我必须在查询中更改哪些内容才能首先获得精确匹配？

例如，任何人在搜索表单中输入快乐歌曲，现在他会得到这个输出..

快乐的孩子

快乐的日子

快乐的人

欢乐歌

快乐站

...

这是我的查询..

SELECT  `artist`, `song`, `genre`, `pass`, `id`
        FROM  `songs`
        JOIN  `artists` USING (`pass`)
        WHERE  MATCH(`artist`) AGAINST("happy song")
          OR  MATCH(`song`, `genre`) AGAINST("happy song")
        ORDER BY  `song` ASC
        LIMIT  0,30
') or die(mysql_error($_connect));```

Thank you!

EDIT 03.07.22
Changed query to:<br>
```$query = mysqli_query($_connect, '
    SELECT  `artist`, `song`, `genre`, `pass`, `id`
        FROM  `songs`
        JOIN  `artists` USING (`pass`)
        WHERE  MATCH(`artist`) AGAINST("happy song")
          OR  MATCH(`song`, `genre`) AGAINST("happy song")
        ORDER BY  `song` = "happy song" DESC
') or die(mysql_error($_connect));```

The output is the same, result with only "song" in the text are above in the result when searched for "happy song"

**EDIT 14.7.22**
**Table Artists:**

CREATE TABLE `artists` (
  `artist` varchar(40) NOT NULL,
  `pass` int(10) UNSIGNED NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 PACK_KEYS=1;

ALTER TABLE `artists`
  ADD PRIMARY KEY (`pass`),
  ADD UNIQUE KEY `uni_que` (`artist`) USING BTREE;
ALTER TABLE `artists` ADD FULLTEXT KEY `full_txt` (`artist`);
COMMIT;

**Table Songs:**

    CREATE TABLE `songs` (
      `song` varchar(80) NOT NULL,
      `tags` varchar(40) NOT NULL,
      `style` varchar(20) NOT NULL,
      `pass` bigint(20) UNSIGNED NOT NULL,
      `id` bigint(20) UNSIGNED NOT NULL
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 PACK_KEYS=1;
    
    ALTER TABLE `songs`
      ADD PRIMARY KEY (`id`);
    ALTER TABLE `songs` ADD FULLTEXT KEY `song` (`song`,`tags`,`style`);
    COMMIT;