這是一個CSDN老帖：
http://community.csdn.net/Expert/FAQ/FAQ_Index.asp?id=170559
我是抱着學習的心態看這個帖子的，下面把握學習結果總結一下。
樓主的問題是這樣的：
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表A：
id name
1 a
2 b
3 c
4 d
5 e
表B(id1,id2都與A表的id關聯，是聯合主外鍵):
id1 id2
1 2
2 3
2 4
3 5
這是一個部門上下級的關係，前面的是上級，後面的下級，我想得到所有部門的列表，按照級別關係寫成完整的字串，結果如下：
id full_name
1 a
2 a/b
3 a/b/c
4 a/b/d
5 a/b/c/d
請問怎麼寫？存儲過程或函數都可以，十分感謝！
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問題的關鍵是把用LEVEL關鍵字把層次關係搞搞清楚。
select level from table_B connect by prior id2=id1 start with id1=0;
LEVEL
----------
1
2
3
4
3
SQL> select level from table_B connect by prior id2=id1 start with id1=1;
LEVEL
----------
1
2
3
2
SQL> select level from table_B connect by prior id2=id1 start with id1=2;
LEVEL
----------
1
2
1
SQL> select level from table_B connect by prior id2=id1 start with id1=3;
LEVEL
----------
1
SQL> select level from table_B connect by prior id2=id1 start with id1=4;
LEVEL
----------
SQL> select level from table_B connect by prior id2=id1 start with id1=5;
LEVEL
----------
可以看出，LEVEL值表示id1領導下人員id2在id1集團所處的層數，被領導者緊跟在by prior後面。
select lpad(id2, level*length(id2), ' ') id,
2 ltrim(sys_connect_by_path(id2,'/'), '/') path
3 from Table_B
4 connect by prior id2=id1
5 start with id1=0
6 /
ID PATH
-------------------------------------------------------------------------------- --------------------------------------------------------------------------------
1 1
2 1/2
3 1/2/3
5 1/2/3/5
4 1/2/4
下面給出完整解答：
DROP TABLE Table_A;
Table dropped
SQL> create table Table_A (id number(4), name varchar2(20));
Table created
SQL> insert into Table_A values(1, 'a');
1 row inserted
SQL> insert into Table_A values(2, 'b');
1 row inserted
SQL> insert into Table_A values(3, 'c');
1 row inserted
SQL> insert into Table_A values(4, 'd');
1 row inserted
SQL> insert into Table_A values(5, 'e');
1 row inserted
SQL> commit;
Commit complete
SQL> DROP TABLE Table_B;
Table dropped
SQL> create table Table_B (id1 number(4), id2 number(4));
Table created
SQL> insert into table_B values(0,1);
1 row inserted
SQL> insert into Table_B values(1,2);
1 row inserted
SQL> insert into Table_B values(2,3);
1 row inserted
SQL> insert into Table_B values(2,4);
1 row inserted
SQL> insert into Table_B values(3,5);
1 row inserted
SQL> commit;
Commit complete
SQL> SELECT id2, ltrim(sys_connect_by_path(NAME, '/'), '/') path
2 from
3 (SELECT B.*, A.NAME
4 FROM Table_B B, Table_A A
5 WHERE B.id2=A.id)
6 connect by prior id2=id1
7 start with id1 = 0
8 ORDER BY id2
9 /
ID2 PATH
----- --------------------------------------------------------------------------------
1 a
2 a/b
3 a/b/c
4 a/b/d
5 a/b/c/e