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获取条件的嵌套集合祖先

[英]Get nested set ancestor where condition

Given a nested set in mysql like this 在mysql中给出这样的嵌套集

+-------------+----------------------+-----+-----+------+
| 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

How can I get an ancestor for eg: "Flash" where condition rank is >=50 AND <= 99 LIMIT 1 sorted by lft DESC. 我如何获得祖先,例如:“ Flash”,其中条件等级> = 50 AND <= 99 LIMIT 1(按lft DESC排序)。 so the result will be 所以结果将是

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

for visualization refer to this link http://mikehillyer.com/media//categories.png 要进行可视化,请参阅此链接http://mikehillyer.com/media//categories.png

sql fiddle is not working for me right now, but you should be able to do something like this. sql小提琴现在对我不起作用,但是您应该可以执行以下操作。

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 SQL小提琴

You dont have enough identifiers to differentiate between ancestors, having just "50" for a ranking doesnt differentiate the records. 您没有足够的标识符来区分祖先,排名只有“ 50”并不能区分记录。

If you did a 如果你做了

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

Would give you "Portable Electronics", "Televisions", and "Electronics", which are all technically ancestors. 会给您“便携式电子产品”,“电视”和“电子产品”,它们都是技术上的祖先。

If you added another column field, ie parent_id then it would make it much easier to build structure from the data. 如果添加了另一个列字段,即parent_id,那么将更容易从数据构建结构。

+-------------+----------------------+-----+-----+------+---------+
| 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

Or implementing a relationship table to hold those data sets 或实现关系表来保存这些数据集

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

I would have added as a comment, but i don't have enough reputation. 我会添加为评论,但是我没有足够的声誉。 hope this answers your question somewhat. 希望这能回答您的问题。

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