PostgreSQL - 选择具有级别的表的所有层次结构

问题描述 投票:1回答:2

我现在有一个问题。我有一个名为这个结构的地方的表:

  • ID
  • PARENT_ID
  • 那么

我想做一个选择,以拥有此表的所有层次结构。有一个小数据的例子:

(1, null, '123 Barclay St')
(2, 1, 'Floor 1')
(3, 1, 'Floor 2')
(4, 1, 'Floor 3')
(5, 2, 'Hall 1')
(6, 2, 'Room 1')
(7, 2, 'Room 2')
(8, 3, 'Room 3')
(9, null, '10 Thames St')

显然,表中的顺序不是这个。

所以我想用我的SELECT(有9行)得到这个结果:

123 Barclay St
   Floor 1
      Hall 1
      Room 1
      Room 2
   Floor 2
      Room 3
   Floor 3
10 Thames St

而不是这个结果(我已经知道如何获得):

10 Thames St
123 Barclay St
   Floor 1
   Floor 2
   Floor 3
      Hall 1
      Room 1
      Room 2
      Room 3

如果你能帮助我,我会提前感谢你。

sql postgresql hierarchical-data recursive-query
2个回答
1
投票

这是一个使用递归CTE的解决方案:

WITH RECURSIVE cte AS (
    SELECT LPAD(id::text, 3, '0') AS marker, '   ' AS buffer,
        id, parent_id, name::text
    FROM yourTable t WHERE parent_id IS NULL
    FROM yourTable t WHERE parent_id IS NULL
    UNION ALL
        SELECT t2.marker || ':' || LPAD(t1.parent_id::text, 3, '0') || ':' ||
            LPAD(t1.id::text, 3, '0') AS marker,
            t2.buffer || '   ', t1.id, t1.parent_id, t2.buffer || t1.name
    FROM yourTable t1
    INNER JOIN cte t2
        ON t1.parent_id = t2.id
)

SELECT name FROM cte ORDER BY marker;

enter image description here

Demo

这里的基本思想是构建路径字符串,该路径字符串跟踪每个节点返回其根节点的完整路径(根节点由parent_idNULL的节点给出)。然后,我们只需在此路径上执行一个ORDER BY即可生成所需的顺序。


1
投票

您尚未提供已经提出的查询。但是 - 据我所知,你想要一个递归树结构。

https://www.db-fiddle.com/f/og5HZDHBhBRmP1cDnqgCBB/1

CREATE TABLE rooms (
  id INTEGER, parent_id INTEGER, name TEXT
);

INSERT INTO rooms VALUES
(1, null, '123 Barclay St'),
(2, 1, 'Floor 1'),
(3, 1, 'Floor 2'),
(4, 1, 'Floor 3'),
(5, 2, 'Hall 1'),
(6, 2, 'Room 1'),
(7, 2, 'Room 2'),
(8, 3, 'Room 3'),
(9, null, '10 Thames St');

和查询:

WITH RECURSIVE tree AS (
    SELECT
        rooms.id,
        rooms.parent_id,
        rooms.name
    FROM
        rooms
    WHERE
        parent_id IS NULL
    UNION ALL
    SELECT
        rooms.id,
        rooms.parent_id,
        rooms.name
    FROM
        tree
        JOIN rooms ON rooms.parent_id = tree.id
)
SELECT
    *
FROM
    tree;

https://www.postgresql.org/docs/current/static/queries-with.html

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