Aaron Bertrand的回答很好，但也有缺陷。它不能准确地将空格作为分隔符处理(就像最初问题中的示例一样)，因为长度函数将空格带在后面。

下面是他的代码，稍微调整了一下，允许使用空格分隔符:

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
    );

我开发了这个，

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

你唯一应该注意的是。'。那@x的末尾就应该在这里。

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

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。

基于纯集的解决方案，使用TVF和递归CTE。您可以将此函数JOIN和APPLY到任何数据集。

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

用法:

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

结果:

value   index
-------------
John    1

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

像这样测试它:

select * from SplitWordList('Hello John Smith')