0

parent_id = null指定された子 ID を持つ子から子へのブランチを取得したい

select *
from company s
start with s.parent_id is null and s.id = 56
connect by prior s.id = s.parent_id

結果は要素ID: 56のフル ブランチです。

選択した子を持つブランチを選択する方法

例：

結果は次のとおりです。

id   parent_id
56     null
57      56
58      57
59      57
60      56
61      60

私が達成したいのは、指定された子ID: 57でそのブランチのみを選択することです

id   parent_id
56     null
57      56
58      57
59      57

without not given child id:
60      56
61      60
4

2 に答える 2

2

これは、ツリーのあらゆる深さで機能し、正規表現や文字列連結を使用していません。理解しやすいはずです：

    with
       all_children_of_57 as 
       (     
          -- all nodes that can be reached by starting a recursive descent from node 57 
          select id
          from company s
          start with s.id = 57
          connect by prior s.id = s.parent_id 
       ),
       all_ancestors_of_57 as
       (
          -- all ANCESTORS that can be reached by walking up the tree (FROM CHILD TO PARENT), 
          -- starting from node 57 (this is a linear recursion)
          select id
          from company s
          start with s.id = 57  
          connect by  s.id = prior s.parent_id  
       ),
       nodes_in_branch as
       (  
          -- we are interested only in nodes extracted from the above two queries
          select * 
          from company
          where id in (select id from all_children_of_57)
             or id in (select id from all_ancestors_of_57)
       )

    -- we do the recursion on the result of nodes_in_branch  
    select *
    from nodes_in_branch s
    start with s.parent_id is null 
    connect by prior s.id = s.parent_id
于 2015-09-03T11:57:01.637 に答える
0

要件を満たす必要があるいくつかの簡単なクエリ:

編集: クエリに追加されたコメント

EDIT2: SYS_CONNECT_BY_PATH の使用を最小限に抑えたため、サイズ制限に達するのが難しくなりました。

    WITH param AS -- this is just our parameter to be used in further subqueries (ID and regex pattern for regexp_like expression)
(
  SELECT
    57 AS id
    ,'(\s?)('||57||')(\s|\W)' AS regex
  FROM
    dual
)
, parent AS -- here we get the inverted hierarchy - searching for the root node of the branch where ID given as parameter exists
(
SELECT
  s.*
FROM
  company s
  ,param
CONNECT BY PRIOR s.parent_id = s.id
START WITH s.id = param.id
)
, children AS -- here we get all the children of the root node we found in parent subquery
(
SELECT
  s.*
  ,CASE
    WHEN LEVEL >= MAX(CASE WHEN s.id = param.id THEN LEVEL ELSE NULL END) OVER (PARTITION BY NULL) THEN 
      SYS_CONNECT_BY_PATH(s.id,' ') 
    ELSE NULL
  END AS path -- this gives us the hierarchical path
  ,LEVEL AS lvl -- this gives us children's levels
  ,MAX(CASE WHEN s.id = param.id THEN LEVEL ELSE NULL END) OVER (PARTITION BY NULL) AS id_level -- this gives the level of the children given as a parameter
FROM
  company s
  ,param
CONNECT BY PRIOR s.id = s.parent_id
START WITH s.id = (SELECT parent.id FROM parent WHERE parent.parent_id IS NULL)
)
-- now we select from the all children hierarchy our desired branch
SELECT
  c.*
FROM
  children c
  ,param
WHERE
  1 = CASE WHEN lvl > c.id_level AND REGEXP_LIKE(c.path, param.regex) THEN 1 -- if current id is of higher level (is a child of our paremeter ID), it must have our parameter ID in it's path
           WHEN lvl = c.id_level AND c.id = param.id THEN 1 -- if the level is equal to our parameter's ID's leve, it has to be our paremeter
           WHEN lvl < c.id_level AND EXISTS(SELECT 1 FROM parent WHERE parent.id = c.id) THEN 1 -- if current level is lower (may be a parent of our parmeter ID) it has to exists in the reverse hierarchy (thus, must be a parent or grandparent etc. of our paremter node)
           ELSE 0
      END

ところで。カルロ・シルナの解決策を読んだ後、彼の方が優れていることがわかりました:)

于 2015-09-03T11:10:06.623 に答える