如何拆分字符串以便我可以访问项目x?

时间:2008-08-05 18:15:48

标签: sql sql-server tsql split

使用SQL Server,如何拆分字符串以便我可以访问项目x?

取一个字符串“Hello John Smith”。如何按空格分割字符串并访问索引1处的项目,该项目应返回“John”?

47 个答案:

答案 0 :(得分:350)

我不相信SQL Server有内置的拆分功能,所以除了UDF之外,我知道的唯一其他答案就是劫持PARSENAME函数:

SELECT PARSENAME(REPLACE('Hello John Smith', ' ', '.'), 2) 

PARSENAME接受一个字符串并将其拆分为句点字符。它需要一个数字作为它的第二个参数,并且该数字指定要返回的字符串的哪个部分(从后到前工作)。

SELECT PARSENAME(REPLACE('Hello John Smith', ' ', '.'), 3)  --return Hello

明显的问题是当字符串已经包含句点时。我仍然认为使用UDF是最好的方法......还有其他建议吗?

答案 1 :(得分:183)

您可以在 SQL User Defined Function to Parse a Delimited String 中找到有用的解决方案(来自The Code Project)。

您可以使用这个简单的逻辑:

Declare @products varchar(200) = '1|20|3|343|44|6|8765'
Declare @individual varchar(20) = null

WHILE LEN(@products) > 0
BEGIN
    IF PATINDEX('%|%', @products) > 0
    BEGIN
        SET @individual = SUBSTRING(@products,
                                    0,
                                    PATINDEX('%|%', @products))
        SELECT @individual

        SET @products = SUBSTRING(@products,
                                  LEN(@individual + '|') + 1,
                                  LEN(@products))
    END
    ELSE
    BEGIN
        SET @individual = @products
        SET @products = NULL
        SELECT @individual
    END
END

答案 2 :(得分:108)

首先,创建一个函数(使用CTE,公共表表达式不需要临时表)

 create function dbo.SplitString 
    (
        @str nvarchar(4000), 
        @separator char(1)
    )
    returns table
    AS
    return (
        with tokens(p, a, b) AS (
            select 
                1, 
                1, 
                charindex(@separator, @str)
            union all
            select
                p + 1, 
                b + 1, 
                charindex(@separator, @str, b + 1)
            from tokens
            where b > 0
        )
        select
            p-1 zeroBasedOccurance,
            substring(
                @str, 
                a, 
                case when b > 0 then b-a ELSE 4000 end) 
            AS s
        from tokens
      )
    GO

然后,将它用作任何表(或修改它以适合您现有的存储过程),就像这样。

select s 
from dbo.SplitString('Hello John Smith', ' ')
where zeroBasedOccurance=1

<强>更新

对于长度超过4000个字符的输入字符串,以前的版本将失败。此版本负责限制:

create function dbo.SplitString 
(
    @str nvarchar(max), 
    @separator char(1)
)
returns table
AS
return (
with tokens(p, a, b) AS (
    select 
        cast(1 as bigint), 
        cast(1 as bigint), 
        charindex(@separator, @str)
    union all
    select
        p + 1, 
        b + 1, 
        charindex(@separator, @str, b + 1)
    from tokens
    where b > 0
)
select
    p-1 ItemIndex,
    substring(
        @str, 
        a, 
        case when b > 0 then b-a ELSE LEN(@str) end) 
    AS s
from tokens
);

GO

用法保持不变。

答案 3 :(得分:56)

这里的大多数解决方案都使用while循环或递归CTE。基于集合的方法将是优越的,我保证:

CREATE FUNCTION [dbo].[SplitString]
    (
        @List NVARCHAR(MAX),
        @Delim VARCHAR(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_objects) AS x
              WHERE Number <= LEN(@List)
              AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim)) = @Delim
          ) AS y
        );

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:

在SQL Server 2016或更高版本上,您应该查看STRING_SPLIT()STRING_AGG()

答案 4 :(得分:37)

您可以利用Number表进行字符串解析。

创建物理数字表:

    create table dbo.Numbers (N int primary key);
    insert into dbo.Numbers
        select top 1000 row_number() over(order by number) from master..spt_values
    go

创建包含1000000行的测试表

    create table #yak (i int identity(1,1) primary key, array varchar(50))

    insert into #yak(array)
        select 'a,b,c' from dbo.Numbers n cross join dbo.Numbers nn
    go

创建功能

    create function [dbo].[ufn_ParseArray]
        (   @Input      nvarchar(4000), 
            @Delimiter  char(1) = ',',
            @BaseIdent  int
        )
    returns table as
    return  
        (   select  row_number() over (order by n asc) + (@BaseIdent - 1) [i],
                    substring(@Input, n, charindex(@Delimiter, @Input + @Delimiter, n) - n) s
            from    dbo.Numbers
            where   n <= convert(int, len(@Input)) and
                    substring(@Delimiter + @Input, n, 1) = @Delimiter
        )
    go

用法(在我的笔记本电脑上输出40s的40m行)

    select * 
    from #yak 
    cross apply dbo.ufn_ParseArray(array, ',', 1)

清理

    drop table dbo.Numbers;
    drop function  [dbo].[ufn_ParseArray]

这里的性能并不令人惊讶,但是调用超过一百万行表的函数并不是最好的选择。如果执行一个字符串拆分多行,我会避免该函数。

答案 5 :(得分:23)

这个问题不是关于字符串拆分方法,而是如何获取第n个元素

这里的所有答案都是使用递归,CTE,多个CHARINDEXREVERSEPATINDEX进行某种字符串拆分,发明函数,调用CLR方法,数字表,CROSS APPLY s ...大多数答案涵盖了许多代码行。

但是 - 如果你真的只想要获得第n个元素的方法 - 这可以作为真正的单行,没有UDF,甚至不是sub-select ...并且作为额外的好处:类型安全

以空格分隔第2部分:

DECLARE @input NVARCHAR(100)=N'part1 part2 part3';
SELECT CAST(N'<x>' + REPLACE(@input,N' ',N'</x><x>') + N'</x>' AS XML).value('/x[2]','nvarchar(max)')

当然您可以使用变量作为分隔符和位置(使用sql:column直接从查询的值中检索位置):

DECLARE @dlmt NVARCHAR(10)=N' ';
DECLARE @pos INT = 2;
SELECT CAST(N'<x>' + REPLACE(@input,@dlmt,N'</x><x>') + N'</x>' AS XML).value('/x[sql:variable("@pos")][1]','nvarchar(max)')

如果您的字符串可能包含禁止字符(尤其是&><中的一个字符串),您仍然可以这样做。首先在字符串上使用FOR XML PATH隐式替换所有禁用字符和拟合转义序列。

如果 - 另外 - 您的分隔符是分号,这是一个非常特殊的情况。在这种情况下,我首先将分隔符替换为&#39; #DLMT#&#39;,并最终将其替换为XML标记:

SET @input=N'Some <, > and &;Other äöü@€;One more';
SET @dlmt=N';';
SELECT CAST(N'<x>' + REPLACE((SELECT REPLACE(@input,@dlmt,'#DLMT#') AS [*] FOR XML PATH('')),N'#DLMT#',N'</x><x>') + N'</x>' AS XML).value('/x[sql:variable("@pos")][1]','nvarchar(max)');

SQL-Server 2016 +

的更新

遗憾的是,开发人员忘记使用STRING_SPLIT返回部分索引。但是,使用SQL-Server 2016+,有OPENJSON

documentation明确指出:

  

当OPENJSON解析JSON数组时,该函数将JSON文本中元素的索引作为键返回。

1,2,3这样的字符串只需要括号:[1,2,3]this is an example这样的字词串必须是["this","is","an"," example"] 这些是非常简单的字符串操作。试试吧:

DECLARE @str VARCHAR(100)='Hello John Smith';

SELECT [value]
FROM OPENJSON('["' + REPLACE(@str,' ','","') + '"]')
WHERE [key]=1 --zero-based!

答案 6 :(得分:21)

这是一个可以做到的UDF。它将返回一个分隔值的表,没有尝试过它的所有场景,但你的例子工作正常。


CREATE FUNCTION SplitString 
(
    -- Add the parameters for the function here
    @myString varchar(500),
    @deliminator varchar(10)
)
RETURNS 
@ReturnTable TABLE 
(
    -- Add the column definitions for the TABLE variable here
    [id] [int] IDENTITY(1,1) NOT NULL,
    [part] [varchar](50) NULL
)
AS
BEGIN
        Declare @iSpaces int
        Declare @part varchar(50)

        --initialize spaces
        Select @iSpaces = charindex(@deliminator,@myString,0)
        While @iSpaces > 0

        Begin
            Select @part = substring(@myString,0,charindex(@deliminator,@myString,0))

            Insert Into @ReturnTable(part)
            Select @part

    Select @myString = substring(@mystring,charindex(@deliminator,@myString,0)+ len(@deliminator),len(@myString) - charindex(' ',@myString,0))


            Select @iSpaces = charindex(@deliminator,@myString,0)
        end

        If len(@myString) > 0
            Insert Into @ReturnTable
            Select @myString

    RETURN 
END
GO

你会这样称呼:


Select * From SplitString('Hello John Smith',' ')

编辑:更新解决方案以处理len&gt; 1的分隔符,如下所示:


select * From SplitString('Hello**John**Smith','**')

答案 7 :(得分:15)

这里我发布了一种简单的解决方法

CREATE FUNCTION [dbo].[split](
          @delimited NVARCHAR(MAX),
          @delimiter NVARCHAR(100)
        ) RETURNS @t TABLE (id INT IDENTITY(1,1), 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


    执行这样的功能

  select * from dbo.split('Hello John Smith',' ')

答案 8 :(得分:10)

在我看来,你们这样做太复杂了。只需创建一个CLR UDF并完成它。

using System;
using System.Data;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;
using System.Collections.Generic;

public partial class UserDefinedFunctions {
  [SqlFunction]
  public static SqlString SearchString(string Search) {
    List<string> SearchWords = new List<string>();
    foreach (string s in Search.Split(new char[] { ' ' })) {
      if (!s.ToLower().Equals("or") && !s.ToLower().Equals("and")) {
        SearchWords.Add(s);
      }
    }

    return new SqlString(string.Join(" OR ", SearchWords.ToArray()));
  }
};

答案 9 :(得分:10)

如何使用stringvalues()声明?

DECLARE @str varchar(max)
SET @str = 'Hello John Smith'

DECLARE @separator varchar(max)
SET @separator = ' '

DECLARE @Splited TABLE(id int IDENTITY(1,1), item varchar(max))

SET @str = REPLACE(@str, @separator, '''),(''')
SET @str = 'SELECT * FROM (VALUES(''' + @str + ''')) AS V(A)' 

INSERT INTO @Splited
EXEC(@str)

SELECT * FROM @Splited

达到了结果集。

id  item
1   Hello
2   John
3   Smith

答案 10 :(得分:9)

我使用了frederic的答案,但这在SQL Server 2005中不起作用

我修改了它,我正在select使用union all并且它正常工作

DECLARE @str varchar(max)
SET @str = 'Hello John Smith how are you'

DECLARE @separator varchar(max)
SET @separator = ' '

DECLARE @Splited table(id int IDENTITY(1,1), item varchar(max))

SET @str = REPLACE(@str, @separator, ''' UNION ALL SELECT ''')
SET @str = ' SELECT  ''' + @str + '''  ' 

INSERT INTO @Splited
EXEC(@str)

SELECT * FROM @Splited

结果集是:

id  item
1   Hello
2   John
3   Smith
4   how
5   are
6   you

答案 11 :(得分:8)

此模式工作正常,您可以概括

Convert(xml,'<n>'+Replace(FIELD,'.','</n><n>')+'</n>').value('(/n[INDEX])','TYPE')
                          ^^^^^                                   ^^^^^     ^^^^

注意字段 INDEX TYPE

让一些表格带有

标识符
sys.message.1234.warning.A45
sys.message.1235.error.O98
....

然后,你可以写

SELECT Source         = q.value('(/n[1])', 'varchar(10)'),
       RecordType     = q.value('(/n[2])', 'varchar(20)'),
       RecordNumber   = q.value('(/n[3])', 'int'),
       Status         = q.value('(/n[4])', 'varchar(5)')
FROM   (
         SELECT   q = Convert(xml,'<n>'+Replace(fieldName,'.','</n><n>')+'</n>')
         FROM     some_TABLE
       ) Q

拆分并铸造所有零件。

答案 12 :(得分:7)

如果您的数据库的兼容级别为130或更高,则可以使用STRING_SPLIT函数和OFFSET FETCH子句按索引获取特定项目。

要获取索引N (基于零)的项目,您可以使用以下代码

SELECT value
FROM STRING_SPLIT('Hello John Smith',' ')
ORDER BY (SELECT NULL)
OFFSET N ROWS
FETCH NEXT 1 ROWS ONLY

要检查compatibility level of your database,请执行以下代码:

SELECT compatibility_level  
FROM sys.databases WHERE name = 'YourDBName';

答案 13 :(得分:6)

另一个通过分隔符功能得到字符串的第n部分:

create function GetStringPartByDelimeter (
    @value as nvarchar(max),
    @delimeter as nvarchar(max),
    @position as int
) returns NVARCHAR(MAX) 
AS BEGIN
    declare @startPos as int
    declare @endPos as int
    set @endPos = -1
    while (@position > 0 and @endPos != 0) begin
        set @startPos = @endPos + 1
        set @endPos = charindex(@delimeter, @value, @startPos)

        if(@position = 1) begin
            if(@endPos = 0)
                set @endPos = len(@value) + 1

            return substring(@value, @startPos, @endPos - @startPos)
        end

        set @position = @position - 1
    end

    return null
end

和用法:

select dbo.GetStringPartByDelimeter ('a;b;c;d;e', ';', 3)

返回:

c

答案 14 :(得分:6)

我正在网上寻找解决方案,以下对我有用。 Ref

你可以这样调用这个函数:

SELECT * FROM dbo.split('ram shyam hari gopal',' ')

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO

CREATE FUNCTION [dbo].[Split](@String VARCHAR(8000), @Delimiter CHAR(1))       
RETURNS @temptable TABLE (items VARCHAR(8000))       
AS       
BEGIN       
    DECLARE @idx INT       
    DECLARE @slice VARCHAR(8000)        
    SELECT @idx = 1       
    IF len(@String)<1 OR @String IS NULL  RETURN       
    WHILE @idx!= 0       
    BEGIN       
        SET @idx = charindex(@Delimiter,@String)       
        IF @idx!=0       
            SET @slice = LEFT(@String,@idx - 1)       
        ELSE       
            SET @slice = @String       
        IF(len(@slice)>0)  
            INSERT INTO @temptable(Items) VALUES(@slice)       
        SET @String = RIGHT(@String,len(@String) - @idx)       
        IF len(@String) = 0 break       
    END   
    RETURN       
END

答案 15 :(得分:5)

试试这个:

CREATE function [SplitWordList]
(
 @list varchar(8000)
)
returns @t table 
(
 Word varchar(50) not null,
 Position int identity(1,1) not null
)
as begin
  declare 
    @pos int,
    @lpos int,
    @item varchar(100),
    @ignore varchar(100),
    @dl int,
    @a1 int,
    @a2 int,
    @z1 int,
    @z2 int,
    @n1 int,
    @n2 int,
    @c varchar(1),
    @a smallint
  select 
    @a1 = ascii('a'),
    @a2 = ascii('A'),
    @z1 = ascii('z'),
    @z2 = ascii('Z'),
    @n1 = ascii('0'),
    @n2 = ascii('9')
  set @ignore = '''"'
  set @pos = 1
  set @dl = datalength(@list)
  set @lpos = 1
  set @item = ''
  while (@pos <= @dl) begin
    set @c = substring(@list, @pos, 1)
    if (@ignore not like '%' + @c + '%') begin
      set @a = ascii(@c)
      if ((@a >= @a1) and (@a <= @z1))  
        or ((@a >= @a2) and (@a <= @z2))
        or ((@a >= @n1) and (@a <= @n2))
      begin
        set @item = @item + @c
      end else if (@item > '') begin
        insert into @t values (@item)
        set @item = ''
      end
    end 
    set @pos = @pos + 1
  end
  if (@item > '') begin
    insert into @t values (@item)
  end
  return
end

像这样测试:

select * from SplitWordList('Hello John Smith')

答案 16 :(得分:5)

以下示例使用递归CTE

更新 18.09.2013

CREATE FUNCTION dbo.SplitStrings_CTE(@List nvarchar(max), @Delimiter nvarchar(1))
RETURNS @returns TABLE (val nvarchar(max), [level] int, PRIMARY KEY CLUSTERED([level]))
AS
BEGIN
;WITH cte AS
 (
  SELECT SUBSTRING(@List, 0, CHARINDEX(@Delimiter,  @List + @Delimiter)) AS val,
         CAST(STUFF(@List + @Delimiter, 1, CHARINDEX(@Delimiter, @List + @Delimiter), '') AS nvarchar(max)) AS stval, 
         1 AS [level]
  UNION ALL
  SELECT SUBSTRING(stval, 0, CHARINDEX(@Delimiter, stval)),
         CAST(STUFF(stval, 1, CHARINDEX(@Delimiter, stval), '') AS nvarchar(max)),
         [level] + 1
  FROM cte
  WHERE stval != ''
  )
  INSERT @returns
  SELECT REPLACE(val, ' ','' ) AS val, [level]
  FROM cte
  WHERE val > ''
  RETURN
END

SQLFiddle上的演示

答案 17 :(得分:3)



    Alter Function dbo.fn_Split
    (
    @Expression nvarchar(max),
    @Delimiter  nvarchar(20) = ',',
    @Qualifier  char(1) = Null
    )
    RETURNS @Results TABLE (id int IDENTITY(1,1), value nvarchar(max))
    AS
    BEGIN
       /* USAGE
            Select * From dbo.fn_Split('apple pear grape banana orange honeydew cantalope 3 2 1 4', ' ', Null)
            Select * From dbo.fn_Split('1,abc,"Doe, John",4', ',', '"')
            Select * From dbo.fn_Split('Hello 0,"&""&&&&', ',', '"')
       */

       -- Declare Variables
       DECLARE
          @X     xml,
          @Temp  nvarchar(max),
          @Temp2 nvarchar(max),
          @Start int,
          @End   int

       -- HTML Encode @Expression
       Select @Expression = (Select @Expression For XML Path(''))

       -- Find all occurences of @Delimiter within @Qualifier and replace with |||***|||
       While PATINDEX('%' + @Qualifier + '%', @Expression) > 0 AND Len(IsNull(@Qualifier, '')) > 0
       BEGIN
          Select
             -- Starting character position of @Qualifier
             @Start = PATINDEX('%' + @Qualifier + '%', @Expression),
             -- @Expression starting at the @Start position
             @Temp = SubString(@Expression, @Start + 1, LEN(@Expression)-@Start+1),
             -- Next position of @Qualifier within @Expression
             @End = PATINDEX('%' + @Qualifier + '%', @Temp) - 1,
             -- The part of Expression found between the @Qualifiers
             @Temp2 = Case When @End < 0 Then @Temp Else Left(@Temp, @End) End,
             -- New @Expression
             @Expression = REPLACE(@Expression,
                                   @Qualifier + @Temp2 + Case When @End < 0 Then '' Else @Qualifier End,
                                   Replace(@Temp2, @Delimiter, '|||***|||')
                           )
       END

       -- Replace all occurences of @Delimiter within @Expression with '</fn_Split><fn_Split>'
       -- And convert it to XML so we can select from it
       SET
          @X = Cast('<fn_Split>' +
                    Replace(@Expression, @Delimiter, '</fn_Split><fn_Split>') +
                    '</fn_Split>' as xml)

       -- Insert into our returnable table replacing '|||***|||' back to @Delimiter
       INSERT @Results
       SELECT
          "Value" = LTRIM(RTrim(Replace(C.value('.', 'nvarchar(max)'), '|||***|||', @Delimiter)))
       FROM
          @X.nodes('fn_Split') as X(C)

       -- Return our temp table
       RETURN
    END

答案 18 :(得分:2)

您可以在SQL中拆分字符串而无需函数:

DECLARE @bla varchar(MAX)
SET @bla = 'BED40DFC-F468-46DD-8017-00EF2FA3E4A4,64B59FC5-3F4D-4B0E-9A48-01F3D4F220B0,A611A108-97CA-42F3-A2E1-057165339719,E72D95EA-578F-45FC-88E5-075F66FD726C'

-- http://stackoverflow.com/questions/14712864/how-to-query-values-from-xml-nodes
SELECT 
    x.XmlCol.value('.', 'varchar(36)') AS val 
FROM 
(
    SELECT 
    CAST('<e>' + REPLACE(@bla, ',', '</e><e>') + '</e>' AS xml) AS RawXml
) AS b 
CROSS APPLY b.RawXml.nodes('e') x(XmlCol);

如果您需要支持任意字符串(使用xml特殊字符)

DECLARE @bla NVARCHAR(MAX)
SET @bla = '<html>unsafe & safe Utf8CharsDon''tGetEncoded ÄöÜ - "Conex"<html>,Barnes & Noble,abc,def,ghi'

-- http://stackoverflow.com/questions/14712864/how-to-query-values-from-xml-nodes
SELECT 
    x.XmlCol.value('.', 'nvarchar(MAX)') AS val 
FROM 
(
    SELECT 
    CAST('<e>' + REPLACE((SELECT @bla FOR XML PATH('')), ',', '</e><e>') + '</e>' AS xml) AS RawXml
) AS b 
CROSS APPLY b.RawXml.nodes('e') x(XmlCol); 

答案 19 :(得分:2)

我知道这是一个古老的问题,但我认为有些人可以从我的解决方案中受益。

select 
SUBSTRING(column_name,1,CHARINDEX(' ',column_name,1)-1)
,SUBSTRING(SUBSTRING(column_name,CHARINDEX(' ',column_name,1)+1,LEN(column_name))
    ,1
    ,CHARINDEX(' ',SUBSTRING(column_name,CHARINDEX(' ',column_name,1)+1,LEN(column_name)),1)-1)
,SUBSTRING(SUBSTRING(column_name,CHARINDEX(' ',column_name,1)+1,LEN(column_name))
    ,CHARINDEX(' ',SUBSTRING(column_name,CHARINDEX(' ',column_name,1)+1,LEN(column_name)),1)+1
    ,LEN(column_name))
from table_name

<强> SQL FIDDLE

<强>优点:

  • 它将所有3个子字符串分隔符分隔为''。
  • 不得使用while循环,因为它会降低性能。
  • 不需要Pivot,因为所有生成的子字符串都将显示在 一行

<强>限制:

  • 必须知道总数没有。 of space(子串)。

注意:解决方案可以提供最多N个子字符串。

为了克服限制,我们可以使用以下ref

但是上述solution不能在表中使用(Actaully我无法使用它)。

我希望这个解决方案可以帮助一个人。

更新:如果是记录&gt;使用 LOOPS 50000 明智,因为它会降低效果

答案 20 :(得分:2)

几乎所有其他答案拆分代码都在替换正在拆分的字符串,这会浪费CPU周期并执行不必要的内存分配。

我在这里介绍了一个更好的方法来进行字符串拆分:http://www.digitalruby.com/split-string-sql-server/

以下是代码:

SET NOCOUNT ON

-- You will want to change nvarchar(MAX) to nvarchar(50), varchar(50) or whatever matches exactly with the string column you will be searching against
DECLARE @SplitStringTable TABLE (Value nvarchar(MAX) NOT NULL)
DECLARE @StringToSplit nvarchar(MAX) = 'your|string|to|split|here'
DECLARE @SplitEndPos int
DECLARE @SplitValue nvarchar(MAX)
DECLARE @SplitDelim nvarchar(1) = '|'
DECLARE @SplitStartPos int = 1

SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)

WHILE @SplitEndPos > 0
BEGIN
    SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, (@SplitEndPos - @SplitStartPos))
    INSERT @SplitStringTable (Value) VALUES (@SplitValue)
    SET @SplitStartPos = @SplitEndPos + 1
    SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)
END

SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, 2147483647)
INSERT @SplitStringTable (Value) VALUES(@SplitValue)

SET NOCOUNT OFF

-- You can select or join with the values in @SplitStringTable at this point.

答案 21 :(得分:1)

SQL Server 2016 开始,我们 string_split

DECLARE @string varchar(100) = 'Richard, Mike, Mark'

SELECT value FROM string_split(@string, ',')

答案 22 :(得分:1)

这是一个函数,它将完成问题的分裂字符串和访问项目X的目标:

CREATE FUNCTION [dbo].[SplitString]
(
   @List       VARCHAR(MAX),
   @Delimiter  VARCHAR(255),
   @ElementNumber INT
)
RETURNS VARCHAR(MAX)
AS
BEGIN

       DECLARE @inp VARCHAR(MAX)
       SET @inp = (SELECT REPLACE(@List,@Delimiter,'_DELMTR_') FOR XML PATH(''))

       DECLARE @xml XML
       SET @xml = '<split><el>' + REPLACE(@inp,'_DELMTR_','</el><el>') + '</el></split>'

       DECLARE @ret VARCHAR(MAX)
       SET @ret = (SELECT
              el = split.el.value('.','varchar(max)')
       FROM  @xml.nodes('/split/el[string-length(.)>0][position() = sql:variable("@elementnumber")]') split(el))

       RETURN @ret

END

用法:

SELECT dbo.SplitString('Hello John Smith', ' ', 2)

结果:

John

答案 23 :(得分:1)

Aaron Bertrand的回答很棒,但有缺陷。由于长度函数剥离尾随空格,因此它不能准确地处理空格作为分隔符(如原始问题中的示例)。

以下是他的代码,通过一个小的调整来允许空格分隔符:

CREATE FUNCTION [dbo].[SplitString]
(
    @List NVARCHAR(MAX),
    @Delim VARCHAR(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_objects) AS x
          WHERE Number <= LEN(@List)
          AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim+'x')-1) = @Delim
      ) AS y
    );

答案 24 :(得分:1)

我知道它已经很晚了,但是我最近有这个要求,并提出了以下代码。我没有选择使用用户定义的功能。希望这会有所帮助。

SELECT 
    SUBSTRING(
                SUBSTRING('Hello John Smith' ,0,CHARINDEX(' ','Hello John Smith',CHARINDEX(' ','Hello John Smith')+1)
                        ),CHARINDEX(' ','Hello John Smith'),LEN('Hello John Smith')
            )

答案 25 :(得分:1)

使用带有递归TVF的{​​{1}}的纯基于集合的解决方案。您可以CTEJOIN将此功能添加到任何数据集中。

APPLY

用法:

create function [dbo].[SplitStringToResultSet] (@value varchar(max), @separator char(1))
returns table
as return
with r as (
    select value, cast(null as varchar(max)) [x], -1 [no] from (select rtrim(cast(@value as varchar(max))) [value]) as j
    union all
    select right(value, len(value)-case charindex(@separator, value) when 0 then len(value) else charindex(@separator, value) end) [value]
    , left(r.[value], case charindex(@separator, r.value) when 0 then len(r.value) else abs(charindex(@separator, r.[value])-1) end ) [x]
    , [no] + 1 [no]
    from r where value > '')

select ltrim(x) [value], [no] [index] from r where x is not null;
go

结果:

select *
from [dbo].[SplitStringToResultSet]('Hello John Smith', ' ')
where [index] = 1;

答案 26 :(得分:1)

解析姓氏和名字的简单解决方案

DECLARE @Name varchar(10) = 'John Smith'

-- Get First Name
SELECT SUBSTRING(@Name, 0, (SELECT CHARINDEX(' ', @Name)))

-- Get Last Name
SELECT SUBSTRING(@Name, (SELECT CHARINDEX(' ', @Name)) + 1, LEN(@Name))

就我而言(还有许多其他事情……),我有一个名字和姓氏列表,以一个空格分隔。可以直接在select语句中使用它来解析名字和姓氏。

-- i.e. Get First and Last Name from a table of Full Names
SELECT SUBSTRING(FullName, 0, (SELECT CHARINDEX(' ', FullName))) as FirstName,
SUBSTRING(FullName, (SELECT CHARINDEX(' ', FullName)) + 1, LEN(FullName)) as LastName,
From FullNameTable

答案 27 :(得分:1)

使用STRING_SPLIT的现代方法需要SQL Server 2016及更高版本。

DECLARE @string varchar(100) = 'Hello John Smith'

SELECT
    ROW_NUMBER() OVER (ORDER BY value) AS RowNr,
    value
FROM string_split(@string, ' ')

结果:

RowNr   value
1       Hello
2       John
3       Smith

现在可以从行号中获取第n个元素。

答案 28 :(得分:0)

使用SQL Server 2016及更高版本。使用此代码修剪TRIM字符串,忽略NULL值,并以正确的顺序应用行索引。它也可以使用空格分隔符:

DECLARE @STRING_VALUE NVARCHAR(MAX) = 'one, two,,three, four,     five'

SELECT ROW_NUMBER() OVER (ORDER BY R.[index]) [index], R.[value] FROM
(
    SELECT
        1 [index], NULLIF(TRIM([value]), '') [value] FROM STRING_SPLIT(@STRING_VALUE, ',') T
    WHERE
        NULLIF(TRIM([value]), '') IS NOT NULL
) R

答案 29 :(得分:0)

这是基于字符串,位置和定界符

创建函数fnx_splitstring(@stringToSplit VARCHAR(MAX),@ Position int,@ SpecialChar char(1))
        返回@returnList表([名称] [nvarchar](500))
  AS
    开始
         SET @stringToSplit = @stringToSplit + @SpecialChar
         宣告@name NVARCHAR(255)
         声明@pos INT
         DECLARE @i int
         SET @i = 0
         WHILE CHARINDEX(@SpecialChar,@stringToSplit)> 0
         开始
               SET @i = @i +1
               SELECT @pos = CHARINDEX(@SpecialChar,@stringToSplit)
               SELECT @name = SUBSTRING(@stringToSplit,1,@ pos-1)
               如果@i = @位置
                    开始
                        插入@returnList
                        选择@name
                        返回
                    结束
                SELECT @stringToSplit = SUBSTRING(@ stringToSplit,@ pos + 1,LEN(@stringToSplit)-@ pos)
         结束
         返回
    结束

像这样测试 SELECT * from fnx_splitstring('V4686 / V4686-H-AW-60.25',2,'-')

答案 30 :(得分:0)

这是我为了在字符串中获取特定标记而做的事情。 (在MSSQL 2008中测试)

首先,创建以下功能:(找到:here

CREATE FUNCTION dbo.SplitStrings_Moden
(
   @List NVARCHAR(MAX),
   @Delimiter NVARCHAR(255)
)
RETURNS TABLE
WITH SCHEMABINDING AS
RETURN
  WITH E1(N)        AS ( SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 
                         UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 
                         UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1),
       E2(N)        AS (SELECT 1 FROM E1 a, E1 b),
       E4(N)        AS (SELECT 1 FROM E2 a, E2 b),
       E42(N)       AS (SELECT 1 FROM E4 a, E2 b),
       cteTally(N)  AS (SELECT 0 UNION ALL SELECT TOP (DATALENGTH(ISNULL(@List,1))) 
                         ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E42),
       cteStart(N1) AS (SELECT t.N+1 FROM cteTally t
                         WHERE (SUBSTRING(@List,t.N,1) = @Delimiter OR t.N = 0))
  SELECT Item = SUBSTRING(@List, s.N1, ISNULL(NULLIF(CHARINDEX(@Delimiter,@List,s.N1),0)-s.N1,8000))
    FROM cteStart s;

create FUNCTION dbo.getToken
(
@List NVARCHAR(MAX),
@Delimiter NVARCHAR(255),
@Pos int
)
RETURNS varchar(max)
as 
begin
declare @returnValue varchar(max);
select @returnValue = tbl.Item from (
select ROW_NUMBER() over (order by (select null)) as id, * from dbo.SplitStrings_Moden(@List, @Delimiter)
) as tbl
where tbl.id = @Pos
return @returnValue
end

那么你就可以这样使用它:

select dbo.getToken('1111_2222_3333_', '_', 1)

返回1111

答案 31 :(得分:0)

我意识到这是一个非常老的问题,但是从SQL Server 2016开始,有一些用于解析JSON数据的函数可用于专门解决OP的问题-无需拆分字符串或使用用户定义的函数。要访问位于分隔字符串的特定索引处的项目,请使用JSON_VALUE函数。但是,需要使用格式正确的JSON数据:字符串必须用双引号"括起来,定界符必须是逗号,,整个字符串都用方括号[]括起来。

DECLARE @SampleString NVARCHAR(MAX) = '"Hello John Smith"';
--Format as JSON data.
SET @SampleString = '[' + REPLACE(@SampleString, ' ', '","') + ']';
SELECT 
    JSON_VALUE(@SampleString, '$[0]') AS Element1Value,
    JSON_VALUE(@SampleString, '$[1]') AS Element2Value,
    JSON_VALUE(@SampleString, '$[2]') AS Element3Value;

输出

Element1Value         Element2Value       Element3Value
--------------------- ------------------- ------------------------------
Hello                 John                Smith

(1 row affected)

答案 32 :(得分:0)

CREATE FUNCTION [dbo].[fnSplitString] 
( 
    @string NVARCHAR(MAX), 
    @delimiter CHAR(1) 
) 
RETURNS @output TABLE(splitdata NVARCHAR(MAX) 
) 
BEGIN 
    DECLARE @start INT, @end INT 
    SELECT @start = 1, @end = CHARINDEX(@delimiter, @string) 
    WHILE @start < LEN(@string) + 1 BEGIN 
        IF @end = 0  
            SET @end = LEN(@string) + 1

        INSERT INTO @output (splitdata)  
        VALUES(SUBSTRING(@string, @start, @end - @start)) 
        SET @start = @end + 1 
        SET @end = CHARINDEX(@delimiter, @string, @start)

    END 
    RETURN 
END

并使用它

select *from dbo.fnSplitString('Querying SQL Server','')

答案 33 :(得分:0)

好吧,我的并不是那么简单,但这里是我用来将逗号分隔的输入变量拆分成单个值的代码,并将其放入表变量中。我确定您可以稍微修改它以基于空格进行拆分,然后针对该表变量执行基本的SELECT查询以获得结果。

-- Create temporary table to parse the list of accounting cycles.
DECLARE @tblAccountingCycles table
(
    AccountingCycle varchar(10)
)

DECLARE @vchAccountingCycle varchar(10)
DECLARE @intPosition int

SET @vchAccountingCycleIDs = LTRIM(RTRIM(@vchAccountingCycleIDs)) + ','
SET @intPosition = CHARINDEX(',', @vchAccountingCycleIDs, 1)

IF REPLACE(@vchAccountingCycleIDs, ',', '') <> ''
BEGIN
    WHILE @intPosition > 0
    BEGIN
        SET @vchAccountingCycle = LTRIM(RTRIM(LEFT(@vchAccountingCycleIDs, @intPosition - 1)))
        IF @vchAccountingCycle <> ''
        BEGIN
            INSERT INTO @tblAccountingCycles (AccountingCycle) VALUES (@vchAccountingCycle)
        END
        SET @vchAccountingCycleIDs = RIGHT(@vchAccountingCycleIDs, LEN(@vchAccountingCycleIDs) - @intPosition)
        SET @intPosition = CHARINDEX(',', @vchAccountingCycleIDs, 1)
    END
END

这个概念几乎是一样的。另一种方法是利用SQL Server 2005本身的.NET兼容性。实际上,您可以在.NET中编写一个简单的方法来分割字符串,然后将其作为存储过程/函数公开。

答案 34 :(得分:0)

我开了这个,

declare @x nvarchar(Max) = 'ali.veli.deli.';
declare @item nvarchar(Max);
declare @splitter char='.';

while CHARINDEX(@splitter,@x) != 0
begin
    set @item = LEFT(@x,CHARINDEX(@splitter,@x))
    set @x    = RIGHT(@x,len(@x)-len(@item) )
     select @item as item, @x as x;
end

唯一需要关注的是点&#39;。&#39; @x的那一端总是应该在那里。

答案 35 :(得分:0)

CREATE TABLE test(
    id int,
    adress varchar(100)
);
INSERT INTO test VALUES(1, 'Ludovic Aubert, 42 rue de la Victoire, 75009, Paris, France'),(2, 'Jose Garcia, 1 Calle de la Victoria, 56500 Barcelona, Espana');

SELECT id, value, COUNT(*) OVER (PARTITION BY id) AS n, ROW_NUMBER() OVER (PARTITION BY id ORDER BY (SELECT NULL)) AS rn, adress
FROM test
CROSS APPLY STRING_SPLIT(adress, ',')

enter image description here

答案 36 :(得分:0)

虽然类似于josejuan的基于xml的回答,但我发现只处理xml路径一次,然后旋转效率会稍微提高:

select ID,
    [3] as PathProvidingID,
    [4] as PathProvider,
    [5] as ComponentProvidingID,
    [6] as ComponentProviding,
    [7] as InputRecievingID,
    [8] as InputRecieving,
    [9] as RowsPassed,
    [10] as InputRecieving2
    from
    (
    select id,message,d.* from sysssislog cross apply       ( 
          SELECT Item = y.i.value('(./text())[1]', 'varchar(200)'),
              row_number() over(order by y.i) as rn
          FROM 
          ( 
             SELECT x = CONVERT(XML, '<i>' + REPLACE(Message, ':', '</i><i>') + '</i>').query('.')
          ) AS a CROSS APPLY x.nodes('i') AS y(i)
       ) d
       WHERE event
       = 
       'OnPipelineRowsSent'
    ) as tokens 
    pivot 
    ( max(item) for [rn] in ([3],[4],[5],[6],[7],[8],[9],[10]) 
    ) as data

在8:30跑了

select id,
tokens.value('(/n[3])', 'varchar(100)')as PathProvidingID,
tokens.value('(/n[4])', 'varchar(100)') as PathProvider,
tokens.value('(/n[5])', 'varchar(100)') as ComponentProvidingID,
tokens.value('(/n[6])', 'varchar(100)') as ComponentProviding,
tokens.value('(/n[7])', 'varchar(100)') as InputRecievingID,
tokens.value('(/n[8])', 'varchar(100)') as InputRecieving,
tokens.value('(/n[9])', 'varchar(100)') as RowsPassed
 from
(
    select id, Convert(xml,'<n>'+Replace(message,'.','</n><n>')+'</n>') tokens
         from sysssislog 
       WHERE event
       = 
       'OnPipelineRowsSent'
    ) as data

在9:20跑了

答案 37 :(得分:0)

declare @strng varchar(max)='hello john smith'
select (
    substring(
        @strng,
        charindex(' ', @strng) + 1,
        (
          (charindex(' ', @strng, charindex(' ', @strng) + 1))
          - charindex(' ',@strng)
        )
    ))

答案 38 :(得分:0)

如果有人想只获得一部分文本可以使用

从fromSplitStringSep中选择*('Word1 wordr2 word3','')

CREATE function [dbo].[SplitStringSep] 
(
    @str nvarchar(4000), 
    @separator char(1)
)
returns table
AS
return (
    with tokens(p, a, b) AS (
        select 
        1, 
        1, 
        charindex(@separator, @str)
        union all
        select
            p + 1, 
            b + 1, 
            charindex(@separator, @str, b + 1)
        from tokens
        where b > 0
        )
        select
            p-1 zeroBasedOccurance,
            substring(
                @str, 
                a, 
                case when b > 0 then b-a ELSE 4000 end) 
            AS s
        from tokens
  )

答案 39 :(得分:0)

具有剧烈疼痛的递归CTE解决方案,test it

MS SQL Server 2008架构设置

create table Course( Courses varchar(100) );
insert into Course values ('Hello John Smith');

查询1

with cte as
   ( select 
        left( Courses, charindex( ' ' , Courses) ) as a_l,
        cast( substring( Courses, 
                         charindex( ' ' , Courses) + 1 , 
                         len(Courses ) ) + ' ' 
              as varchar(100) )  as a_r,
        Courses as a,
        0 as n
     from Course t
    union all
      select 
        left(a_r, charindex( ' ' , a_r) ) as a_l,
        substring( a_r, charindex( ' ' , a_r) + 1 , len(a_R ) ) as a_r,
        cte.a,
        cte.n + 1 as n
    from Course t inner join cte 
         on t.Courses = cte.a and len( a_r ) > 0

   )
select a_l, n from cte
--where N = 1

<强> Results

|    A_L | N |
|--------|---|
| Hello  | 0 |
|  John  | 1 |
| Smith  | 2 |

答案 40 :(得分:0)

建立@NothingsImpossible解决方案,或者更确切地说,评论最多投票的答案(在接受的答案之下),我发现以下快速而肮脏的解决方案满足了我自己的需求 - 它具有仅在SQL域内的好处。

给出一个字符串“first; second; third; 4th; 5th”,比方说,我想获得第三个标记。只有当我们知道字符串将包含多少个标记时才有效 - 在这种情况下它是5.所以我的行动方式是将最后两个标记切掉(内部查询),然后将前两个标记切掉(外部查询)

我知道这很丑陋并且涵盖了我所处的特定条件,但我发布它以防有人发现它有用。欢呼声

select 
    REVERSE(
        SUBSTRING(
            reverse_substring, 
            0, 
            CHARINDEX(';', reverse_substring)
        )
    ) 
from 
(
    select 
        msg,
        SUBSTRING(
            REVERSE(msg), 
            CHARINDEX(
                ';', 
                REVERSE(msg), 
                CHARINDEX(
                    ';',
                    REVERSE(msg)
                )+1
            )+1,
            1000
        ) reverse_substring
    from 
    (
        select 'first;second;third;fourth;fifth' msg
    ) a
) b

答案 41 :(得分:0)

如果子字符串不包含重复项,则可以使用以下内容:

WITH testdata(string) AS (
    SELECT 'a' UNION ALL
    SELECT 'a b' UNION ALL
    SELECT 'a b c' UNION ALL
    SELECT 'a b c d'
)
SELECT *
FROM testdata
CROSS APPLY (
    SELECT value AS substring
         , ROW_NUMBER() OVER(ORDER BY CHARINDEX(' ' + value + ' ', ' ' + string + ' ')) AS n
    FROM STRING_SPLIT(string, ' ')
) AS substrings
WHERE n = 1

STRING_SPLIT生成子字符串,但不提供子字符串的 index 。您可以使用CHARINDEX生成索引号,只要子字符串是唯一的,它就会正确。对于a b b ca b c c d e等,它将失败。

答案 42 :(得分:-1)

如果您在splitting string using SQL上查看以下SQL教程,则会发现许多函数可用于在SQL Server上拆分给定的字符串

例如, SplitAndReturnNth UDF函数可用于使用分隔符分割文本并返回第N个段作为函数的输出

select dbo.SplitAndReturnNth('Hello John Smith',' ',2)

enter image description here

答案 43 :(得分:-1)

一个简单的优化算法:

ALTER FUNCTION [dbo].[Split]( @Text NVARCHAR(200),@Splitor CHAR(1) )
RETURNS @Result TABLE ( value NVARCHAR(50)) 
AS
BEGIN
    DECLARE @PathInd INT
    Set @Text+=@Splitor
    WHILE LEN(@Text) > 0
    BEGIN
        SET @PathInd=PATINDEX('%'+@Splitor+'%',@Text)
        INSERT INTO  @Result VALUES(SUBSTRING(@Text, 0, @PathInd))
        SET @Text= SUBSTRING(@Text, @PathInd+1, LEN(@Text))
    END
        RETURN 
END

答案 44 :(得分:-1)

我一直在使用vzczc的回答使用递归cte's一段时间,但是想要更新它以处理可变长度分隔符,并且还要处理带有前导和滞后“分隔符”的字符串,例如当你有一个csv文件时记录如:

<强> “鲍勃”, “史密斯”, “桑尼维尔”, “CA”

或当您处理六部分fqn时,如下所示。我广泛使用这些来记录subject_fqn以进行审计,错误处理等。而parsename只处理四个部分:

[netbios_name].[machine_name].[instance].[database].[schema].[table].[column]

这是我的更新版本,感谢vzczc的原始帖子!

select * from [utility].[split_string](N'"this"."string"."gets"."split"."and"."removes"."leading"."and"."trailing"."quotes"', N'"."', N'"', N'"');

select * from [utility].[split_string](N'"this"."string"."gets"."split"."but"."leaves"."leading"."and"."trailing"."quotes"', N'"."', null, null);

select * from [utility].[split_string](N'[netbios_name].[machine_name].[instance].[database].[schema].[table].[column]', N'].[', N'[', N']');

create function [utility].[split_string] ( 
  @input       [nvarchar](max) 
  , @separator [sysname] 
  , @lead      [sysname] 
  , @lag       [sysname]) 
returns @node_list table ( 
  [index]  [int] 
  , [node] [nvarchar](max)) 
  begin 
      declare @separator_length [int]= len(@separator) 
              , @lead_length    [int] = isnull(len(@lead), 0) 
              , @lag_length     [int] = isnull(len(@lag), 0); 
      -- 
      set @input = right(@input, len(@input) - @lead_length); 
      set @input = left(@input, len(@input) - @lag_length); 
      -- 
      with [splitter]([index], [starting_position], [start_location]) 
           as (select cast(@separator_length as [bigint]) 
                      , cast(1 as [bigint]) 
                      , charindex(@separator, @input) 
               union all 
               select [index] + 1 
                      , [start_location] + @separator_length 
                      , charindex(@separator, @input, [start_location] + @separator_length) 
               from   [splitter] 
               where  [start_location] > 0) 
      -- 
      insert into @node_list 
                  ([index],[node]) 
        select [index] - @separator_length                   as [index] 
               , substring(@input, [starting_position], case 
                                                            when [start_location] > 0 
                                                                then 
                                                              [start_location] - [starting_position] 
                                                            else 
                                                              len(@input) 
                                                        end) as [node] 
        from   [splitter]; 
      -- 
      return; 
  end; 
go 

答案 45 :(得分:-1)

这是我可以帮助某人的解决方案。修改了Jonesinator上面的答案。

如果我有一串分隔的INT值,并希望返回一个INT表(我可以加入)。例如'1,20,3,343,44,6,8765'

创建UDF:

IF OBJECT_ID(N'dbo.ufn_GetIntTableFromDelimitedList', N'TF') IS NOT NULL
    DROP FUNCTION dbo.[ufn_GetIntTableFromDelimitedList];
GO

CREATE FUNCTION dbo.[ufn_GetIntTableFromDelimitedList](@String NVARCHAR(MAX),                 @Delimiter CHAR(1))

RETURNS @table TABLE 
(
    Value INT NOT NULL
)
AS 
BEGIN
DECLARE @Pattern NVARCHAR(3)
SET @Pattern = '%' + @Delimiter + '%'
DECLARE @Value NVARCHAR(MAX)

WHILE LEN(@String) > 0
    BEGIN
        IF PATINDEX(@Pattern, @String) > 0
        BEGIN
            SET @Value = SUBSTRING(@String, 0, PATINDEX(@Pattern, @String))
            INSERT INTO @table (Value) VALUES (@Value)

            SET @String = SUBSTRING(@String, LEN(@Value + @Delimiter) + 1, LEN(@String))
        END
        ELSE
        BEGIN
            -- Just the one value.
            INSERT INTO @table (Value) VALUES (@String)
            RETURN
        END
    END

RETURN
END
GO

然后得到表格结果:

SELECT * FROM dbo.[ufn_GetIntTableFromDelimitedList]('1,20,3,343,44,6,8765', ',')

1
20
3
343
44
6
8765

在连接声明中:

SELECT [ID], [FirstName]
FROM [User] u
JOIN dbo.[ufn_GetIntTableFromDelimitedList]('1,20,3,343,44,6,8765', ',') t ON u.[ID] = t.[Value]

1    Elvis
20   Karen
3    David
343  Simon
44   Raj
6    Mike
8765 Richard

如果要返回NVARCHAR列表而不是INT,则只需更改表定义:

RETURNS @table TABLE 
(
    Value NVARCHAR(MAX) NOT NULL
)

答案 46 :(得分:-1)

这是一个SQL UDF,可以拆分字符串并抓取某个部分。

create FUNCTION [dbo].[udf_SplitParseOut]
(
    @List nvarchar(MAX),
    @SplitOn nvarchar(5),
    @GetIndex smallint
)  
returns varchar(1000)
AS  

BEGIN

DECLARE @RtnValue table 
(

    Id int identity(0,1),
    Value nvarchar(MAX)
) 


    DECLARE @result varchar(1000)

    While (Charindex(@SplitOn,@List)>0)
    Begin
        Insert Into @RtnValue (value)
        Select Value = ltrim(rtrim(Substring(@List,1,Charindex(@SplitOn,@List)-1)))
        Set @List = Substring(@List,Charindex(@SplitOn,@List)+len(@SplitOn),len(@List))
    End

    Insert Into @RtnValue (Value)
    Select Value = ltrim(rtrim(@List))

    select @result = value from @RtnValue where ID = @GetIndex

    Return @result
END