mysql搜索标题，描述和多行标签（关于条件）

Question

我想实现类似mysql search title, description and multi rows tag的搜索。

这是我的桌子：

书籍：

+----+-----------------------+-------------+
| id | name                  | description | 
+----+-----------------------+-------------+
|  1 | Me Before You         | [TEXT]      |
|  2 | How To Win Friends... | [TEXT]      |
|  3 | The Girl on the Train | [TEXT]      |
|  4 | After You             | [TEXT]      |
|  5 | We Were Liars         | [TEXT]      |
+----+-----------------------+-------------+

标签：

+----+-----------------------+
| id | tag                   |
+----+-----------------------+
|  1 | romance               |
|  2 | thriller              |
|  3 | fantasy               |
|  4 | science fiction       |
|  5 | drama                 |
|  6 | friends               |
+----+-----------------------+

书籍标签：

+---------+--------+
| book_id | tag_id |
+---------+--------+
|       1 |      1 |
|       1 |      3 |
|       2 |      3 |
|       3 |      3 |
|       3 |      5 |
|       4 |      1 |
|       4 |      5 |
|       4 |      6 |
|       5 |      2 |
|       5 |      6 |
+---------+--------+

以下是一些示例搜索和所需结果：

'romance'       -> books 1, 4
'friends'       -> books 2, 4, 5
'friends win'   -> books 2
'fantasy'       -> books 2, 3
'fantasy train' -> books 3

在构建 SQL 查询之前，函数会同时检查每个给定的关键字，如果它甚至是一个标签。例如，在这种情况下，我的问题是：

• 案例：3 /
• 关键词：朋友赢 /
• 标签：朋友

查询：

SELECT SQL_CALC_FOUND_ROWS 
    b.id, b.name, 
    MATCH(b.name) AGAINST('*friends* *win*' IN BOOLEAN MODE) as name_score,
    MATCH(t.tag) AGAINST('friends' IN BOOLEAN MODE)as tag_score
FROM 
    books b
LEFT JOIN 
    books_tags bt ON bt.book_id = b.id
LEFT JOIN 
    tags t ON t.id = bt.tag_id 
WHERE  
    MATCH(b.name) AGAINST('*friends win*' IN BOOLEAN MODE)
    OR MATCH(t.tag) AGAINST('friends' IN BOOLEAN MODE)
GROUP BY 
    b.id
ORDER BY 
    name_score DESC, (tag_score + name_score) DESC

结果：

array (size=3)
  0 => string '2' (length=1)
  1 => string '4' (length=1)
  2 => string '5' (length=1)

在这种情况下，关键字“朋友”已经与标题匹配，因此必须减少条件，不应再搜索标签。我该如何解决？

Answer 1

好的，最后我写了一个函数，它为我生成了一个有效的查询。该功能非常复杂，取决于多个用户输入。 解决方案是连接 t.tag 和 b.name。这就是我的 WHERE 条件的样子，它对我来说很好：

... WHERE ap.active='yes'
AND (LOWER(CONCAT_WS(' ', IF(LENGTH(t.tag), t.tag, NULL), IF(LENGTH(b.name), b.name, NULL) )) REGEXP 'friends'
AND LOWER(CONCAT_WS(' ', IF(LENGTH(t.tag), t.tag, NULL), IF(LENGTH(b.name), b.name, NULL) )) REGEXP 'win')
GROUP BY b.id

Answer 2

这就是你所追求的吗？...

数据集

DROP TABLE IF EXISTS books;

CREATE TABLE books
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,name VARCHAR(50) NOT NULL
,FULLTEXT(name)
) ENGINE = MyISAM;

INSERT INTO books VALUES
(1,'Me Before You'),
(2,'How To Win Friends...'),
(3,'The Girl on the Train'),
(4,'After You'),
(5,'We Were Liars');

DROP TABLE IF EXISTS tags;

CREATE TABLE tags
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,tag VARCHAR(20) NOT NULL
);

INSERT INTO tags VALUES
(1,'romance'),
(2,'thriller'),
(3,'fantasy'),
(4,'science fiction'),
(5,'drama'),
(6,'friends');

DROP TABLE IF EXISTS books_tags;

CREATE TABLE books_tags
(book_id INT NOT NULL
,tag_id INT NOT NULL
,PRIMARY KEY(book_id,tag_id)
);

INSERT INTO books_tags VALUES
(1,1),
(1,3),
(2,3),
(3,3),
(3,5),
(4,1),
(4,5),
(4,6),
(5,2),
(5,6);

查询和结果

SELECT DISTINCT b.*
              , MATCH(b.name) AGAINST('*friends* *win*' IN BOOLEAN MODE) name_score
           FROM books b 
           LEFT 
           JOIN books_tags bt 
             ON bt.book_id = b.id 
           LEFT 
           JOIN tags t 
             ON t.id = bt.tag_id 
            AND t.tag IN ('friends','win') 
          WHERE t.id IS NULL;

+----+-----------------------+------------+
| id | name                  | name_score |
+----+-----------------------+------------+
|  1 | Me Before You         |          0 |
|  2 | How To Win Friends... |          1 |
|  3 | The Girl on the Train |          0 |
|  4 | After You             |          0 |
|  5 | We Were Liars         |          0 |
+----+-----------------------+------------+