, n) - n)))
FROM n
WHERE SUBSTRING(@Delim + @List, n, LEN(@Delim)) = @Delim
) AS y
);
但是您必须将OPTION (MAXRECURSION 0)（或MAXRECURSION <longest possible string length if < 32768>）附加到外部查询，以避免字符串的递归错误＆gt; 100个字符。如果这也不是一个好的选择，那么请参阅评论中指出的this answer。
有关分割函数的更多信息，为什么（并证明）while循环和递归CTE不能缩放，以及更好的替代方法，如果分割来自应用程序层的字符串：
http://www.sqlperformance.com/2012/07/t-sql-queries/split-strings
http://www.sqlperformance.com/2012/08/t-sql-queries/splitting-strings-now-with-less-t-sql
https://sqlblog.org/2010/07/07/splitting-a-list-of-integers-another-roundup
答案 2 :(得分：36)
最后等待已经在 SQL Server 2016 中引入了分裂字符串函数：STRING_SPLIT
select * From STRING_SPLIT ('a,b', ',') cs
所有其他分割字符串的方法，如XML，Tally表，while循环等等，都被这个STRING_SPLIT函数所震撼。
以下是一篇性能比较优秀的文章：Performance Surprises and Assumptions : STRING_SPLIT
答案 3 :(得分：19)
最简单的方法是使用XML格式。
<强> 1。将字符串转换为不带表格的行
<强> QUERY
DECLARE @String varchar(100) = 'String1,String2,String3'
-- To change ',' to any other delimeter, just change ',' to your desired one
DECLARE @Delimiter CHAR = ','
SELECT LTRIM(RTRIM(Split.a.value('.', 'VARCHAR(100)'))) 'Value'
FROM
(
SELECT CAST ('<M>' + REPLACE(@String, @Delimiter, '</M><M>') + '</M>' AS XML) AS Data
) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a)
<强> RESULT
x---------x
| Value |
x---------x
| String1 |
| String2 |
| String3 |
x---------x
<强> 2。转换为表中具有每个CSV行ID的行
消息来源
x-----x--------------------------x
| Id | Value |
x-----x--------------------------x
| 1 | String1,String2,String3 |
| 2 | String4,String5,String6 |
x-----x--------------------------x
<强> QUERY
-- To change ',' to any other delimeter, just change ',' before '</M><M>' to your desired one
DECLARE @Delimiter CHAR = ','
SELECT ID,LTRIM(RTRIM(Split.a.value('.', 'VARCHAR(100)'))) 'Value'
FROM
(
SELECT ID,CAST ('<M>' + REPLACE(VALUE, @Delimiter, '</M><M>') + '</M>' AS XML) AS Data
FROM TABLENAME
) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a)
<强> RESULT
x-----x----------x
| Id | Value |
x-----x----------x
| 1 | String1 |
| 1 | String2 |
| 1 | String3 |
| 2 | String4 |
| 2 | String5 |
| 2 | String6 |
x-----x----------x
答案 4 :(得分：10)
我需要一种快速的方法来摆脱邮政编码中的+4。
UPDATE #Emails
SET ZIPCode = SUBSTRING(ZIPCode, 1, (CHARINDEX('-', ZIPCODE)-1))
WHERE ZIPCode LIKE '%-%'
没有proc ...没有UDF ...只是一个紧凑的小内联命令，它做了它必须做的事情。不花哨，不优雅。
根据需要更改分隔符等，它可以用于任何事情。
答案 5 :(得分：8)
如果你更换
WHILE CHARINDEX(',', @stringToSplit) > 0
与
WHILE LEN(@stringToSplit) > 0
你可以在while循环之后消除最后一次插入！
CREATE FUNCTION dbo.splitstring ( @stringToSplit VARCHAR(MAX) )
RETURNS
@returnList TABLE ([Name] [nvarchar] (500))
AS
BEGIN
DECLARE @name NVARCHAR(255)
DECLARE @pos INT
WHILE LEN(@stringToSplit) > 0
BEGIN
SELECT @pos = CHARINDEX(',', @stringToSplit)
if @pos = 0
SELECT @pos = LEN(@stringToSplit)
SELECT @name = SUBSTRING(@stringToSplit, 1, @pos-1)
INSERT INTO @returnList
SELECT @name
SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, LEN(@stringToSplit)-@pos)
END
RETURN
END
答案 6 :(得分：3)
使用某种循环（迭代）的字符串拆分的所有函数都有不良的性能。它们应该用基于集合的解决方案替换。
此代码执行得非常好。
CREATE FUNCTION dbo.SplitStrings
(
@List NVARCHAR(MAX),
@Delimiter NVARCHAR(255)
)
RETURNS TABLE
WITH SCHEMABINDING
AS
RETURN
(
SELECT Item = y.i.value('(./text())[1]', 'nvarchar(4000)')
FROM
(
SELECT x = CONVERT(XML, '<i>'
+ REPLACE(@List, @Delimiter, '</i><i>')
+ '</i>').query('.')
) AS a CROSS APPLY x.nodes('i') AS y(i)
);
GO
答案 7 :(得分：2)
我稍微修改了Andy Robinson的功能。现在您只能从返回表中选择所需的部分：
CREATE FUNCTION dbo.splitstring ( @stringToSplit VARCHAR(MAX) )
RETURNS
@returnList TABLE ([numOrder] [tinyint] , [Name] [nvarchar] (500)) AS
BEGIN
DECLARE @name NVARCHAR(255)
DECLARE @pos INT
DECLARE @orderNum INT
SET @orderNum=0
WHILE CHARINDEX('.', @stringToSplit) > 0
BEGIN
SELECT @orderNum=@orderNum+1;
SELECT @pos = CHARINDEX('.', @stringToSplit)
SELECT @name = SUBSTRING(@stringToSplit, 1, @pos-1)
INSERT INTO @returnList
SELECT @orderNum,@name
SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, LEN(@stringToSplit)-@pos)
END
SELECT @orderNum=@orderNum+1;
INSERT INTO @returnList
SELECT @orderNum, @stringToSplit
RETURN
END
Usage:
SELECT Name FROM dbo.splitstring('ELIS.YD.CRP1.1.CBA.MDSP.T389.BT') WHERE numOrder=5
答案 8 :(得分：2)
如果您需要针对具有最少代码的常见案例的快速临时解决方案，那么这个递归CTE双线程将执行此操作：
DECLARE @s VARCHAR(200) = ',1,2,,3,,,4,,,,5,'
;WITH
a AS (SELECT i=-1, j=0 UNION ALL SELECT j, CHARINDEX(',', @s, j + 1) FROM a WHERE j > i),
b AS (SELECT SUBSTRING(@s, i+1, IIF(j>0, j, LEN(@s)+1)-i-1) s FROM a WHERE i >= 0)
SELECT * FROM b
将此作为独立语句使用，或者只是将上述CTE添加到任何查询中，您就可以将结果表b与其他人一起加入，以便在任何其他表达式中使用。
如果您添加一个计数器，您将获得一个位置索引以及列表：
DECLARE @s VARCHAR(200) = '1,2333,344,4'
;WITH
a AS (SELECT n=0, i=-1, j=0 UNION ALL SELECT n+1, j, CHARINDEX(',', @s, j+1) FROM a WHERE j > i),
b AS (SELECT n, SUBSTRING(@s, i+1, IIF(j>0, j, LEN(@s)+1)-i-1) s FROM a WHERE i >= 0)
SELECT * FROM b;
结果：
n s
1 1
2 2333
3 344
4 4
答案 9 :(得分：2)
这里有一个正确的版本，但我认为添加一点容错以防它们有一个尾随逗号以及它可以使它不是作为一个函数而是作为一个更大的一部分使用它会很好一段代码。以防您只使用一次并且不需要功能。这也适用于整数（这是我需要的），因此您可能需要更改数据类型。
DECLARE @StringToSeperate VARCHAR(10)
SET @StringToSeperate = '1,2,5'
--SELECT @StringToSeperate IDs INTO #Test
DROP TABLE #IDs
CREATE TABLE #IDs (ID int)
DECLARE @CommaSeperatedValue NVARCHAR(255) = ''
DECLARE @Position INT = LEN(@StringToSeperate)
--Add Each Value
WHILE CHARINDEX(',', @StringToSeperate) > 0
BEGIN
SELECT @Position = CHARINDEX(',', @StringToSeperate)
SELECT @CommaSeperatedValue = SUBSTRING(@StringToSeperate, 1, @Position-1)
INSERT INTO #IDs
SELECT @CommaSeperatedValue
SELECT @StringToSeperate = SUBSTRING(@StringToSeperate, @Position+1, LEN(@StringToSeperate)-@Position)
END
--Add Last Value
IF (LEN(LTRIM(RTRIM(@StringToSeperate)))>0)
BEGIN
INSERT INTO #IDs
SELECT SUBSTRING(@StringToSeperate, 1, @Position)
END
SELECT * FROM #IDs
答案 10 :(得分：2)
我最近不得不写这样的东西。这是我提出的解决方案。它适用于任何分隔符字符串，我认为它会稍微好一些：
CREATE FUNCTION [dbo].[SplitString]
( @string nvarchar(4000)
, @delim nvarchar(100) )
RETURNS
@result TABLE
( [Value] nvarchar(4000) NOT NULL
, [Index] int NOT NULL )
AS
BEGIN
DECLARE @str nvarchar(4000)
, @pos int
, @prv int = 1
SELECT @pos = CHARINDEX(@delim, @string)
WHILE @pos > 0
BEGIN
SELECT @str = SUBSTRING(@string, @prv, @pos - @prv)
INSERT INTO @result SELECT @str, @prv
SELECT @prv = @pos + LEN(@delim)
, @pos = CHARINDEX(@delim, @string, @pos + 1)
END
INSERT INTO @result SELECT SUBSTRING(@string, @prv, 4000), @prv
RETURN
END
答案 11 :(得分：2)
使用CTE的解决方案，如果有人需要它（除了我，显然这样做，这就是我写它的原因）。
declare @StringToSplit varchar(100) = 'Test1,Test2,Test3';
declare @SplitChar varchar(10) = ',';
with StringToSplit as (
select
ltrim( rtrim( substring( @StringToSplit, 1, charindex( @SplitChar, @StringToSplit ) - 1 ) ) ) Head
, substring( @StringToSplit, charindex( @SplitChar, @StringToSplit ) + 1, len( @StringToSplit ) ) Tail
union all
select
ltrim( rtrim( substring( Tail, 1, charindex( @SplitChar, Tail ) - 1 ) ) ) Head
, substring( Tail, charindex( @SplitChar, Tail ) + 1, len( Tail ) ) Tail
from StringToSplit
where charindex( @SplitChar, Tail ) > 0
union all
select
ltrim( rtrim( Tail ) ) Head
, '' Tail
from StringToSplit
where charindex( @SplitChar, Tail ) = 0
and len( Tail ) > 0
)
select Head from StringToSplit
答案 12 :(得分：1)
Personnaly我使用这个功能：
ALTER FUNCTION [dbo].[CUST_SplitString]
(
@String NVARCHAR(4000),
@Delimiter NCHAR(1)
)
RETURNS TABLE
AS
RETURN
(
WITH Split(stpos,endpos)
AS(
SELECT 0 AS stpos, CHARINDEX(@Delimiter,@String) AS endpos
UNION ALL
SELECT endpos+1, CHARINDEX(@Delimiter,@String,endpos+1)
FROM Split
WHERE endpos > 0
)
SELECT 'Id' = ROW_NUMBER() OVER (ORDER BY (SELECT 1)),
'Data' = SUBSTRING(@String,stpos,COALESCE(NULLIF(endpos,0),LEN(@String)+1)-stpos)
FROM Split
)
答案 13 :(得分：1)
这是一个可以使用patindex在模式上拆分的版本，这是对上面帖子的简单改编。我有一个案例需要拆分包含多个分隔符字符的字符串。
alter FUNCTION dbo.splitstring ( @stringToSplit VARCHAR(1000), @splitPattern varchar(10) )
RETURNS
@returnList TABLE ([Name] [nvarchar] (500))
AS
BEGIN
DECLARE @name NVARCHAR(255)
DECLARE @pos INT
WHILE PATINDEX(@splitPattern, @stringToSplit) > 0
BEGIN
SELECT @pos = PATINDEX(@splitPattern, @stringToSplit)
SELECT @name = SUBSTRING(@stringToSplit, 1, @pos-1)
INSERT INTO @returnList
SELECT @name
SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, LEN(@stringToSplit)-@pos)
END
INSERT INTO @returnList
SELECT @stringToSplit
RETURN
END
select * from dbo.splitstring('stringa/stringb/x,y,z','%[/,]%');
结果看起来像这样
stringa stringb X ÿ ž
答案 14 :(得分：1)
我根据要求Here开发了一个双分割器（需要两个分割字符）。可能在此线程中看到与字符串拆分相关的查询引用最多的一些值。
CREATE FUNCTION uft_DoubleSplitter
(
-- Add the parameters for the function here
@String VARCHAR(4000),
@Splitter1 CHAR,
@Splitter2 CHAR
)
RETURNS @Result TABLE (Id INT,MId INT,SValue VARCHAR(4000))
AS
BEGIN
DECLARE @FResult TABLE(Id INT IDENTITY(1, 1),
SValue VARCHAR(4000))
DECLARE @SResult TABLE(Id INT IDENTITY(1, 1),
MId INT,
SValue VARCHAR(4000))
SET @String = @String+@Splitter1
WHILE CHARINDEX(@Splitter1, @String) > 0
BEGIN
DECLARE @WorkingString VARCHAR(4000) = NULL
SET @WorkingString = SUBSTRING(@String, 1, CHARINDEX(@Splitter1, @String) - 1)
--Print @workingString
INSERT INTO @FResult
SELECT CASE
WHEN @WorkingString = '' THEN NULL
ELSE @WorkingString
END
SET @String = SUBSTRING(@String, LEN(@WorkingString) + 2, LEN(@String))
END
IF ISNULL(@Splitter2, '') != ''
BEGIN
DECLARE @OStartLoop INT
DECLARE @OEndLoop INT
SELECT @OStartLoop = MIN(Id),
@OEndLoop = MAX(Id)
FROM @FResult
WHILE @OStartLoop <= @OEndLoop
BEGIN
DECLARE @iString VARCHAR(4000)
DECLARE @iMId INT
SELECT @iString = SValue+@Splitter2,
@iMId = Id
FROM @FResult
WHERE Id = @OStartLoop
WHILE CHARINDEX(@Splitter2, @iString) > 0
BEGIN
DECLARE @iWorkingString VARCHAR(4000) = NULL
SET @IWorkingString = SUBSTRING(@iString, 1, CHARINDEX(@Splitter2, @iString) - 1)
INSERT INTO @SResult
SELECT @iMId,
CASE
WHEN @iWorkingString = '' THEN NULL
ELSE @iWorkingString
END
SET @iString = SUBSTRING(@iString, LEN(@iWorkingString) + 2, LEN(@iString))
END
SET @OStartLoop = @OStartLoop + 1
END
INSERT INTO @Result
SELECT MId AS PrimarySplitID,
ROW_NUMBER() OVER (PARTITION BY MId ORDER BY Mid, Id) AS SecondarySplitID ,
SValue
FROM @SResult
END
ELSE
BEGIN
INSERT INTO @Result
SELECT Id AS PrimarySplitID,
NULL AS SecondarySplitID,
SValue
FROM @FResult
END
RETURN
<强>用法：
--FirstSplit
SELECT * FROM uft_DoubleSplitter('ValueA=ValueB=ValueC=ValueD==ValueE&ValueA=ValueB=ValueC===ValueE&ValueA=ValueB==ValueD===','&',NULL)
--Second Split
SELECT * FROM uft_DoubleSplitter('ValueA=ValueB=ValueC=ValueD==ValueE&ValueA=ValueB=ValueC===ValueE&ValueA=ValueB==ValueD===','&','=')
可能的用法（获取每个拆分的第二个值）：
SELECT fn.SValue
FROM uft_DoubleSplitter('ValueA=ValueB=ValueC=ValueD==ValueE&ValueA=ValueB=ValueC===ValueE&ValueA=ValueB==ValueD===', '&', '=')AS fn
WHERE fn.mid = 2
答案 15 :(得分：1)
在禁用字符的情况下，常用的XML元素方法会中断。这是一种将此方法用于任何类型字符的方法，即使使用分号作为分隔符。
诀窍是，首先使用SELECT SomeString AS [*] FOR XML PATH('')来正确转义所有禁用的字符。这就是为什么我将分隔符替换为魔术值以避免将;作为分隔符的麻烦。
DECLARE @Dummy TABLE (ID INT, SomeTextToSplit NVARCHAR(MAX))
INSERT INTO @Dummy VALUES
(1,N'A&B;C;D;E, F')
,(2,N'"C" & ''D'';<C>;D;E, F');
DECLARE @Delimiter NVARCHAR(10)=';'; --special effort needed (due to entities coding with "&code;")!
WITH Casted AS
(
SELECT *
,CAST(N'<x>' + REPLACE((SELECT REPLACE(SomeTextToSplit,@Delimiter,N'§§Split$me$here§§') AS [*] FOR XML PATH('')),N'§§Split$me$here§§',N'</x><x>') + N'</x>' AS XML) AS SplitMe
FROM @Dummy
)
SELECT Casted.ID
,x.value(N'.',N'nvarchar(max)') AS Part
FROM Casted
CROSS APPLY SplitMe.nodes(N'/x') AS A(x)
结果
ID Part
1 A&B
1 C
1 D
1 E, F
2 "C" & 'D'
2 <C>
2 D
2 E, F
答案 16 :(得分：0)
简单
DECLARE @String varchar(100) = '11,21,84,85,87'
SELECT * FROM TB_PAPEL WHERE CD_PAPEL IN (SELECT value FROM STRING_SPLIT(@String, ','))
-- EQUIVALENTE
SELECT * FROM TB_PAPEL WHERE CD_PAPEL IN (11,21,84,85,87)
答案 17 :(得分：0)
我通过将值包装到元素中来进行xml路由（M，但一切正常）：
declare @v nvarchar(max) = '100,201,abcde'
select
a.value('.', 'varchar(max)')
from
(select cast('<M>' + REPLACE(@v, ',', '</M><M>') + '</M>' AS XML) as col) as A
CROSS APPLY A.col.nodes ('/M') AS Split(a)
答案 18 :(得分：0)
这是基于安迪·罗伯逊（Andy Robertson）的回答，我需要除逗号以外的定界符。
CREATE FUNCTION dbo.splitstring ( @stringToSplit nvarchar(MAX), @delim nvarchar(max))
RETURNS
@returnList TABLE ([value] [nvarchar] (MAX))
AS
BEGIN
DECLARE @value NVARCHAR(max)
DECLARE @pos INT
WHILE CHARINDEX(@delim, @stringToSplit) > 0
BEGIN
SELECT @pos = CHARINDEX(@delim, @stringToSplit)
SELECT @value = SUBSTRING(@stringToSplit, 1, @pos - 1)
INSERT INTO @returnList
SELECT @value
SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos + LEN(@delim), LEN(@stringToSplit) - @pos)
END
INSERT INTO @returnList
SELECT @stringToSplit
RETURN
END
GO
并使用它：
SELECT * FROM dbo.splitstring('test1 test2 test3', ' ');
（在SQL Server 2008 R2上测试）
编辑：正确的测试代码
答案 19 :(得分：0)
在对@AviG表示应有的尊重的同时，这是他配备的无错误版本的函数，可完全返回所有令牌。
IF EXISTS (SELECT * FROM sys.objects WHERE type = 'TF' AND name = 'TF_SplitString')
DROP FUNCTION [dbo].[TF_SplitString]
GO
-- =============================================
-- Author: AviG
-- Amendments: Parameterize the delimeter and included the missing chars in last token - Gemunu Wickremasinghe
-- Description: Tabel valued function that Breaks the delimeted string by given delimeter and returns a tabel having split results
-- Usage
-- select * from [dbo].[TF_SplitString]('token1,token2,,,,,,,,token969',',')
-- 969 items should be returned
-- select * from [dbo].[TF_SplitString]('4672978261,4672978255',',')
-- 2 items should be returned
-- =============================================
CREATE FUNCTION dbo.TF_SplitString
( @stringToSplit VARCHAR(MAX) ,
@delimeter char = ','
)
RETURNS
@returnList TABLE ([Name] [nvarchar] (500))
AS
BEGIN
DECLARE @name NVARCHAR(255)
DECLARE @pos INT
WHILE LEN(@stringToSplit) > 0
BEGIN
SELECT @pos = CHARINDEX(@delimeter, @stringToSplit)
if @pos = 0
BEGIN
SELECT @pos = LEN(@stringToSplit)
SELECT @name = SUBSTRING(@stringToSplit, 1, @pos)
END
else
BEGIN
SELECT @name = SUBSTRING(@stringToSplit, 1, @pos-1)
END
INSERT INTO @returnList
SELECT @name
SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, LEN(@stringToSplit)-@pos)
END
RETURN
END
答案 20 :(得分：0)
基于递归cte的解决方案
declare @T table (iden int identity, col1 varchar(100));
insert into @T(col1) values
('ROOT/South America/Lima/Test/Test2')
, ('ROOT/South America/Peru/Test/Test2')
, ('ROOT//South America/Venuzuala ')
, ('RtT/South America / ')
, ('ROOT/South Americas// ');
declare @split char(1) = '/';
select @split as split;
with cte as
( select t.iden, case when SUBSTRING(REVERSE(rtrim(t.col1)), 1, 1) = @split then LTRIM(RTRIM(t.col1)) else LTRIM(RTRIM(t.col1)) + @split end as col1, 0 as pos , 1 as cnt
from @T t
union all
select t.iden, t.col1 , charindex(@split, t.col1, t.pos + 1), cnt + 1
from cte t
where charindex(@split, t.col1, t.pos + 1) > 0
)
select t1.*, t2.pos, t2.cnt
, ltrim(rtrim(SUBSTRING(t1.col1, t1.pos+1, t2.pos-t1.pos-1))) as bingo
from cte t1
join cte t2
on t2.iden = t1.iden
and t2.cnt = t1.cnt+1
and t2.pos > t1.pos
order by t1.iden, t1.cnt;
答案 21 :(得分：0)
/ *
回答T-SQL split string
基于Andy Robinson和AviG的答案 增强功能参考：LEN function not including trailing spaces in SQL Server
此“文件”应作为markdown文件和SQL文件有效
```
* /
CREATE FUNCTION dbo.splitstring ( --CREATE OR ALTER
@stringToSplit NVARCHAR(MAX)
) RETURNS @returnList TABLE ([Item] NVARCHAR (MAX))
AS BEGIN
DECLARE @name NVARCHAR(MAX)
DECLARE @pos BIGINT
SET @stringToSplit = @stringToSplit + ',' -- this should allow entries that end with a `,` to have a blank value in that "column"
WHILE ((LEN(@stringToSplit+'_') > 1)) BEGIN -- `+'_'` gets around LEN trimming terminal spaces. See URL referenced above
SET @pos = COALESCE(NULLIF(CHARINDEX(',', @stringToSplit),0),LEN(@stringToSplit+'_')) -- COALESCE grabs first non-null value
SET @name = SUBSTRING(@stringToSplit, 1, @pos-1) --MAX size of string of type nvarchar is 4000
SET @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, 4000) -- With SUBSTRING fn (MS web): "If start is greater than the number of characters in the value expression, a zero-length expression is returned."
INSERT INTO @returnList SELECT @name --additional debugging parameters below can be added
-- + ' pos:' + CAST(@pos as nvarchar) + ' remain:''' + @stringToSplit + '''(' + CAST(LEN(@stringToSplit+'_')-1 as nvarchar) + ')'
END
RETURN
END
GO
/ *
```
测试用例：请参阅上面引用为“增强功能”的网址
SELECT *,LEN(Item+'_')-1 'L' from splitstring('a,,b')
Item | L
--- | ---
a | 1
| 0
b | 1
SELECT *,LEN(Item+'_')-1 'L' from splitstring('a,,')
Item | L
--- | ---
a | 1
| 0
| 0
SELECT *,LEN(Item+'_')-1 'L' from splitstring('a,, ')
Item | L
--- | ---
a | 1
| 0
| 1
SELECT *,LEN(Item+'_')-1 'L' from splitstring('a,, c ')
Item | L
--- | ---
a | 1
| 0
c | 3
* /
答案 22 :(得分：0)
这是一个可以作为函数使用的示例，也可以在程序中使用相同的逻辑。 --SELECT *来自[dbo] .fn_SplitString;
CREATE FUNCTION [dbo].[fn_SplitString]
(@CSV VARCHAR(MAX), @Delimeter VARCHAR(100) = ',')
RETURNS @retTable TABLE
(
[value] VARCHAR(MAX) NULL
)AS
BEGIN
DECLARE
@vCSV VARCHAR (MAX) = @CSV,
@vDelimeter VARCHAR (100) = @Delimeter;
IF @vDelimeter = ';'
BEGIN
SET @vCSV = REPLACE(@vCSV, ';', '~!~#~');
SET @vDelimeter = REPLACE(@vDelimeter, ';', '~!~#~');
END;
SET @vCSV = REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(@vCSV, '&', '&'), '<', '<'), '>', '>'), '''', '''), '"', '"');
DECLARE @xml XML;
SET @xml = '<i>' + REPLACE(@vCSV, @vDelimeter, '</i><i>') + '</i>';
INSERT INTO @retTable
SELECT
x.i.value('.', 'varchar(max)') AS COLUMNNAME
FROM @xml.nodes('//i')AS x(i);
RETURN;
END;
答案 23 :(得分：0)
ALTER FUNCTION [dbo].func_split_string
(
@input as varchar(max),
@delimiter as varchar(10) = ";"
)
RETURNS @result TABLE
(
id smallint identity(1,1),
csv_value varchar(max) not null
)
AS
BEGIN
DECLARE @pos AS INT;
DECLARE @string AS VARCHAR(MAX) = '';
WHILE LEN(@input) > 0
BEGIN
SELECT @pos = CHARINDEX(@delimiter,@input);
IF(@pos<=0)
select @pos = len(@input)
IF(@pos <> LEN(@input))
SELECT @string = SUBSTRING(@input, 1, @pos-1);
ELSE
SELECT @string = SUBSTRING(@input, 1, @pos);
INSERT INTO @result SELECT @string
SELECT @input = SUBSTRING(@input, @pos+len(@delimiter), LEN(@input)-@pos)
END
RETURN
END
答案 24 :(得分：0)
您可以使用此功能：
CREATE FUNCTION SplitString
(
@Input NVARCHAR(MAX),
@Character CHAR(1)
)
RETURNS @Output TABLE (
Item NVARCHAR(1000)
)
AS
BEGIN
DECLARE @StartIndex INT, @EndIndex INT
SET @StartIndex = 1
IF SUBSTRING(@Input, LEN(@Input) - 1, LEN(@Input)) <> @Character
BEGIN
SET @Input = @Input + @Character
END
WHILE CHARINDEX(@Character, @Input) > 0
BEGIN
SET @EndIndex = CHARINDEX(@Character, @Input)
INSERT INTO @Output(Item)
SELECT SUBSTRING(@Input, @StartIndex, @EndIndex - 1)
SET @Input = SUBSTRING(@Input, @EndIndex + 1, LEN(@Input))
END
RETURN
END
GO
答案 25 :(得分：0)
这更加狭窄。当我这样做时，我通常有一个以逗号分隔的唯一ID列表（INT或BIGINT），我想将其作为一个表进行转换，以用作另一个具有INT或BIGINT主键的表的内连接。我希望返回一个内联表值函数，以便我可以实现最有效的连接。
示例用法是：
DECLARE @IDs VARCHAR(1000);
SET @IDs = ',99,206,124,8967,1,7,3,45234,2,889,987979,';
SELECT me.Value
FROM dbo.MyEnum me
INNER JOIN dbo.GetIntIdsTableFromDelimitedString(@IDs) ids ON me.PrimaryKey = ids.ID
我从http://sqlrecords.blogspot.com/2012/11/converting-delimited-list-to-table.html窃取了这个想法，将其更改为内联表值并转换为INT。
create function dbo.GetIntIDTableFromDelimitedString
(
@IDs VARCHAR(1000) --this parameter must start and end with a comma, eg ',123,456,'
--all items in list must be perfectly formatted or function will error
)
RETURNS TABLE AS
RETURN
SELECT
CAST(SUBSTRING(@IDs,Nums.number + 1,CHARINDEX(',',@IDs,(Nums.number+2)) - Nums.number - 1) AS INT) AS ID
FROM
[master].[dbo].[spt_values] Nums
WHERE Nums.Type = 'P'
AND Nums.number BETWEEN 1 AND DATALENGTH(@IDs)
AND SUBSTRING(@IDs,Nums.number,1) = ','
AND CHARINDEX(',',@IDs,(Nums.number+1)) > Nums.number;
GO
答案 26 :(得分：-3)
最简单的方法：
它甚至可以在快递版中使用:)。