如何使用Oracle CONNECT BY獲取鏈接到某個值的層次結構中的所有值

Question

關系模型是

1   3   
 \ / \
  2   4
   \
    7   5     8
     \ /     /
      6     9

表是：

select 2 child, 1 father from dual
union all
select 2 child, 3 father from dual
union all
select 4 child, 3 father from dual
union all
select 7 child, 2 father from dual
union all
select 6 child, 5 father from dual
union all
select 6 child, 7 father from dual
union all
select 9 child, 8 father from dual

如何獲得與CHILD或FATHER = 2值鏈接的所有值？

肯定是

1,2,3,4,5,6,7

並不是

8,9

因為它不鏈接到值2。

如何通過使用CONNECT BY語句實現此目的？ 謝謝。

ps這個解決方案離我很近，但不適用於我的模型：

使用oracle connect by查找鄰接表模型中的所有節點

數據庫版本-10.2.0.5.0

用Oracle連接模型

因此，近似策略可能是這樣的（例如，以node = 7開頭）：

步驟1（方向=向上）

select t1.father,connect_by_root father as root,connect_by_isleaf from 
(my_table) t1
start with father=7
connect by prior father = child 

結果是7,2,1,3，其中1,3是高級根（isleaf = 1）

步驟2（取得1,3方向的路線=下）

select t1.child,connect_by_root father as root,connect_by_isleaf from 
(my_table) t1
start with father=1
connect by father = prior child 

結果是2,7,6，其中6是低級根（isleaf = 1）

select t1.child,connect_by_root father as root,connect_by_isleaf from 
(my_table) t1
start with father=3
connect by father = prior child

結果是2,7,6,4，其中6,4是低級根（isleaf = 1）

第3步（獲取6,4個方向的路線=向上）

select t1.father,connect_by_root father as root,connect_by_isleaf from 
(my_table) t1
start with child=6
connect by prior father = child 

結果是5,7,2,1,3，其中5,1,3是高級根（isleaf = 1）是我找到了node = 5的結果嗎？

然后我必須將方向更改為向下..然后再次向上..然后再次向下..

但是如何將所有這些步驟整合到一個選擇中呢？ 對於初學者來說很難。 請幫幫我。

Answer 1

對於您的輸出，您不需要定向圖形，因此將反向鏈接添加到所有現有鏈接。 這就是我在子查詢“ bi”中所做的事情。 然后，您可以使用Nocyle通過查詢進行連接。

    with h as (
                     SELECT 2 child, 1 father FROM dual
                     UNION ALL
                     SELECT 2 child, 3 father FROM dual
                     UNION ALL
                     SELECT 4 child, 3 father FROM dual
                     UNION ALL
                     SELECT 7 child, 2 father FROM dual
                     UNION ALL
                     SELECT 6 child, 5 father FROM dual
                     UNION ALL
                     SELECT 6 child, 7 father FROM dual
                     UNION ALL
                     SELECT 9 child, 8 father FROM dual
            ),
    bi as (select * from h union all select father , child from h )     
    select distinct father from bi
    start with child = 2
    connect by nocycle
    prior father = child

我在查詢中使用“ with”符號，以提高可讀性。

Answer 2

這是您要的：

SELECT     child
FROM       (
             SELECT 2 child, 1 father FROM dual
             UNION ALL
             SELECT 2 child, 3 father FROM dual
             UNION ALL
             SELECT 4 child, 3 father FROM dual
             UNION ALL
             SELECT 7 child, 2 father FROM dual
             UNION ALL
             SELECT 6 child, 5 father FROM dual
             UNION ALL
             SELECT 6 child, 7 father FROM dual
             UNION ALL
             SELECT 9 child, 8 father FROM dual
           )
START WITH father IN ( SELECT  father
                       FROM     
                       (
                                   SELECT 2 child, 1 father FROM dual
                                   UNION ALL
                                   SELECT 2 child, 3 father FROM dual
                                   UNION ALL
                                   SELECT 4 child, 3 father FROM dual
                                   UNION ALL
                                   SELECT 7 child, 2 father FROM dual
                                   UNION ALL
                                   SELECT 6 child, 5 father FROM dual
                                   UNION ALL
                                   SELECT 6 child, 7 father FROM dual
                                   UNION ALL
                                   SELECT 9 child, 8 father FROM dual
                        )
                       START WITH child = 2
                       CONNECT BY PRIOR father = child)
CONNECT BY PRIOR child = father
UNION
SELECT     father
FROM       (
SELECT 2 child, 1 father FROM dual
UNION ALL
SELECT 2 child, 3 father FROM dual
UNION ALL
SELECT 4 child, 3 father FROM dual
UNION ALL
SELECT 7 child, 2 father FROM dual
UNION ALL
SELECT 6 child, 5 father FROM dual
UNION ALL
SELECT 6 child, 7 father FROM dual
UNION ALL
SELECT 9 child, 8 father FROM dual
          )
START WITH child IN (
SELECT     child
FROM       (
             SELECT 2 child, 1 father FROM dual
             UNION ALL
             SELECT 2 child, 3 father FROM dual
             UNION ALL
             SELECT 4 child, 3 father FROM dual
             UNION ALL
             SELECT 7 child, 2 father FROM dual
             UNION ALL
             SELECT 6 child, 5 father FROM dual
             UNION ALL
             SELECT 6 child, 7 father FROM dual
             UNION ALL
             SELECT 9 child, 8 father FROM dual
           )
        START WITH child = 2
          CONNECT BY PRIOR child = father)


CONNECT BY PRIOR father = child;

該查詢包含兩個塊。 第一個塊提取2個所有父親的所有孩子的父親（參數），第二個塊（使用聯合以避免重復）提取2個所有孩子的所有父親的父親。

Answer 3

如果我理解正確，則可以使用connect_by_root函數，如下所示：

select hier.child , hier.father
from 
(
  select t.* , connect_by_root(father) top_father
  from 
  (
    select 2 child, 1 father from dual
    union all
    select 2 child, 3 father from dual
    union all
    select 4 child, 3 father from dual
    union all
    select 7 child, 2 father from dual
    union all
    select 6 child, 5 father from dual
    union all
    select 6 child, 7 father from dual
    union all
    select 9 child, 8 father from dual
  ) t
  connect by t.child = prior t.father
) hier
where 2 in (hier.child , hier.top_father);