SQL Server 2005 中的树形数据处理示例-1

王朝mssql·作者佚名  2006-01-31
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SQL Server 2005 中的树形数据处理示例

-- 创建测试数据

if exists (select * from dbo.sysobjects where id = object_id(N'[tb]') and OBJECTPROPERTY(id, N'IsUserTable') = 1)

drop table [tb]

GO

-- 示例数据

create table [tb]([id] int PRIMARY KEY,[pid] int,name nvarchar(20))

INSERT [tb] SELECT 1,0,N'中国'

UNION ALL SELECT 2,0,N'美国'

UNION ALL SELECT 3,0,N'加拿大'

UNION ALL SELECT 4,1,N'北京'

UNION ALL SELECT 5,1,N'上海'

UNION ALL SELECT 6,1,N'江苏'

UNION ALL SELECT 7,6,N'苏州'

UNION ALL SELECT 8,7,N'常熟'

UNION ALL SELECT 9,6,N'南京'

UNION ALL SELECT 10,6,N'无锡'

UNION ALL SELECT 11,2,N'纽约'

UNION ALL SELECT 12,2,N'旧金山'

GO

-- 查询指定id的所有子

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_cid]') and xtype in (N'FN', N'IF', N'TF'))

drop function [dbo].[f_cid]

GO

-- =====================================================

-- 查询指定id的所有子

-- 邹建 2005-07(引用请保留此信息)

-- 调用示例

/*--调用(查询所有的子)

SELECT A.*,层次=B.[level]

FROM [tb] A,f_cid(2)B

WHERE A.[id]=B.[id]

--*/

-- =====================================================

CREATE FUNCTION f_cid(@id int)

RETURNS TABLE

AS

RETURN(

WITH ctb([id],[level])

AS(

SELECT [id],1 FROM [tb]

WHERE [pid]=@id

UNION ALL

SELECT A.[id],B.[level]+1

FROM [tb] A,ctb B

WHERE A.[pid]=B.[id])

SELECT * FROM ctb

--如果只显示最明细的子(下面没有子),则将上面这句改为下面的

-- SELECT * FROM ctb A

-- WHERE NOT EXISTS(

-- SELECT 1 FROM [tb] WHERE [pid]=A.[id])

)

GO

--调用(查询所有的子)

SELECT A.*,层次=B.[level]

FROM [tb] A,f_cid(2)B

WHERE A.[id]=B.[id]

GO

-- 查询指定id的所有父

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_pid]') and xtype in (N'FN', N'IF', N'TF'))

drop function [dbo].[f_pid]

GO

-- =====================================================

-- 查询指定id的所有父

-- 邹建 2005-07(引用请保留此信息)

-- 调用示例

/*--调用(查询所有的父)

SELECT A.*,层次=B.[level]

FROM [tb] A,[f_pid](2)B

WHERE A.[id]=B.[id]

--*/

-- =====================================================

CREATE FUNCTION [f_pid](@id int)

RETURNS TABLE

AS

RETURN(

WITH ptb([id],[level])

AS(

SELECT [pid],1 FROM [tb]

WHERE [id]=@id

AND [pid]<>0

UNION ALL

SELECT A.[pid],B.[level]+1

FROM [tb] A,ptb B

WHERE A.[id]=B.[id]

AND [pid]<>0)

SELECT * FROM ptb

)

GO

--调用(查询所有的父)

SELECT A.*,层次=B.[level]

FROM [tb] A,[f_pid](7)B

WHERE A.[id]=B.[id]

GO

-- 树形分级显示

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_id]') and xtype in (N'FN', N'IF', N'TF'))

drop function [dbo].[f_id]

GO

-- =====================================================

-- 级别及排序字段(树形分级显示)

-- 邹建 2005-07(引用请保留此信息)

-- 调用示例

/*--调用实现树形显示

--调用函数实现分级显示

SELECT N'|'+REPLICATE('-',B.[level]*4)+A.name

FROM [tb] A,f_id()B

WHERE a.[id]=b.[id]

ORDER BY b.sid

--当然,这个也可以根本不用写函数,直接排序即可

WITH stb([id],[level],[sid])

AS(

SELECT [id],1,CAST(RIGHT(10000+[id],4) as varchar(8000))

FROM [tb]

WHERE [pid]=0

UNION ALL

SELECT A.[id],B.[level]+1,B.sid+RIGHT(10000+A.[id],4)

FROM [tb] A,stb B

WHERE A.[pid]=B.[id])

SELECT N'|'+REPLICATE('-',B.[level]*4)+A.name

FROM [tb] A,stb B

WHERE a.[id]=b.[id]

ORDER BY b.sid

--*/

-- =====================================================

CREATE FUNCTION f_id()

RETURNS TABLE

AS

RETURN(

WITH stb([id],[level],[sid])

AS(

SELECT [id],1,CAST(RIGHT(10000+[id],4) as varchar(8000))

FROM [tb]

WHERE [pid]=0

UNION ALL

SELECT A.[id],B.[level]+1,B.sid+RIGHT(10000+A.[id],4)

FROM [tb] A,stb B

WHERE A.[pid]=B.[id])

SELECT * FROM stb

)

GO

--调用函数实现分级显示

SELECT N'|'+REPLICATE('-',B.[level]*4)+A.name

FROM [tb] A,f_id()B

WHERE a.[id]=b.[id]

ORDER BY b.sid

GO

 
 
 
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