use test set nocount on if object_id ( ' Dept ' , ' U ' ) is not null drop table Dept go create table Dept(ID int ,ParentID int ,Name varchar ( 20 )) insert into Dept select 1 , 0 , ' AA ' insert into Dept select 2 , 1 , ' BB ' insert into
use test
set nocount on
if object_id('Dept','U') isnotnull
drop table Dept
go
create table Dept(ID int,ParentID int,Name varchar(20))
insert into Dept select 1,0,'AA'
insert into Dept select 2,1,'BB'
insert into Dept select 3,1,'CC'
insert into Dept select 4,2,'DD'
insert into Dept select 5,3,'EE'
insert into Dept select 6,0,'FF'
insert into Dept select 7,6,'GG'
insert into Dept select 8,7,'HH'
insert into Dept select 9,7,'II'
insert into Dept select 10,7,'JJ'
insert into Dept select 11,9,'KK'
go
SELECT*FROM Dept;
--查询树状结构某节点的上级所有根节点。
with cte_root(ID,ParentID,NAME)
as
(
--起始条件
select ID,ParentID,NAME from Dept where ID='1'--查询条件
union all
--递归条件
select a.ID,a.ParentID,a.NAME from Dept a
innerjoin cte_root b --执行递归，这里就要理解下了
on a.ID=b.ParentID --根据基础表条件查询子节点(a.ID)，通过CTE递归找到其父节点(b.ParentID)。
) --可以和下面查询子节点的cte_child对比。
select*from cte_root ;
--查询树状结构某节点下的所有子节点。
with cte_child(ID,ParentID,NAME)
as
(
--起始条件
select ID,ParentID,NAME
from Dept
where Name ='II'--列出父节点查询条件
union all
--递归条件
select a.ID,a.ParentID,a.NAME
from Dept a
inner join
cte_child b
on a.ID=b.ParentID --根据查询到的父节点(a.Parent)，通过CTE递归查询出其子节点(b.ID)
)
select*from cte_child --可以改变之前的查询条件'II'再测试结果
ID ParentID Name
----------- ----------- --------------------
10 AA
21 BB
31 CC
42 DD
53 EE
60 FF
76 GG
87 HH
97 II
107 JJ
119 KK
ID ParentID NAME
----------- ----------- --------------------
97 II
76 GG
60 FF
ID ParentID NAME
----------- ----------- --------------------
97 II
119 KK
