获取嵌套集合祖先 where 条件

Question

给定 mysql 中的嵌套集，如下所示

+-------------+----------------------+-----+-----+------+
| category_id | name                 | lft | rgt | rank |
+-------------+----------------------+-----+-----+------+
|           1 | ELECTRONICS          |   1 |  20 |  99
|           2 | TELEVISIONS          |   2 |   9 |  50
|           3 | TUBE                 |   3 |   4 |  25
|           4 | LCD                  |   5 |   6 |  25
|           5 | PLASMA               |   7 |   8 |  25
|           6 | PORTABLE ELECTRONICS |  10 |  19 |  50
|           7 | MP3 PLAYERS          |  11 |  14 |  25
|           8 | FLASH                |  12 |  13 |  10
|           9 | CD PLAYERS           |  15 |  16 |  25
|          10 | 2 WAY RADIOS         |  17 |  18 |  20

我怎样才能获得祖先，例如："Flash"，其中条件排名 >=50 AND <= 99 LIMIT 1 按 lft DESC 排序。所以结果将是

+-------------+----------------------+-----+-----+------+
| category_id | name                 | lft | rgt | rank |
+-------------+----------------------+-----+-----+------+
|           6 | PORTABLE ELECTRONICS |  10 |  19 |  50

可视化参考这个link http://mikehillyer.com/media//categories.png

Answer 1

您没有足够的标识符来区分祖先，只有“50”的排名并不能区分记录。

如果你做了

SELECT * FROM [SET] WHERE RANK >= 50 AND RANK <= 99

会给你 "Portable Electronics"、"Televisions" 和 "Electronics"，它们在技术上都是祖先。

如果您添加了另一个列字段，即 parent_id，则可以更轻松地从数据构建结构。

+-------------+----------------------+-----+-----+------+---------+
| category_id | name                 | lft | rgt | rank |Parent_id|
+-------------+----------------------+-----+-----+------+---------+
|           1 | ELECTRONICS          |   1 |  20 |  99  | -
|           2 | TELEVISIONS          |   2 |   9 |  50  | 1
|           3 | TUBE                 |   3 |   4 |  25  | 2
|           4 | LCD                  |   5 |   6 |  25  | 2
|           5 | PLASMA               |   7 |   8 |  25  | 2
|           6 | PORTABLE ELECTRONICS |  10 |  19 |  50  | 1
|           7 | MP3 PLAYERS          |  11 |  14 |  25  | 6
|           8 | FLASH                |  12 |  13 |  10  | 6
|           9 | CD PLAYERS           |  15 |  16 |  25  | 6
|          10 | 2 WAY RADIOS         |  17 |  18 |  20  | 6

或者实现一个关系table来保存那些数据集

+------+--------+
| Item | Parent |
+------+--------+
|  2   |  1     |
|  3   |  2     |

我会添加为评论，但我没有足够的声誉。希望这能在一定程度上回答您的问题。

Answer 2

sql fiddle 现在不适合我，但你应该可以做这样的事情。

SELECT  
    c2.* 
FROM
    Categories as c1
    INNER JOIN Categories as c2 ON 
            c1.`lft` >= c2.lft AND c1.`lft` <= c2.`rgt` 
            AND c2.`RANK` >=50 AND c2.`RANK` <= 99 
WHERE 
    c1.`category_id` = 8
ORDER BY 
    c2.`lft` DESC
LIMIT 1;

SQL Fiddle

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