Question

我有两张桌子：

表pages：

|id |title      |
+---+-----------+
|1  |Foxes      |
|2  |Pet dogs   |

表keywords：

|id |page_id |phrase                |
+---+--------+----------------------+
|1  |1       |quick brown fox jumps |
|2  |1       |lazy dog              |
|3  |2       |caring for a dog      |
|4  |2       |pet dog               |

我目前正在使用此查询：

SELECT pages.title, MATCH (phrase) AGAINST ('pet dog') as relevance
FROM pages
INNER JOIN keywords
ON pages.id = keywords.page_id
WHERE MATCH (phrase) AGAINST ('pet dog')
ORDER BY relevance DESC;

我得到了这个结果：

|title |relevance            |
+------+---------------------+
|Dogs  |0.3780859112739563   |
|Foxes |0.015609688125550747 |
|Dogs  |0.015609688125550747 |

但我希望每个title只有一个relevance，因此每个适用于relevance的短语的title应该像这样加在一起：

|title |relevance            |
+------+---------------------+
|dogs  |0.3936955994         |
|foxes |0.015609688125550747 |

Answer 1

在结果集中添加GROUP BY title语句，因此相同title的记录将汇总到一个组中。
调用SUM函数获取每个组relevance的总和。

因此，此查询可能会帮助您：

SELECT title, SUM(relevance)
FROM
(
  SELECT pages.title, MATCH (phrase) AGAINST ('pet dog') as relevance
  FROM pages
  INNER JOIN keywords
  ON pages.id = keywords.page_id
  WHERE MATCH (phrase) AGAINST ('pet dog')
  ORDER BY relevance DESC
)
GROUP BY title;

Answer 2

您需要按标题分组并总结相关性分数。以下查询解决了您的问题：

SELECT title, SUM(MATCH (phrase) AGAINST ('pet dog' IN BOOLEAN MODE)) as relevance
FROM pages AS p
INNER JOIN keywords AS k ON p.id = k.page_id
GROUP BY title
HAVING relevance > 0
ORDER BY relevance DESC;

参见 DEMO

从全文搜索MATCH中添加分数

2 个答案: