Question

我有以下查询，我希望只返回与label_name“Ultra Music”匹配的记录，但是它会返回任何label_name：

SELECT  `chart_rating` . * , SUM( chart_rating.rating_score ) AS total_score, release.release_label
FROM  `chart_rating` 
INNER JOIN  `release` ON  `chart_rating`.`rating_release_id` =  `release`.`id` 
WHERE MATCH (
release.release_label
)
AGAINST (
"Ultra Music"
IN BOOLEAN MODE
)
GROUP BY  `rating_release_id` 
ORDER BY  `total_score` DESC

我对release_artist有一个几乎完全相同的查询，它完美无缺。

此查询有什么问题？

Answer 1

您可以尝试使用Addition Subtraction Multiplication Division Exponents Square roots Sin Cos Basic Loops, Conditionals, and relational operators.，

删除space

然后添加REPLACE

+UltraMusic

希望这有帮助。

Answer 2

为什么不使用简单的比较。

SELECT  `chart_rating` . * , SUM( chart_rating.rating_score ) AS total_score, release.release_label
FROM  `chart_rating` 
INNER JOIN  `release` ON  `chart_rating`.`rating_release_id` =  `release`.`id` 
WHERE release.release_label = "Ultra Music"
GROUP BY  `rating_release_id` 
ORDER BY  `total_score` DESC

如果您确实需要全文搜索，where子句可能如下所示：MATCH (release.release_label) AGAINST ("+Ultra +Music" IN BOOLEAN MODE)

Answer 3

默认情况下（当既未指定+也未指定时），该单词是可选的，但包含它的行的等级更高。这模仿了没有IN BOOLEAN MODE修饰符的MATCH（）... AGAINST（）的行为。

MySQL Ref - Bool

糟糕的mySQL查询

3 个答案: