Oracle SQL：如何获取树中根节点和任何节点之间的第 n 个节点优化中

如何解决Oracle SQL：如何获取树中根节点和任何节点之间的第 n 个节点优化中

我试图从给定节点开始从根中找到第 N 个节点。如果给定节点小于或等于从根开始的第 N 个节点，则返回给定节点。我的查询有效，但速度很慢，我对此并不满意。关于如何重构它以提高性能的任何想法？

test_data
---------
unique_id | lookup_acct_code 
'1'         'leaf-1'
'2'         'stem-2'
'3'         'branch-1'
'4'         'trunk-2'
'5'         'root-1'

linked_accounts
---------------
acct_code | parent_code
'leaf-1'    'stem-1'
'stem-1'    'twig-1'
'twig-1'    'stick-1'
'stick-1'   'branch-1'
'branch-1'  'trunk-1'
'trunk-1'   'root-1'
'root-1'    NULL
'leaf-2'    'stem-2'
'stem-2'    'twig-2'
'twig-2'    'stick-2'
'stick-2'   'branch-2'
'branch-2'  'trunk-2'
'trunk-2'   'root-2'
'root-2'    NULL

SELECT unique_id,(SELECT acct_code
  FROM (SELECT acct_code,ROW_NUMBER() OVER (ORDER BY level desc) rn
        FROM linked_accounts
        CONNECT BY acct_code = PRIOR parent_code
        START WITH acct_code = lookup_acct_code)
  WHERE rn <= 3
  ORDER BY rn DESC
  FETCH FIRST 1 ROWS ONLY)    
FROM test_data;

/* Correct output
1   branch-1
2   branch-2
3   branch-1
4   trunk-2
5   root-1
*/

解决方法

您可以首先从根生成值直到第 n^th 个节点（使用递归子查询分解子句而不是分层查询），然后加入而不是使用相关子查询:

WITH from_roots ( acct_code,depth,nth_from_root ) AS (
  SELECT acct_code,1,acct_code
  FROM   linked_accounts
  WHERE  parent_code IS NULL
UNION ALL
  SELECT l.acct_code,depth + 1,CASE
         WHEN depth >= 3
         THEN nth_from_root
         ELSE l.acct_code
         END
  FROM   linked_accounts l
         INNER JOIN from_roots f
         ON ( f.acct_code = l.parent_code )
)
SELECT unique_id,nth_from_root
FROM   from_roots f
       INNER JOIN test_data d
       ON ( d.lookup_acct_code = f.acct_code )

或者，您可以调整查询以使用 MAX(LEVEL) OVER () - LEVEL + 1 直接计算深度：

SELECT unique_id,( SELECT acct_code
         FROM   (
           SELECT acct_code,MAX(LEVEL) OVER () - LEVEL + 1 AS depth
           FROM   linked_accounts
           CONNECT BY acct_code = PRIOR parent_code
           START WITH acct_code = lookup_acct_code
         )
         WHERE depth <= 3
         ORDER BY depth DESC
         FETCH FIRST 1 ROW ONLY
       ) AS nth_node
FROM   test_data;

对于您的示例数据：

CREATE TABLE test_data ( unique_id,lookup_acct_code ) AS
SELECT '1','leaf-1'   FROM DUAL UNION ALL
SELECT '2','stem-2'   FROM DUAL UNION ALL
SELECT '3','branch-1' FROM DUAL UNION ALL
SELECT '4','trunk-2'  FROM DUAL UNION ALL
SELECT '5','root-1'   FROM DUAL;

CREATE TABLE linked_accounts ( acct_code,parent_code ) AS
SELECT 'leaf-1','stem-1'   FROM DUAL UNION ALL
SELECT 'stem-1','twig-1'   FROM DUAL UNION ALL
SELECT 'twig-1','stick-1'  FROM DUAL UNION ALL
SELECT 'stick-1','branch-1' FROM DUAL UNION ALL
SELECT 'branch-1','trunk-1'  FROM DUAL UNION ALL
SELECT 'trunk-1','root-1'   FROM DUAL UNION ALL
SELECT 'root-1',NULL       FROM DUAL UNION ALL
SELECT 'leaf-2','stem-2'   FROM DUAL UNION ALL
SELECT 'stem-2','twig-2'   FROM DUAL UNION ALL
SELECT 'twig-2','stick-2'  FROM DUAL UNION ALL
SELECT 'stick-2','branch-2' FROM DUAL UNION ALL
SELECT 'branch-2','trunk-2'  FROM DUAL UNION ALL
SELECT 'trunk-2','root-2'   FROM DUAL UNION ALL
SELECT 'root-2',NULL       FROM DUAL;

两个输出：

UNIQUE_ID | NTH_FROM_ROOT
:-------- | :------------
1         | branch-1     
2         | branch-2     
3         | branch-1     
4         | trunk-2      
5         | root-1       

您需要在更大的数据集上分析所有解决方案，以查看哪个解决方案的性能更高。

dbfiddle here

,

您可以尝试通过使用 CONNECT_BY_ROOT 函数来获取您已开始的 unique_id 来避免标量子查询。并根据构建树后找到的最大级别进行相同的过滤。试试这个：

with hier as (
  select
    /*Find unique ID of subtree*/
    connect_by_root(f.unique_id) as unique_id,connect_by_root(la.acct_code) as start_node,la.acct_code,/*Find max depth of subtree*/
    max(level) over(partition by connect_by_root(f.unique_id)) as max_lvl,/*Find current reversed depth of node*/
    level as lvl
  from linked_accounts la
      left join test_data f
        on la.acct_code = f.lookup_acct_code
  start with f.unique_id is not null
  connect by prior la.parent_code = la.acct_code
)
select
  unique_id,case max_lvl
    when lvl
    /*Select our start node for short subtree*/
    then start_node
    /*And node al level 3 otherwise*/
    else acct_code
  end as node_name
from hier
/*Select the node on level 3 from the root*/
where max_lvl - lvl = 2
  /*Or the last node in subtree for the depth < 3*/
  or (max_lvl <= 2 and lvl = max_lvl)
order by 1 asc

UNIQUE_ID | NODE_NAME
:-------- | :--------
1         | branch-1 
2         | branch-2 
3         | branch-1 
4         | trunk-2  
5         | root-1   

dbfiddle here