SQL 樹查詢 - 使用遞歸和聯合。 如何按第二個值排序

Question

我有一個具有樹狀結構的類別表（見下圖）。 對於快速的單個查詢，我有以下 SQL 來獲取完整的樹。

WITH RECURSIVE search_tree(id, name, path, position) AS 
(
    SELECT id, name, ARRAY[id], position
    FROM table_name
    WHERE id = id

    UNION ALL

    SELECT table_name.id, table_name.name, path || table_name.id, 
           table_name.position
    FROM search_tree
    JOIN table_name ON table_name.parent_id = search_tree.id
    WHERE NOT table_name.id = ANY(path)
)
SELECT id, name, path, position
FROM search_tree 
ORDER BY path

此查詢結果如下表

id  | name         | path    | position
----+--------------+---------+-------------
  1 | Cat Pictures | {1}.    |.    0
  2 | Funny        | {1,2}.  |.    0
  3 | LOLCats      | {1,2,3} |.    1
  4 | Animated     | {1,2,4} |.    2
  5 | Classic      | {1,2,5} |.    0
  6 | Renaissance  | {1,6}   |.    1

所以根據路徑排序效果很好。 但我需要的是根據列 position 的順序，如果路徑級別是同一級別（如 id 2 和 4，以及 3、4、5）。

因此 ID 的順序是

ids: 1, 6, 2, 5, 3, 4

如何更改我的 SQL 聲明以反映該訂單？

Answer 1

可以這樣實現https://www.db-fiddle.com/f/rpFiPjKSwHW88C4cp6o9Rm/0

with recursive search_tree(id, parentPath, name) as (
    select id, cast(row_number() over(order by pos) as text), name
    from objects
    where parent_id = 0
union all
    select o.id, concat(search_tree.parentPath, ',', cast(row_number() over(order by o.pos) as text)), o.name
    from search_tree
    join objects as o on search_tree.id = o.parent_id
)
select *
from search_tree
order by parentPath;