oracle树形父子关系递归检索

1、初始化数据

id parent_id

1

2 1

3 1

4 2

5 2

6 4

7 4

8

9 8
create table connect_table (ID int, PARENT_ID int);
insert into connect_table (ID, PARENT_ID)

values (1, null);
insert into connect_table (ID, PARENT_ID)

values (2, 1);
insert into connect_table (ID, PARENT_ID)

values (3, 1);
insert into connect_table (ID, PARENT_ID)

values (4, 2);
insert into connect_table (ID, PARENT_ID)

values (5, 2);
insert into connect_table (ID, PARENT_ID)

values (6, 4);
insert into connect_table (ID, PARENT_ID)

values (7, 4);
insert into connect_table (ID, PARENT_ID)

values (8, null);
insert into connect_table (ID, PARENT_ID)

values (9, 8);
2、分析
这个表是2棵树，2个根节点分别是 id=1 or id =8
connect by prior id=parent_id -----自根节点开始便利到叶子节点
start with id =? ----选一个根，则查询出该root的所有分支和叶子
3、适用场合举例
查询id=8是否为id=1的分支或叶子，是返回y，否返回n。
select t.*, level from connect_table t connect by prior id =parent_id start with id=1; ---显示level
select decode(count(1), 0, 'n', 'y')

from dual

where 8 in (select id

from connect_table t

connect by prior id = parent_id

start with id =1);