我在网上寻找解决方案，下面的工作对我来说。 Ref。

然后像这样调用函数:

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

首先，创建一个函数(使用CTE，公共表表达式不再需要临时表)

 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，像这样使用它作为任何表(或修改它以适应现有存储的proc)。

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

更新

以前的版本将失败的输入字符串长度超过4000个字符。这个版本考虑到了以下限制:

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

用法不变。

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

你可以在SQL用户定义函数解析带分隔符的字符串中找到有用的解决方案(来自代码项目)。

你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

我在网上寻找解决方案，下面的工作对我来说。 Ref。

然后像这样调用函数:

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

好吧，我的代码并不那么简单，但下面是我用来将逗号分隔的输入变量分割为单个值，并将其放入表变量中的代码。我相信您可以稍微修改一下，根据空格进行分割，然后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

概念是差不多的。另一种选择是利用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，将字符串分割，然后将其作为存储过程/函数公开。