这是此查询的完整示例数据集，没有修剪任何节点与搜索字符串匹配的树：
级别父 ID 文本 --------------------------------------------- 0 0 1 顶层 1 1 2 富 1 1 3 其他 1 1 4 英尺 0 0 7 顶层2 1 7 8 秒级 1 7 9 另一个二级
如果用户搜索“foo”，我需要返回以下内容：
0 0 1 顶层 1 1 2 富 1 1 4 英尺
实际情况稍微复杂一些（即我想要返回的树中的三个级别），但这抓住了问题。在英语中，返回与搜索字符串匹配的节点的祖先树，该搜索字符串从文本列上的匹配节点开始，并返回所有祖先。
我是 Oracle 的新手（至少最近是这样）并尝试添加到 CONNECT BY 子句但没有任何成功 - 总是返回以下内容：
1 1 2 富 1 1 4 英尺
PS - 关于此的 oracle 文档和示例暗示 CONNECT_BY_ROOT 将捕获祖先，但它似乎所做的只是返回顶级（ROOT）值。
要从下往上遍历，重要的是CONNECT BY PRIOR)之后的值的顺序，order by用于反转输出（因为根是 foo）并distinct删除重复的顶层值：
SELECT DISTINCT LEVEL, id, text
FROM t1
CONNECT BY PRIOR parent = id
START WITH text = 'foo'
ORDER BY LEVEL DESC
注意：如果您将孩子添加到 foo 并切换 CONNCT BY PRIOR id = parent 您将获得孩子
如果您想查看整个层次结构，您可以找到树的顶部（通过查找没有父级的行）
SELECT id
FROM t1
WHERE parent=0
CONNECT BY PRIOR parent = id
START WITH text = 'foo'
然后将其用作 START WITH id（并反转外部查询中树遍历的顺序，id = parent）：
SELECT DISTINCT LEVEL, id, text
FROM t1
CONNECT BY PRIOR id = parent
START WITH id IN
(
SELECT id
FROM t1
WHERE parent=0
CONNECT BY PRIOR parent = id
START WITH text = 'foo'
)
ORDER BY LEVEL DESC
Depending upon your use of the LEVEL column (as per my comment).
Info on Oracle Hierarchical Queries: http://download.oracle.com/docs/cd/B19306_01/server.102/b14200/queries003.htm
This returns what you ask for if LEVEL is the Oracle pseudocolumn:
WITH t AS (SELECT 0 AS parent,
1 AS id,
'toplevel' AS text FROM DUAL
UNION
SELECT 1 AS parent,
2 AS id,
'foo' AS text FROM DUAL
UNION
SELECT 1 AS parent,
3 AS id,
'sumthin else' AS text FROM DUAL
UNION
SELECT 1 AS parent,
4 AS id,
'foo' AS text FROM DUAL
UNION
SELECT 0 AS parent,
7 AS id,
'toplevel2' AS text FROM DUAL
UNION
SELECT 7 AS parent,
8 AS id,
'secondlevel' AS text FROM DUAL
UNION
SELECT 7 AS parent,
9 AS id,
'anothersecondlevel' AS text FROM DUAL
)
SELECT UNIQUE
level,
parent,
id,
text
FROM t
START WITH text = 'foo'
CONNECT BY PRIOR parent = id
ORDER BY parent;
Returns:
LEVEL PARENT ID TEXT
2 0 1 toplevel
1 1 2 foo
1 1 4 foo
If LEVEL is a column in your table then:
WITH t AS (SELECT 0 AS tlevel,
0 AS parent,
1 AS id,
'toplevel' AS text FROM DUAL
UNION
SELECT 1 AS tlevel,
1 AS parent,
2 AS id,
'foo' AS text FROM DUAL
UNION
SELECT 1 AS tlevel,
1 AS parent,
3 AS id,
'sumthin else' AS text FROM DUAL
UNION
SELECT 1 AS tlevel,
1 AS parent,
4 AS id,
'foo' AS text FROM DUAL
UNION
SELECT 0 AS tlevel,
0 AS parent,
7 AS id,
'toplevel2' AS text FROM DUAL
UNION
SELECT 1 AS tlevel,
7 AS parent,
8 AS id,
'secondlevel' AS text FROM DUAL
UNION
SELECT 1 AS tlevel,
7 AS parent,
9 AS id,
'anothersecondlevel' AS text FROM DUAL
)
SELECT UNIQUE
tlevel,
parent,
id,
text
FROM t
START WITH text = 'foo'
CONNECT BY PRIOR parent = id
ORDER BY parent;
Returns:
TLEVEL PARENT ID TEXT
0 0 1 toplevel
1 1 2 foo
1 1 4 foo
Hope it helps...
我对“仅返回完整树的 Oracle 分层查询”部分的看法是：
SELECT Parent_ID
, Child_ID
, INITIAL_Parent_ID
, child_level
, INITIAL_Parent_ID || children_CHAIN AS CONNECTION_CHAIN
FROM ( SELECT CONNECT_BY_ROOT Parent_ID AS INITIAL_Parent_ID
, Parent_ID
, Child_ID
, LEVEL AS child_level
, SYS_CONNECT_BY_PATH(Child_ID, '/') AS children_chain
FROM REF_DECOMMISSIONS
CONNECT BY NOCYCLE Parent_ID = PRIOR Child_ID
)
WHERE INITIAL_Parent_ID NOT IN (SELECT Child_ID FROM REF_DECOMMISSIONS)
;
要将树过滤为包含特定子项的树，您需要在最后一个 WHERE 中添加另一个条件以进一步过滤 INITIAL_Parent_ID。查询将变为：
WITH ROOT_TREES AS
( SELECT Parent_ID
, Child_ID
, INITIAL_Parent_ID
, child_level
, INITIAL_Parent_ID || children_CHAIN AS CONNECTION_CHAIN
FROM ( SELECT CONNECT_BY_ROOT Parent_ID AS INITIAL_Parent_ID
, Parent_ID
, Child_ID
, LEVEL AS child_level
, SYS_CONNECT_BY_PATH(Child_ID, '/') AS children_chain
FROM REF_DECOMMISSIONS
CONNECT BY NOCYCLE Parent_ID = PRIOR Child_ID
)
WHERE INITIAL_Parent_ID NOT IN (SELECT Child_ID FROM REF_DECOMMISSIONS)
)
SELECT *
FROM root_trees
WHERE INITIAL_Parent_ID IN (SELECT INITIAL_Parent_ID
FROM root_trees
WHERE Child_ID = 123)
;