MySQL 函数 : rank table by most similar attributes

Question

我有一个包含产品 ID 和 关键字 的表格，如下所示:

+------------+------------------+------+-----+---------+----------------+
| Field      | Type             | Null | Key | Default | Extra          |
+------------+------------------+------+-----+---------+----------------+
| id         | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| product_id | int(10) unsigned | YES  | MUL | NULL    |                |
| keyword    | varchar(255)     | YES  |     | NULL    |                |
+------------+------------------+------+-----+---------+----------------+

此表仅存储产品 ID 以及与这些产品关联的关键字。例如，它可能包含:

+----+------------+---------+
| id | product_id | name    |
+----+------------+---------+
|  1 |         1  | soft    |
|  2 |         1  | red     |
|  3 |         1  | leather |
|  4 |         2  | cloth   |
|  5 |         2  | red     |
|  6 |         2  | new     |
|  7 |         3  | soft    |
|  8 |         3  | red     |
|  9 |         4  | blue    |
+----+------------+---------+

换句话说:

• 产品 1 是柔软的红色皮革。
• 产品 2 是布料，红色，全新。
• 产品3是红色的，柔软的，
• 产品 4 是蓝色的。

我需要一些方法来获取产品 ID，并取回按常用关键字数量排序的产品 ID 列表

例如，如果我传入 product_id 1，我希望返回:

+----+-------+------------+
| product_id | matches    |
+------------+------------+
|     3      | 2          | (product 3 has two common keywords with product 1)
|     2      | 1          | (product 2 has one common keyword with product 1)
|     4      | 0          | (product 4 has no common keywords with product 1)
+------------+------------+

Answer 1

一个选项使用带有条件聚合的自右外连接来计算匹配名称之间的数量，例如产品 ID 1 和所有其他产品 ID:

SELECT t2.product_id,
       SUM(CASE WHEN t1.name IS NOT NULL THEN 1 ELSE 0 END) AS matches
FROM yourTable t1
RIGHT JOIN yourTable t2
    ON t1.name = t2.name AND
       t1.product_id = 1
WHERE t2.product_id <> 1
GROUP BY t2.product_id
ORDER BY t2.product_id

点击下面的链接获取正在运行的演示:

SQLFiddle