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php - MySQL - 如何访问表中级别 2 中的所有节点

转载 作者:搜寻专家 更新时间:2023-10-30 22:11:12 26 4
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我正在为具有嵌套类别的电子商务网站创建数据库,并且我正在使用修改后的预购遍历算法。我的问题是如何访问第 2 级的所有节点,即 ArticlesPortfolioContact

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最佳答案

The article没有明确告诉您如何从一个级别获取所有节点。但如果你仔细阅读它,它会告诉你如何做更多 -> 获取每个类别的深度计数。然后您所要做的就是按该深度进行过滤。

SELECT node.name, (COUNT(parent.name) - 1) AS depth
FROM nested_category AS node, nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
HAVING depth = 1
ORDER BY node.lft;

编辑(发生了什么):

为了使用 nested_category 表的 lftrgt 列,我们应该选择该表两次。

SELECT *
FROM nested_category AS node, nested_category AS parent

如果检查此查询,您会发现对于 nested_category 中的每一行,我们都会再次选择所有行。所以我们现在想要的是从第一个表(我们称为 AS 节点)中删除所有行,这些行不是它们的 parent 的子行。这就是我们使用 WHERE node.lft BETWEEN parent.lft AND parent.rgt

进行过滤的原因

我想提一下这个查询:

SELECT *
FROM nested_category AS node, nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
ORDER BY node.lft;

等于

SELECT *
FROM nested_category AS node
LEFT JOIN nested_category AS parent ON (node.lft BETWEEN parent.lft AND parent.rgt)
ORDER BY node.lft;

所以现在我们有所有 child 和他们的 parent + 1(由于我们的过滤方式,每个 child 都属于自己)

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

下一步 - 获取深度计数。为了做到这一点,我们必须按每个 child 分组(示例使用 GROUP BY node.name 但它也可以在 node.category_id 上完成并计算数量parents - 每组 1 个 (COUNT(parent.name) - 1) AS depth(也可以使用 parent.category_id )

这样做

SELECT node.*, (COUNT(parent.category_id) - 1) AS depth
FROM nested_category AS node
LEFT JOIN nested_category AS parent ON (node.lft BETWEEN parent.lft AND parent.rgt)
GROUP BY node.category_id
ORDER BY node.lft;

我们明白了

+-------------+----------------------+-----+-----+-------+
| category_id | name | lft | rgt | depth |
+-------------+----------------------+-----+-----+-------+
| 1 | ELECTRONICS | 1 | 20 | 0 |
| 2 | TELEVISIONS | 2 | 9 | 1 |
| 3 | TUBE | 3 | 4 | 2 |
| 4 | LCD | 5 | 6 | 2 |
| 5 | PLASMA | 7 | 8 | 2 |
| 6 | PORTABLE ELECTRONICS | 10 | 19 | 1 |
| 7 | MP3 PLAYERS | 11 | 14 | 2 |
| 8 | FLASH | 12 | 13 | 3 |
| 9 | CD PLAYERS | 15 | 16 | 2 |
| 10 | 2 WAY RADIOS | 17 | 18 | 2 |
+-------------+----------------------+-----+-----+-------+

现在是最后一步,我们只需要这些记录,深度 = 1(HAVING depth = 1HAVING 在这里使用是因为它在聚合之后应用(因此它可以过滤聚合))

SELECT node.*, (COUNT(parent.category_id) - 1) AS depth
FROM nested_category AS node
LEFT JOIN nested_category AS parent ON (node.lft BETWEEN parent.lft AND parent.rgt)
GROUP BY node.category_id
HAVING depth = 1
ORDER BY node.lft;

+-------------+----------------------+-----+-----+-------+
| category_id | name | lft | rgt | depth |
+-------------+----------------------+-----+-----+-------+
| 2 | TELEVISIONS | 2 | 9 | 1 |
| 6 | PORTABLE ELECTRONICS | 10 | 19 | 1 |
+-------------+----------------------+-----+-----+-------+

我希望它现在更清楚了。如果我犯了一些错误,再次为我糟糕的英语感到抱歉。

关于php - MySQL - 如何访问表中级别 2 中的所有节点,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28020029/

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