获取Oracle分层查询中的所有子路径答案

【问题标题】：Get all subpaths in a hierarchical query in Oracle获取Oracle分层查询中的所有子路径
【发布时间】：2020-04-30 14:29:57
【问题描述】：

我有一个包含销售部门列表的表格，如下所示：

+-------+---------------+-----------+
|DEPT_ID|DEPT_NAME      |DEPT_PARENT|
+-------+---------------+-----------+
|5500   |World          |           |
|5510   |Region 1       |5500       |
|5511   |Cell 1 Region 1|5510       |
|5512   |Cell 2 Region 1|5510       |
|5513   |Cell 3 Region 1|5510       |
|5514   |Cell 4 Region 1|5510       |
|5515   |Cell 5 Region 1|5510       |
|5520   |Region 2       |5500       |
|5521   |Cell 1 Region 2|5520       |
|5522   |Cell 2 Region 2|5520       |
|5530   |Region 3       |5500       |
|5531   |Cell 1 Region 3|5530       |
|5532   |Cell 2 Region 3|5530       |
|5540   |Region 4       |5500       |
|5533   |Cell 1 Region 4|5540       |
|5534   |Cell 2 Region 4|5533       |
|5590   |Region 5       |5500       |
|5591   |Cell 1 Region 5|5590       |
+-------+---------------+-----------+

我需要一个返回所有可能子路径的查询。所以前三行应该如下：

5500 5510
5500 5511
5510 5511

因此对于每个可能的子路径，它将返回路径的第一个和最后一个部门。这样做很容易获得路径：

    SELECT d.*, LTRIM (SYS_CONNECT_BY_PATH (dept_id, '-'), '-') AS PATH
      FROM depts d
START WITH dept_parent IS NULL
CONNECT BY PRIOR dept_id = dept_parent

但是我怎样才能获得所有可能的子路径？

【问题讨论】：

标签： sql oracle hierarchy hierarchical-data

【解决方案1】：

使用CONNECT_BY_ROOT 查找路径的起点。

Oracle 设置：

CREATE TABLE depts ( DEPT_ID, DEPT_NAME, DEPT_PARENT ) AS
SELECT 5500, 'World',           NULL FROM DUAL UNION ALL
SELECT 5510, 'Region 1',        5500 FROM DUAL UNION ALL
SELECT 5511, 'Cell 1 Region 1', 5510 FROM DUAL UNION ALL
SELECT 5512, 'Cell 2 Region 1', 5510 FROM DUAL UNION ALL
SELECT 5513, 'Cell 3 Region 1', 5510 FROM DUAL UNION ALL
SELECT 5514, 'Cell 4 Region 1', 5510 FROM DUAL UNION ALL
SELECT 5515, 'Cell 5 Region 1', 5510 FROM DUAL UNION ALL
SELECT 5520, 'Region 2',        5500 FROM DUAL UNION ALL
SELECT 5521, 'Cell 1 Region 2', 5520 FROM DUAL UNION ALL
SELECT 5522, 'Cell 2 Region 2', 5520 FROM DUAL UNION ALL
SELECT 5530, 'Region 3',        5500 FROM DUAL UNION ALL
SELECT 5531, 'Cell 1 Region 3', 5530 FROM DUAL UNION ALL
SELECT 5532, 'Cell 2 Region 3', 5530 FROM DUAL UNION ALL
SELECT 5540, 'Region 4',        5500 FROM DUAL UNION ALL
SELECT 5533, 'Cell 1 Region 4', 5540 FROM DUAL UNION ALL
SELECT 5534, 'Cell 2 Region 4', 5533 FROM DUAL UNION ALL
SELECT 5590, 'Region 5',        5500 FROM DUAL UNION ALL
SELECT 5591, 'Cell 1 Region 5', 5590 FROM DUAL;

查询：

SELECT CONNECT_BY_ROOT( dept_parent ) AS ancestor,
       dept_id,
       SYS_CONNECT_BY_PATH( dept_parent, '>' ) || '>' || dept_id AS path
FROM   depts
START WITH dept_parent IS NOT NULL
CONNECT BY PRIOR dept_id = dept_parent;

输出：

祖先 |部门 ID |小路 --------: | ------: | :-------------------- 5500 | 5510 | >5500>5510 5500 | 5511 | >5500>5510>5511 5500 | 5512 | >5500>5510>5512 5500 | 5513 | >5500>5510>5513 5500 | 5514 | >5500>5510>5514 5500 | 5515 | >5500>5510>5515 5500 | 5520 | >5500>5520 5500 | 5521 | >5500>5520>5521 5500 | 5522 | >5500>5520>5522 5500 | 5530 | >5500>5530 5500 | 5531 | >5500>5530>5531 5500 | 5532 | >5500>5530>5532 5500 | 5540 | >5500>5540 5500 | 5533 | >5500>5540>5533 5500 | 5534 | >5500>5540>5533>5534 5500 | 5590 | >5500>5590 5500 | 5591 | >5500>5590>5591 5510 | 5511 | >5510>5511 5510 | 5512 | >5510>5512 5510 | 5513 | >5510>5513 5510 | 5514 | >5510>5514 5510 | 5515 | >5510>5515 5520 | 5521 | >5520>5521 5520 | 5522 | >5520>5522 5530 | 5531 | >5530>5531 5530 | 5532 | >5530>5532 5533 | 5534 | >5533>5534 5540 | 5533 | >5540>5533 5540 | 5534 | >5540>5533>5534 5590 | 5591 | >5590>5591

db小提琴here

【讨论】：

【解决方案2】：

我了解到您想要生成一个包含层次结构中所有可能路径的闭包表。

这是一个使用标准递归查询来实现此目的的解决方案。我还添加了一个存储关系深度的列，因为这些信息在闭包表中通常很有用。

with cte(node_id, dept_id, dept_parent, lvl) as (
    select dept_id node_id, dept_id, dept_parent, 0 lvl from dept
    union all 
    select c.node_id, d.dept_id, d.dept_parent, c.lvl + 1
    from cte c
    inner join dept d on d.dept_id = c.dept_parent
)
select dept_id ancestor, node_id node, lvl 
from cte 
where lvl > 0
order by node, ancestor

这个 demo on DB Fiddle 与您的示例数据生成 30 行，其前 10 行是：

祖先 |节点 |层积层 --------: | ---: | --: 5500 | 5510 | 1 5500 | 5511 | 2 5510 | 5511 | 1 5500 | 5512 | 2 5510 | 5512 | 1 5500 | 5513 | 2 5510 | 5513 | 1 5500 | 5514 | 2 5510 | 5514 | 1 5500 | 5515 | 2

【讨论】：