Question

我有一个我无法解决的树/祖先/查询问题：

我有一个包含菜单数据的表格和一个包含菜单所有祖先的表格：

table menu               table ancestors
+-----+------------+--------+     +---------+--------------+-------+
| id  |      title | active |     | menu_id |  ancestor_id | level |
+-----+------------+--------+     +---------+--------------+-------+
|   1 |       Home |      0 |     |       1 |            0 |     0 |
|   2 |       News |      0 |     |       2 |            1 |     1 |
|   3 |        Foo |      0 |     |       3 |            2 |     2 |
|   4 |        Bar |      1 |     |       3 |            1 |     1 |
|   5 |  Downloads |      1 |     |       4 |            3 |     3 |
+-----+------------+--------+     |       4 |            2 |     2 |
                                  |       4 |            1 |     1 |
                                  |       5 |            1 |     1 |
                                  +---------+--------------+-------+

我可以轻松地通过以下方式获取所有活动菜单条目及其祖先：

 SELECT menu.id, menu.title, GROUP_CONCAT(ancestors.ancestor_id) as ancestors
FROM menu, ancestors
WHERE menu.active = 1
GROUP BY (menu.id);

 +----+-----------+----------+
 | id |     title |ancestors |
 +----+-----------+----------+
 |  4 |       Bar | 3,2,1    | 
 |  5 | Downloads | 1        |
 +----+-----------+----------+

但是，我怎样才能获得树所需的所有祖先呢？在我的结果中，我需要输入Foo和News，以便我得到一个一致的树。它应该是这样的：

 +----+-----------+----------+
 | id |     title |ancestors |
 +----+-----------+----------+
 |  2 |      News | 1        | 
 |  3 |       Foo | 2,1      | 
 |  4 |       Bar | 3,2,1    | 
 |  5 | Downloads | 1        |
 +----+-----------+----------+

查询是如何形成的？

Answer 1

当我这样做时，我对ancestors表的结构略有不同。我存储level而不是pathlength。还为每个菜单项存储一行以指向自身，路径长度为0。

+---------+--------------+------------+
| menu_id |  ancestor_id | pathlength |
+---------+--------------+------------+
|       1 |            1 |          0 |
|       2 |            2 |          0 |
|       3 |            3 |          0 |
|       4 |            4 |          0 |
|       5 |            5 |          0 |
|       2 |            1 |          1 |
|       3 |            2 |          2 |
|       3 |            1 |          1 |
|       4 |            3 |          3 |
|       4 |            2 |          2 |
|       4 |            1 |          1 |
|       5 |            1 |          1 |
+---------+--------------+------------+

这些“反身”条目允许您将活动菜单项集合加入闭包表。将级别更改为pathlength允许您从祖先集中排除反身条目。

现在您可以查询作为“活动”菜单项的祖先的所有菜单项，包括活动菜单项本身：

SELECT a2.menu_id, m2.title, GROUP_CONCAT(a2.ancestor_id) AS ancestors
FROM menu m1
JOIN ancestors a1 ON (m1.id = a1.menu_id)
JOIN ancestors a2 ON (a1.ancestor_id = a2.menu_id AND a2.pathlength > 0)
JOIN menu m2 ON (a2.menu_id = m2.id)
WHERE m1.active = 1
GROUP BY a2.menu_id;

结果：

+---------+-----------+-----------+
| menu_id | title     | ancestors |
+---------+-----------+-----------+
|       2 | News      | 1         | 
|       3 | Foo       | 2,1       | 
|       4 | Bar       | 3,2,1     | 
|       5 | Downloads | 1         | 
+---------+-----------+-----------+

从MySQL中的菜单表中获取一致的菜单树数据

1 个答案: