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SQL 遍历父子关系表(二叉树)获得所有子节点 所有父节点

發(fā)布時間:2025/7/14 35 豆豆
生活随笔 收集整理的這篇文章主要介紹了 SQL 遍历父子关系表(二叉树)获得所有子节点 所有父节点 小編覺得挺不錯的,現(xiàn)在分享給大家,幫大家做個參考.

--建立測試環(huán)境
Create Table A
(IDInt,
?fatherIDInt,
?NameVarchar(10)
)
Insert A Select 1,??????? NULL,?????? 'tt'
Union All Select 2,??????? 1,????????? 'aa'
Union All Select 3,??????? 1,????????? 'bb'
Union All Select 4,??????? 2,????????? 'cc'
Union All Select 5,??????? 2,????????? 'gg'
Union All Select 6,??????? 4,????????? 'yy'
Union All Select 7,??????? 4,????????? 'jj'
Union All Select 8,??????? 7,?????????? 'll'
Union All Select 9,??????? NULL,? 'uu'
Union All Select 10,?????? 9,???????? 'oo'
GO
--建立函數(shù)

//取字子節(jié)點
Create Function GetChildren(@ID Int)
Returns @Tree Table (ID Int, fatherID Int, Name Varchar(10))
As
Begin
Insert @Tree Select ID, fatherID, Name From A Where fatherID = @ID
While @@Rowcount > 0
Insert @Tree Select A.ID, A.fatherID, A.Name From A A Inner Join @Tree B On A.fatherID = B.ID And A.ID Not In (Select ID From @Tree)
Return
End
GO

//取父節(jié)點

set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
GO
ALTER Function [dbo].[GetParent](@ID Int)
Returns @Tree Table (ID Int, fatherID Int, NameVar Varchar(10))
As
Begin
Insert @Tree Select ID, fatherID, NameVar From A Where ID = @ID
While @@Rowcount > 0
Insert @Tree Select A.ID, A.fatherID, A.NameVar From A A Inner Join @Tree B On A.ID = B.fatherID And A.ID Not In (Select ID From @Tree)
Return
End

--測試
Select * From dbo.GetChildren(1)

Select * From dbo.GetParent(9)
GO


轉(zhuǎn)載于:https://www.cnblogs.com/aaa6818162/archive/2009/05/11/1453877.html

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