2

我有一棵使用物化路径的树。

该表类似于：

+-------------+--------+----------+-------+
| path        | depth  | children | score |
+-------------+--------+----------+-------+
| 0001        | 1      | 3        | 5     |
| 00010001    | 2      | 0        | -3    |
| 00010002    | 2      | 0        | 27    |
| 00010003    | 2      | 0        | 10    |
| 0002        | 1      | 2        | 12    |
| 00020001    | 2      | 0        | 0     |
| 00020002    | 2      | 1        | 3     |
| 00020002001 | 3      | 0        | 1     |
+-------------+--------+----------+-------+

我想score在保持树结构的同时按列排序。
重要的是孩子在父母之下。

+-------------+--------+----------+-------+
| path        | depth  | children | score |
+-------------+--------+----------+-------+
| 0002        | 1      | 2        | 12    |
| 00020002    | 2      | 1        | 3     |
| 00020002001 | 3      | 0        | 1     |
| 00020001    | 2      | 0        | 0     |
| 0001        | 1      | 3        | 5     |
| 00010002    | 2      | 0        | 27    |
| 00010003    | 2      | 0        | 10    |
| 00010001    | 2      | 0        | -3    |
+-------------+--------+----------+-------+

该path列仅在数据库中使用，因此不必是顺序的。

我目前用来对树进行排序以便构建它的 SQL：

SELECT path, depth, children, score FROM mytable ORDER BY path ASC
4

1 回答 1

1

您将需要一个递归查询和一个窗口函数。它看起来类似于：

with recursive
ordered_tree as (
select tree.*,
       array[row_number() over w] as score_path
from   tree
where  depth = 1
window w as (order by tree.score desc)
union all
select tree.*,
       parent.score_path || array[row_number() over w] as score_path
from   tree
join   ordered_tree as parent on parent.id = tree.parent_id
window w as (partition by tree.parent_id order by tree.score desc)
)
select *
from   ordered_tree
order by score_path

注意：如果你的集合很大，上面会很慢......

于 2011-06-25T10:36:56.280 回答