化解32层以上嵌套,树型结构的递归实现方法,给大家参考一下!
前一镇,写了两个递归的树型结构处理函数:
http://expert.csdn.net/Expert/topic/1343/1343007.xml?temp=.730694
后来有朋友提出32层以上嵌套的一个展BOM的实例,没有办法用递归实现,特想了个办法,实现32层以上树型结构的递归方法。现在特将以前的那个函数进行了改进,具体如下:
---------------------------------表及函数脚本
if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[TreeClass]') and OBJECTPROPERTY(id, N'IsUserTable') = 1)
drop table [dbo].[TreeClass]
GO
CREATE TABLE [dbo].[TreeClass] (
[TC_id] [int] IDENTITY (1, 1) NOT NULL ,
[TC_PID] [int] NOT NULL ,
[TC_OtherTypeID] [varchar] (8000) COLLATE Chinese_PRC_CI_AS NULL ,
[TC_Name] [varchar] (50) COLLATE Chinese_PRC_CI_AS NOT NULL
) ON [PRIMARY]
GO
ALTER TABLE [dbo].[TreeClass] WITH NOCHECK ADD
CONSTRAINT [PK_TreeClass] PRIMARY KEY CLUSTERED
(
[TC_id]
) ON [PRIMARY]
GO
if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[FN_32GetSubClass]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[FN_32GetSubClass]
GO
if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[FN_32GetTopClass]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[FN_32GetTopClass]
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS OFF
GO
CREATE FUNCTION FN_32GetSubClass (@InputId int,@IdStr varchar(8000)='',@LevelCount int=-1)
/*
参数: @InputId,被搜索子类的ID
@IdStr,一个特殊参数,用于在递归中传数据,注意:调用函数时一定要传入‘’空值
@LevelCount 用于判断是不是递归调用的开始层
*/
RETURNS Varchar(8000)
AS
BEGIN
Declare @TC_ID int,@TC_PID int,@StartLevel int,@Id32 int
if @LevelCount=-1
begin
set @StartLevel=@@NESTLEVEL
set @LevelCount=@StartLevel
end
else
set @StartLevel=-1
If @IdStr='' Set @IdStr=''''+cast(@InputId as varchar)+''''
DECLARE TreeClass CURSOR local FOR --定义游标
SELECT TC_Id,TC_PID
FROM TreeClass
where TC_PID=@InputId
OPEN TreeClass
FETCH NEXT FROM TreeClass
INTO @TC_ID,@TC_PID
WHILE @@FETCH_STATUS = 0 --循环游标,即循环当前类的弟一级子类
BEGIN
select @IdStr=@IdStr+','+''''+cast(@tC_ID as varchar)+''''
if @@NESTLEVEL<32
set @IdStr=dbo.FN_32GetSubClass (@TC_ID,@IdStr,@LevelCount) --递归,自己调用自己。
else
set @IdStr='['+cast(@tC_ID as varchar)+']'+@IdStr
FETCH NEXT FROM TreeClass
INTO @tC_ID,@TC_PID
End
CLOSE TreeClass
DEALLOCATE TreeClass
while @StartLevel=@@NESTLEVEL and charindex(']',@IdStr)>0
begin
set @Id32=substring(@IdStr,2,charindex(']',@IdStr)-2)
set @IdStr=dbo.FN_32GetSubClass (@Id32,@IdStr,@LevelCount)
set @IdStr=replace(@IdStr,'['+cast(@Id32 as varchar)+']','')
end
Return @IdStr
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS OFF
GO
CREATE FUNCTION FN_32GetTopClass (@InputId int,@IdStr varchar(8000)='',@type int=0,@LevelCount int=-1)
RETURNS Varchar(8000)
AS
BEGIN
Declare @TC_ID int,@TC_PID int,@StartLevel int,@Id32 int
if @LevelCount=-1
begin
set @StartLevel=@@NESTLEVEL
set @LevelCount=@StartLevel
end
else
set @StartLevel=-1
DECLARE TreeClass CURSOR local FOR
SELECT TC_Id,TC_PID
FROM TreeClass
where TC_ID=@InputId
OPEN TreeClass
FETCH NEXT FROM TreeClass
INTO @TC_ID,@TC_PID
WHILE @@FETCH_STATUS = 0
BEGIN
if @type=1
begin
if @IdStr<>'' select @IdStr=','+@IdStr
select @IdStr=''''+cast(@tC_ID as varchar)+''''+@IdStr
end
else
if @TC_PID=0 select @IdStr=cast(@tC_ID as varchar)
if @@NESTLEVEL<32
select @IdStr=dbo.FN_32GetTopClass (@TC_PID,@IdStr,@type,@LevelCount)
else
set @IdStr=@IdStr+'['+cast(@tC_ID as varchar)+']'
FETCH NEXT FROM TreeClass
INTO @tC_ID,@TC_PID
End
CLOSE TreeClass
DEALLOCATE TreeClass
while @StartLevel=@@NESTLEVEL and charindex(']',@IdStr)>0
begin
set @Id32=substring(@IdStr,charindex('[',@Idstr)+1,charindex(']',@IdStr)-1-charindex('[',@Idstr))
set @IdStr=dbo.FN_32GetTopClass (@Id32,@IdStr,@type,@LevelCount)
set @IdStr=replace(@IdStr,'['+cast(@Id32 as varchar)+']','')
end
Return @IdStr
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO
--------------------------例子表数据,将下面数据写入TXT文档,DTS导入。
TC_id,TC_PID,TC_OtherTypeID,TC_Name
1,0,,宇宙
2,1,,银河系
3,2,,太阳系
4,3,,地球
5,4,,亚洲
6,5,,中国
7,6,,省
8,7,,江苏
9,8,,苏州
10,9,,公园
11,10,,苏州乐园
12,11,,山
13,12,,石头
14,13,,树
15,14,,枝
17,15,,叶
18,17,,虫
21,18,,眼睛
22,21,,眼球
24,22,,泪
25,24,,水
26,25,,氧
28,26,,A
30,28,,B
31,30,,C
32,31,,D
33,32,,E
34,33,,F
35,34,,G
36,35,,H
37,36,,I
38,37,,L
39,38,,M
40,39,,N
41,40,,O
42,41,,P
43,42,,Q
44,43,,R
45,44,,S
46,45,,T
47,46,,U
48,47,,V
49,48,,W
50,49,,X
51,50,,Y
52,51,,Z
53,52,,a
54,53,,b
55,54,,c
56,55,,d
57,56,,e
58,57,,f
59,58,,g
60,59,,h
61,60,,i
62,61,,j
63,62,,k
64,63,,l
65,64,3,m
66,65,,n
67,66,,o
68,67,,p
69,68,,q
70,69,,r
71,70,,s
72,71,,t
73,72,,u
74,73,,v
75,74,,w
76,75,,x
77,76,,y
78,77,,z
---------------------------应用实例
--得到ID为1的所有下层类别ID串
declare @aa varchar(8000)
set @aa=dbo.FN_32GetSubClass(1,default,default)
print @aa
--查询ID为1的所有下层记录
declare @aa varchar(8000)
set @aa=dbo.FN_32GetSubClass(1,default,default)
print @aa
exec('SELECT * from treeclass where TC_id in ('+ @aa +')')
方法二,
select * from treeclass where charindex(''''+cast(TC_id as varchar)+'''',dbo.fn_32getsubclass(1,default,default))>0
方法二有一个问题,在我这边运行非常的慢,不知道为什么,有没有高手解释一下,谢谢!!
--得到ID为78顶层ID
declare @aa varchar(8000)
set @aa=dbo.fn_32gettopclass(78,default,default,default)
print @aa
--得到提供ID:78所在枝的所有ID
declare @aa varchar(8000)
set @aa=dbo.FN_32GetSubClass(dbo.fn_32gettopclass(78,default,default,default),default,default)
print @aa
--得到当前ID到顶层的ID串
declare @aa varchar(8000)
set @aa=dbo.fn_32gettopclass(78,default,1,default)
print @aa
---按照树顺序排序,
select * from treeclass order by dbo.fn_gettopclass(tc_id,'',1)
原来的函数可以实现,现在的函数不行,因为32以后的ID不是连续的在返回结果中,现在是,先得到所有32层以前的,再逐个处理超过32层的分枝,所以需要这个功能的话需要适当修改函数,当然我想是很容易实现的。