我现在有一个问题。我有一个名为这个结构的地方的表:
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。
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;
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_id
为NULL
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。
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