Oracle 返回具有所有组合的树

Question

我需要一个返回组内所有女性的查询。

示例数据表（Oracle 11g）：

with t as (
  select 'K' as GROUP_NAME, 1 as ELEMENT_ID from dual
  union
  select 'K' as GROUP_NAME, 2 as ELEMENT_ID from dual
  union
  select 'K' as GROUP_NAME, 3 as ELEMENT_ID from dual
  union
  select 'L' as GROUP_NAME, 1 as ELEMENT_ID from dual
  union
  select 'L' as GROUP_NAME, 2 as ELEMENT_ID from dual
  union
  select 'L' as GROUP_NAME, 3 as ELEMENT_ID from dual
  union
  select 'P' as GROUP_NAME, 1 as ELEMENT_ID from dual
  union
  select 'P' as GROUP_NAME, 2 as ELEMENT_ID from dual
  union
  select 'P' as GROUP_NAME, 3 as ELEMENT_ID from dual
)
select * from t

K   1
K   2
K   3
L   1
L   2
L   3
P   1
P   2
P   3

我需要生成一个包含所有组合的树，而不是在组中重复它们（组 K、L、P）

一棵树不能包含一组树，即不能有这样一个分支的树：K1->L1->K2

K1
 +-- L1
   +-- P1
   +-- P2
   +-- P3
 +-- L2
   +-- P1
   +-- P2
   +-- P3
 +-- L3
   +-- P1
   +-- P2
   +-- P3
K2
 +-- L1
   +-- P1
   +-- P2
   +-- P3
 +-- L2
   +-- P1
   +-- P2
   +-- P3
 +-- L3
   +-- P1
   +-- P2
   +-- P3
K3
 +-- L1
   +-- P1
   +-- P2
   +-- P3
 +-- L2
   +-- P1
   +-- P2
   +-- P3
 +-- L3
   +-- P1
   +-- P2
   +-- P3
L1
  +-- K1
    +-- P1
    +-- P2
    +-- P3
  +-- K2
   ....

Answer 1

我讨厌回答这样的问题，因为 SO不是代码编写服务。

但这是一个简单的问题，我认为解决这个问题需要付出更多的努力，而不是仅仅回答，特别是因为您确实提供了测试数据和预期结果。

SELECT ltrim(sys_connect_by_path(group_name || element_id,'-'),'-') FROM t
START WITH t.group_name = 'K'
CONNECT BY ( t.group_name = 'L' AND PRIOR t.group_name = 'K' ) OR ( t.group_name = 'P' 
AND PRIOR t.group_name = 'L');

 +-----------+ | TREE_PATH | +-----------+ | K1 | | K1-L1 | | K1-L1-P1 | | K1-L1-P2 | | K1-L1-P3 | | K1-L2 | | K1-L2-P1 | | K1-L2-P2 | | K1-L2-P3 | | K1-L3 | | K1-L3-P1 | | K1-L3-P2 | | K1-L3-P3 | | K2 | | K2-L1 | | K2-L1-P1 | | K2-L1-P2 | | K2-L1-P3 | | K2-L2 | | K2-L2-P1 | | K2-L2-P2 | | K2-L2-P3 | | K2-L3 | | K2-L3-P1 | | K2-L3-P2 | | K2-L3-P3 | | K3 | | K3-L1 | | K3-L1-P1 | | K3-L1-P2 | | K3-L1-P3 | | K3-L2 | | K3-L2-P1 | | K3-L2-P2 | | K3-L2-P3 | | K3-L3 | | K3-L3-P1 | | K3-L3-P2 | | K3-L3-P3 | +-----------+

如果您不想要一棵树而只想要所有组合，那将是“交叉连接”。 像这样：

select * from t k CROSS JOIN t l CROSS JOIN t p
where k.group_name = 'K'
AND l.group_name = 'L'
and p.group_name = 'P'

 +------------+------------+--------------+--------------+--------------+--------------+ | GROUP_NAME | ELEMENT_ID | GROUP_NAME_1 | ELEMENT_ID_1 | GROUP_NAME_2 | ELEMENT_ID_2 | +------------+------------+--------------+--------------+--------------+--------------+ | K | 1 | L | 1 | P | 1 | | K | 1 | L | 1 | P | 2 | | K | 1 | L | 1 | P | 3 | | K | 1 | L | 2 | P | 1 | | K | 1 | L | 2 | P | 2 | | K | 1 | L | 2 | P | 3 | | K | 1 | L | 3 | P | 1 | | K | 1 | L | 3 | P | 2 | | K | 1 | L | 3 | P | 3 | | K | 2 | L | 1 | P | 1 | | K | 2 | L | 1 | P | 2 | | K | 2 | L | 1 | P | 3 | | K | 2 | L | 2 | P | 1 | | K | 2 | L | 2 | P | 2 | | K | 2 | L | 2 | P | 3 | | K | 2 | L | 3 | P | 1 | | K | 2 | L | 3 | P | 2 | | K | 2 | L | 3 | P | 3 | | K | 3 | L | 1 | P | 1 | | K | 3 | L | 1 | P | 2 | | K | 3 | L | 1 | P | 3 | | K | 3 | L | 2 | P | 1 | | K | 3 | L | 2 | P | 2 | | K | 3 | L | 2 | P | 3 | | K | 3 | L | 3 | P | 1 | | K | 3 | L | 3 | P | 2 | | K | 3 | L | 3 | P | 3 | +------------+------------+--------------+--------------+--------------+--------------+