Oracle - start with connect by 树操作
准备工作:建表、插数据
create table tprior
( parentid number(10),
subid number(10)
);
insert into tprior values ( 1, 2 );
insert into tprior values ( 1, 3 );
insert into tprior values ( 2, 4 );
insert into tprior values ( 2, 5 );
insert into tprior values ( 3, 6 );
insert into tprior values ( 3, 7 );
insert into tprior values ( 5, 8 );
insert into tprior values ( 5, 9 );
insert into tprior values ( 7, 10 );
insert into tprior values ( 7, 11 );
insert into tprior values ( 10, 12 );
insert into tprior values ( 10, 13 );
commit;
表数据概况、B-Tree 结构
select * from tprior;
- 表数据
- 对应的 B-Tree 结构:
SQL
select parentid, subid, level
from tprior
start with subid = 7
connect by subid = prior parentid
order by level;
- start with subid = 7,首先找出 level 1 记录 (parentid, 7) 即 (3, 7),作为起始点
- connect by subid = prior parentid,表示从子节点往父节点方向遍历,即 level 1 记录 (3, 7) 中的父节点 3,作为 level 2 记录的子节点,查得 level 2 记录 (1, 3)
- 一直查到根节点结束
select parentid, subid, level
from tprior
start with subid = 7
connect by prior subid = parentid
order by level;
- start with subid = 7,level 1 记录 (3, 7)
- connect by prior subid = parentid,从父节点向子节点方向遍历,即 level 1 记录 (3, 7) 中的子节点 7,作为 level 2 记录的父节点,查得 level 2 记录 (7, 10)、(7, 11)
- 以此类推,一直查到没有子节点的节点为止
select parentid, subid, level
from tprior
start with parentid = 7
connect by subid = prior parentid
order by level;
- start with parentid = 7,level 1 记录 (7, 10)、(7, 11)
- connect by subid = prior parentid,子节点向节点方向遍历,上一组数据的父节点 7,作为 level 2 记录的子节点,查得 (3, 7)、(3, 7) {为什么 level 2 是 2 组数据呢?因为 level 1 记录有 2 个}
- 以此类推
select parentid, subid, level
from tprior
start with parentid = 7
connect by prior subid = parentid
order by level;
- start with parentid = 7,level 1 记录 (7, subid)、(7, subid) 即 (7, 10)、(7, 11)
- connect with prior parentid = subid,子节点向父节点遍历,level 1 记录的父节点 10、11,作为 level 2 记录的子节点,查得 (10, 13)、(10, 12)
- 以此类推
select parentid, subid, level
from tprior
where parentid > 3
start with subid = 12
connect by subid = prior parentid
order by level;
- SQL 执行顺序是:先执行 start with,再执行 connect by,最后执行 where。
- 结论:where 只是树结构的修剪,不改变树的层次结构。
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