set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
go
-- =============================================
-- Author: netcorner
-- Create date: 3/10/2008
-- Description: Split string variant with type of ntext
-- =============================================
Create FUNCTION [dbo].[split]
(
@text ntext,
@delimiter char(1)
)
RETURNS @arrayTable TABLE(Idx bigint, [Value] nvarchar(200))
AS
BEGIN
DECLARE @splitlen int
DECLARE @tmp varchar(1024)
SET @splitlen = 4000
DECLARE @Idx int SET @Idx = 0
-- 定义取子串的起始位置
DECLARE @textsplit bigint
SET @textsplit = 1
WHILE( @textsplit <= DATALENGTH(@text) )
BEGIN
-- 由于许多字符串处理函数无法用于ntext数据类型
-- 所以需要循环按批处理ntext字符串，一批取出
-- 个字符放入nvarchar(4000)类型的变量中.
DECLARE @string nvarchar(4000)
SELECT @string = SUBSTRING(@text,@textsplit,@splitlen)
-- 能够取出满个字符
IF LEN(@string) = @splitlen
BEGIN
-- 确保取出的个字符是完整的由分隔符隔开的字符串组合
DECLARE @lastcomma int
SELECT @lastcomma = CHARINDEX(@delimiter,REVERSE(@string),1)
-- 最后一个分隔符后面的字符串不完整，应抛弃
IF @lastcomma > 0
BEGIN
SELECT @string = SUBSTRING(@string,1,@splitlen - @lastcomma)
-- 设置下一次从@text取字符的起始位置
SELECT @textsplit = @textsplit + @splitlen - @lastcomma + 1
END
-- 最后一个分隔符后面的字符串完整.
ELSE
BEGIN
SELECT @textsplit = @textsplit + @splitlen + 1
END
END
-- 取出不满个字符
ELSE
BEGIN
SELECT @textsplit = @textsplit + @splitlen + 1
END
-- 解析@string，取出以分隔符为界限的子字符串
DECLARE @i1 int SET @i1 = 1
DECLARE @i2 int SET @i2 = 1
WHILE @i1 <= LEN(@string)
BEGIN
SET @i2 = CHARINDEX(@delimiter,@string,@i1)
IF @i2 = 0
SET @i2 = LEN(@string) + 1
set @tmp=SUBSTRING(@string,@i1,@i2-@i1)
if(@tmp='')
INSERT @arrayTable (Idx) select @Idx
else
begin
INSERT @arrayTable (Idx, Value)
SELECT @Idx, @tmp
end
SET @i1 = @i2 + 1
SET @Idx = @Idx + 1
END
END
RETURN
END