递归CTE解决方案与服务器疼痛，测试它

MS SQL Server 2008模式设置:

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

结果:

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

通过delimeter函数得到字符串的n个部分:

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

以及用法:

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

返回:

c

如果数据库的兼容性级别为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的项。

要获得索引N(从零开始)的项，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

要检查数据库的兼容性级别，执行以下代码:

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

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')

以下是我的解决方案，可能会对某些人有所帮助。修改以上Jonesinator的回答。

如果我有一个带分隔符的INT值字符串，并希望返回一个INT表(然后我可以加入)。如。44岁的1,3343 6,8765年

创建一个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join语句中:

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

如果你想返回一个nvarchar列表而不是int，那么只需更改表定义:

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

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