MySQL实现嵌套集合模型

时间  2019-11-06
标签 mysql 实现 嵌套 集合 模型 栏目 MySQL 繁體版
原文   原文链接

MySQL实现嵌套集合模型

译文主要是介绍如何用MySQL来存储嵌套集合数据。在其中会增长一些本身的理解，也会删除掉一些自认为无用的废话。
这篇文章主要讲的是嵌套集合模型，因此邻接表不是本文的重点，简单略过就好。node

也许这是原文地址，由于我也不知道这是否是原文。mysql

介绍

什么是分层数据？
算法

相似于树形结构，除了根节点和叶子节点外，全部节点都有用一个父节点和多个子节点。sql

那么，在MySQL中如何处理分层数据呢？函数

原文中介绍了两种分层结构模型：邻接表模型和嵌套集合模型。测试

邻接表模型(The Adjacency List Model)

首先，创建测试表，导入测试数据，网站

CREATE TABLE category(
        category_id INT AUTO_INCREMENT PRIMARY KEY,
        name VARCHAR(20) NOT NULL,
        parent INT DEFAULT NULL
);

INSERT INTO category VALUES
        (1,'ELECTRONICS',NULL),
        (2,'TELEVISIONS',1),
        (3,'TUBE',2),
        (4,'LCD',2),
        (5,'PLASMA',2),
        (6,'PORTABLE ELECTRONICS',1),
        (7,'MP3 PLAYERS',6),
        (8,'FLASH',7),
        (9,'CD PLAYERS',6),
        (10,'2 WAY RADIOS',6);

SELECT * FROM category ORDER BY category_id;
+-------------+----------------------+--------+
| category_id | name                 | parent |
+-------------+----------------------+--------+
|           1 | ELECTRONICS          |   NULL |
|           2 | TELEVISIONS          |      1 |
|           3 | TUBE                 |      2 |
|           4 | LCD                  |      2 |
|           5 | PLASMA               |      2 |
|           6 | PORTABLE ELECTRONICS |      1 |
|           7 | MP3 PLAYERS          |      6 |
|           8 | FLASH                |      7 |
|           9 | CD PLAYERS           |      6 |
|          10 | 2 WAY RADIOS         |      6 |
+-------------+----------------------+--------+
10 rows in set (0.00 sec)

在邻接表中，全部的数据均拥有一个Parent字段，用来存储它的父节点。当前节点为根节点的话，它的父节点则为NULL。
那么在遍历的时候，可使用递归来实现查询整棵树，从根节点开始，不断寻找子节点（父节点->子节点->父节点->子节点）。3d

检索分层路径

通常须要获取一个分层结构的路径问题，那么code

SELECT t1.name AS lev1, t2.name as lev2, t3.name as lev3, t4.name as lev4
FROM category AS t1
LEFT JOIN category AS t2 ON t2.parent = t1.category_id
LEFT JOIN category AS t3 ON t3.parent = t2.category_id
LEFT JOIN category AS t4 ON t4.parent = t3.category_id
WHERE t1.name = 'ELECTRONICS';

+-------------+----------------------+--------------+-------+
| lev1        | lev2                 | lev3         | lev4  |
+-------------+----------------------+--------------+-------+
| ELECTRONICS | TELEVISIONS          | TUBE         | NULL  |
| ELECTRONICS | TELEVISIONS          | LCD          | NULL  |
| ELECTRONICS | TELEVISIONS          | PLASMA       | NULL  |
| ELECTRONICS | PORTABLE ELECTRONICS | MP3 PLAYERS  | FLASH |
| ELECTRONICS | PORTABLE ELECTRONICS | CD PLAYERS   | NULL  |
| ELECTRONICS | PORTABLE ELECTRONICS | 2 WAY RADIOS | NULL  |
+-------------+----------------------+--------------+-------+
6 rows in set (0.00 sec)

检索叶子节点

SELECT t1.name FROM
category AS t1 LEFT JOIN category as t2
ON t1.category_id = t2.parent
WHERE t2.category_id IS NULL;

+--------------+
| name         |
+--------------+
| TUBE         |
| LCD          |
| PLASMA       |
| FLASH        |
| CD PLAYERS   |
| 2 WAY RADIOS |
+--------------+

检索指定路径

SELECT t1.name AS lev1, t2.name as lev2, t3.name as lev3, t4.name as lev4
FROM category AS t1
LEFT JOIN category AS t2 ON t2.parent = t1.category_id
LEFT JOIN category AS t3 ON t3.parent = t2.category_id
LEFT JOIN category AS t4 ON t4.parent = t3.category_id
WHERE t1.name = 'ELECTRONICS' AND t4.name = 'FLASH';

+-------------+----------------------+-------------+-------+
| lev1        | lev2                 | lev3        | lev4  |
+-------------+----------------------+-------------+-------+
| ELECTRONICS | PORTABLE ELECTRONICS | MP3 PLAYERS | FLASH |
+-------------+----------------------+-------------+-------+
1 row in set (0.01 sec)

邻接表的缺点

在检索路径的过程当中，除了本层外，每一层都会对应一个LEFT JOIN，那么若是层数不定怎么办？或者层数过多？
在删除中间层的节点时，须要同时删除该节点下的全部节点，不然会出现孤立节点。blog

嵌套集合模型Nested Set Model

原文中主要的目的是介绍嵌套集合模型，以下

经过集合的包含关系，嵌套结合模型能够表示分层结构，每个分层能够用一个Set来表示（一个圈），父节点所在的圈包含全部子节点所在的圈。

为了用MySQL来表示集合关系，须要定义连个字段left和right（表示一个集合的范围）。

CREATE TABLE nested_category (
        category_id INT AUTO_INCREMENT PRIMARY KEY,
        name VARCHAR(20) NOT NULL,
        lft INT NOT NULL,
        rgt INT NOT NULL
);

INSERT INTO nested_category VALUES
  (1,'ELECTRONICS',1,20),
  (2,'TELEVISIONS',2,9),
  (3,'TUBE',3,4),
  (4,'LCD',5,6),
  (5,'PLASMA',7,8),
  (6,'PORTABLE ELECTRONICS',10,19),
  (7,'MP3 PLAYERS',11,14),
  (8,'FLASH',12,13),
  (9,'CD PLAYERS',15,16),
  (10,'2 WAY RADIOS',17,18);

SELECT * FROM nested_category ORDER BY category_id;

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

因为left和right是MySQL的保留字，所以，字段名称用lft和rgt代替。每个集合都是从lft开始到rgt结束，也就是集合的两个边界。

在树中也一样适用，

当为树状结构编号时，咱们从左到右，一次一层，赋值按照从左到右的顺序遍历其子节点，这种方法称为先序遍历算法。

检索分层路径

因为子节点的lft值总在父节点的lft和rgt值之间，因此能够经过父节点链接到子节点上来检索整棵树。

SELECT node.name
FROM nested_category AS node,
        nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND parent.name = 'ELECTRONICS'
ORDER BY node.lft;

+----------------------+
| name                 |
+----------------------+
| ELECTRONICS          |
| TELEVISIONS          |
| TUBE                 |
| LCD                  |
| PLASMA               |
| PORTABLE ELECTRONICS |
| MP3 PLAYERS          |
| FLASH                |
| CD PLAYERS           |
| 2 WAY RADIOS         |
+----------------------+</pre>

这个方法并不须要考虑层数，并且不须要考虑节点的rgt。

检索全部叶子节点

因为每个叶子节点的rgt=lft+1，那么只须要这一个条件便可。

SELECT name
FROM nested_category
WHERE rgt = lft + 1;

+--------------+
| name         |
+--------------+
| TUBE         |
| LCD          |
| PLASMA       |
| FLASH        |
| CD PLAYERS   |
| 2 WAY RADIOS |
+--------------+

检索节点路径

再也不须要多个join链接操做。

SELECT parent.name
FROM nested_category AS node,
        nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.name = 'FLASH'
ORDER BY node.lft;

+----------------------+
| name                 |
+----------------------+
| ELECTRONICS          |
| PORTABLE ELECTRONICS |
| MP3 PLAYERS          |
| FLASH                |
+----------------------+

检索节点深度

经过COUNT和GROUP BY函数来获取父节点的个数。

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
ORDER BY node.lft;

+----------------------+-------+
| name                 | depth |
+----------------------+-------+
| ELECTRONICS          |     0 |
| TELEVISIONS          |     1 |
| TUBE                 |     2 |
| LCD                  |     2 |
| PLASMA               |     2 |
| PORTABLE ELECTRONICS |     1 |
| MP3 PLAYERS          |     2 |
| FLASH                |     3 |
| CD PLAYERS           |     2 |
| 2 WAY RADIOS         |     2 |
+----------------------+-------+

甚至能够获得分层的缩进结果，

SELECT CONCAT( REPEAT(' ', COUNT(parent.name) - 1), node.name) AS name
FROM nested_category AS node,
        nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;

+-----------------------+
| name                  |
+-----------------------+
| ELECTRONICS           |
|  TELEVISIONS          |
|   TUBE                |
|   LCD                 |
|   PLASMA              |
|  PORTABLE ELECTRONICS |
|   MP3 PLAYERS         |
|    FLASH              |
|   CD PLAYERS          |
|   2 WAY RADIOS        |
+-----------------------+

检索子树的深度

考虑到检索中须要自链接的node或parent，所以须要增长一个额外的链接来做为子查询来限制子树。

SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
FROM nested_category AS node,
        nested_category AS parent,
        nested_category AS sub_parent,
        (
                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
                AND node.name = 'PORTABLE ELECTRONICS'
                GROUP BY node.name
                ORDER BY node.lft
        )AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
        AND sub_parent.name = sub_tree.name
GROUP BY node.name
ORDER BY node.lft;

+----------------------+-------+
| name                 | depth |
+----------------------+-------+
| PORTABLE ELECTRONICS |     0 |
| MP3 PLAYERS          |     1 |
| FLASH                |     2 |
| CD PLAYERS           |     1 |
| 2 WAY RADIOS         |     1 |
+----------------------+-------+

检索节点的直接子节点

假设一个场景，当用户点击网站上电子产品的一个分类时，将呈现该分类下的产品，同时须要列出全部子分类，并非所有分类。
为了限制显示分类的层数，须要使用HAVING字句，

SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
FROM nested_category AS node,
        nested_category AS parent,
        nested_category AS sub_parent,
        (
                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
                        AND node.name = 'PORTABLE ELECTRONICS'
                GROUP BY node.name
                ORDER BY node.lft
        )AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
        AND sub_parent.name = sub_tree.name
GROUP BY node.name
HAVING depth &lt;= 1
ORDER BY node.lft;

+----------------------+-------+
| name                 | depth |
+----------------------+-------+
| PORTABLE ELECTRONICS |     0 |
| MP3 PLAYERS          |     1 |
| CD PLAYERS           |     1 |
| 2 WAY RADIOS         |     1 |
+----------------------+-------+

增长新节点

上面已经介绍了如何检索结果，那么如何才能增长新的节点呢？

若是但愿在TELEVISIONS和PROTABLE ELECTRONICS节点之间增长一个新的节点，那么新节点的lft和rgt的值应该是10和11，那么全部大于10的节点（新节点右侧的节点）的lft和rgt都应该加2，如上图所示。

LOCK TABLE nested_category WRITE;

SELECT @myRight := rgt FROM nested_category
WHERE name = 'TELEVISIONS';

UPDATE nested_category SET rgt = rgt + 2 WHERE rgt &gt; @myRight;
UPDATE nested_category SET lft = lft + 2 WHERE lft &gt; @myRight;

INSERT INTO nested_category(name, lft, rgt) VALUES('GAME CONSOLES', @myRight + 1, @myRight + 2);

UNLOCK TABLES

若是但愿在叶子节点下增长节点，须要修改下查询语句，

LOCK TABLE nested_category WRITE;

SELECT @myLeft := lft FROM nested_category

WHERE name = '2 WAY RADIOS';

UPDATE nested_category SET rgt = rgt + 2 WHERE rgt &gt; @myLeft;
UPDATE nested_category SET lft = lft + 2 WHERE lft &gt; @myLeft;

INSERT INTO nested_category(name, lft, rgt) VALUES('FRS', @myLeft + 1, @myLeft + 2);

UNLOCK TABLES;```



###删除节点

删除叶子节点比较容易，只须要删除本身，而删除一个中间层节点就须要删除其全部子节点。在这个模型中，全部子节点的节点正好在lft和rgt之间。

LOCK TABLE nested_category WRITE;

SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1
FROM nested_category
WHERE name = 'GAME CONSOLES';

DELETE FROM nested_category WHERE lft BETWEEN @myLeft AND @myRight;

UPDATE nested_category SET rgt = rgt - @myWidth WHERE rgt > @myRight;
UPDATE nested_category SET lft = lft - @myWidth WHERE lft > @myRight;

UNLOCK TABLES;

在某些状况下，只须要删除某个节点，可是并不但愿删除该节点下的子节点数据。
经过把右侧全部节点的左右值-2，当前节点的子节点左右值-1

LOCK TABLE nested_category WRITE;

SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1
FROM nested_category
WHERE name = 'PORTABLE ELECTRONICS';

DELETE FROM nested_category WHERE lft = @myLeft;

UPDATE nested_category SET rgt = rgt - 1, lft = lft - 1 WHERE lft BETWEEN @myLeft AND @myRight;
UPDATE nested_category SET rgt = rgt - 2 WHERE rgt > @myRight;
UPDATE nested_category SET lft = lft - 2 WHERE lft > @myRight;

UNLOCK TABLES;
```

最后的思考

原做者推荐了一本名为《Joe Celko's Trees and Hierarchies in SQL for Smarties》的书籍，该书的做者是SQL领域的大神Joe Celko（嵌套几何模型的创造者）。这本书涵盖了本文中未涉及到的一些高级话题。

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