sql - 获取叶级递归选择

Question

我有一个带有 id,parent_forum_post_id 的表 forumposts，对于给定的 id=1221，我发现它的孩子数。

with recursive all_posts (id, parentid, root_id) as (
    select  t1.id,
            t1.parent_forum_post_id as parentid,
            t1.id as root_id
    from    forumposts t1

    union all

    select  c1.id,
            c1.parent_forum_post_id as parentid,
            p.root_id
    from    forumposts c1
    join    all_posts p on p.id = c1.parent_forum_post_id
)
select      (count(*)-1) as child_count
from        all_posts
where       root_id=1221
group by    root_id;

我现在需要的是完全相反的：对于给定的 id，找出它的级别，这取决于他的父母数量（它是父母，它是父母的父母，直到它在它的 parent_forum_post_id 列中找到 null）。希望这是有道理的。

任何帮助表示赞赏。谢谢。

Answer 1

此查询可以在很大程度上简化为：

WITH RECURSIVE p AS (
    SELECT parent_forum_post_id AS p_id
    FROM   forumposts
    WHERE  id = 1221   

    UNION ALL
    SELECT f.parent_forum_post_id
    FROM   p
    JOIN   forumposts f ON f.id = p.p_id
    )
SELECT count(*) AS level
FROM   posts;

也应该快得多。

Answer 2

WITH recursive
  anticendent
AS
(
  SELECT
    id                       AS post_id,
    parent_forum_post_id     AS anticendent_post_id,
    1                        AS distance
  FROM
    forumposts

  UNION ALL

  SELECT
    anticendent.post_id,
    forumposts.parent_forum_post_id,
    distance + 1
  FROM
    anticendent
  INNER JOIN
    forumposts
      ON forumposts.id = anticendent.anticendent_post_id
)

SELECT
  post_id,
  MAX(distance)  AS level
FROM
  anticendent
GROUP BY
  post_id
WHERE
  post_id = 1221

或者...

SELECT
  *
FROM
  anticendent
WHERE
  post_id = 1221
  AND anticendent_post_id IS NULL

Answer 3

如果我理解正确，您希望特定节点的层次结构深度给定其 id（根为 1 级）。这适用于 postgresql：

with recursive all_posts (id, parentid, node_id) as (
    select  t1.id,
            t1.parent_forum_post_id as parentid,
            t1.id as node_id
    from    forumposts t1

    union all

    select  c1.id,
            c1.parent_forum_post_id as parentid,
            p.node_id
    from    forumposts c1
    join    all_posts p on p.parentid = c1.id
)
select      count(*) as level
from        all_posts
where       node_id=1221   
group by    node_id;