5

the table is very simple,pid means the parent id,and cid means the child id.And there may be more than one trees in the table.So my question is:
knowing several cid,how can we get the root ancestors
here is an example

pid cid
1 2
2 3
3 4
5 6
6 7
7 8

given cid = 4 or cid = 8,I want to get their root ancestors whose pid is 1 ro 5
finally,I'm using an oracle 10g

4

2 回答 2

7

在数据库环境中,顶层的外键很可能是空值,如下所示:

| pid  | cid  |
|------*------|
| null |  2   |
|  2   |  3   |
|  3   |  4   |
| null |  6   |
|  6   |  7   |
|  7   |  8   |

所以我建议使用类似的东西:

select connect_by_root(t1.cid) as startpoint,
       t1.cid                  as rootnode
  from your_table t1
 where connect_by_isleaf = 1
 start with t1.cid in (8, 4)
connect by prior t1.pid = t1.cid;

小提琴

于 2017-07-24T16:43:51.303 回答
4
select 
  t1.cid,
  connect_by_root(t1.pid) as root
from 
  your_table t1
  left join your_table t2
    on t2.cid = t1.pid
where t1.cid in (4, 8)
start with t2.cid is null
connect by t1.pid = prior t1.cid

小提琴

于 2013-03-24T06:42:31.893 回答