问题提出：
一个高级SQL语句问题
假设有一张表，有这样一些数据
a b
1 2
0 2
2 3
3 4
4
5 6
6 7
12 7
7
a是子节点 b是父节点 ,如果b为null 则为一棵树的根节点
根据A与B的这个关系 找出一棵树的结构显示出来 ，并显示出每个节点在书中的层级，将树的结构打印出来
我这边假设树最多有5个层级，SQL如下：
with tt as(
select 1 a,2 b from dual
union all SELECT 0,2 from dual
union all select 2,3 from dual
union all select 3,4 from dual
union all select 4,null from dual
union all select 5,6 from dual
union all select 6,7 from dual
union all select 12,7 from dual
union all select 7,null from dual)
SELECT
a
,tree_lvl
,root
,tree_lvl_path
,(CASE
WHEN tree_lvl >1 THEN SUBSTR(tree_lvl_path,1,instr(tree_lvl_path,',',1,1)-1 )
ELSE tree_lvl_path
END) COL01
,(CASE
WHEN tree_lvl <2 THEN NULL
WHEN tree_lvl =2 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,1)+1 )
WHEN tree_lvl >2 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,1)+1,instr(tree_lvl_path,',',1,2)- instr(tree_lvl_path,',',1,1)-1 )
END) COL02
,(CASE
WHEN tree_lvl <3 THEN NULL
WHEN tree_lvl =3 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,2)+1 )
WHEN tree_lvl >3 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,2)+1,instr(tree_lvl_path,',',1,2)- instr(tree_lvl_path,',',1,1)-1 )
END) COL03
,(CASE
WHEN tree_lvl <4 THEN NULL
WHEN tree_lvl =4 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,3)+1 )
WHEN tree_lvl >4 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,3)+1,instr(tree_lvl_path,',',1,3)- instr(tree_lvl_path,',',1,2)-1 )
END) COL04
,(CASE
WHEN tree_lvl <5 THEN NULL
WHEN tree_lvl =5 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,4)+1 )
WHEN tree_lvl >5 THEN SUBSTR(tree_lvl_path,instr(tree_lvl_path,',',1,4)+1,instr(tree_lvl_path,',',1,5)- instr(tree_lv
在现代的多用户多任务系统中，必然会出现多个用户同时访问共享的某个对象，这个对象可能是表，行，或者内存结构，为了解决多个用户并发性访问带来的数据的安全性，完整性及一致性问题，必须要有一种机制，来使对这些共享资源的并发性访问串行化，oracle中的锁就可以提供这样的功能，当事务在对某个对象� ......
