Was muss ich in der Abfrage ändern, um überhaupt eine exakte Übereinstimmung zu erhalten?

Wenn zum Beispiel jemand Fröhliches Lied in das Suchformular eingibt, erhält er nun diese Ausgabe..

Glückliches Kind

Schöne Tage

Fröhliche Menschen

Fröhliches Lied

Fröhlicher Bahnhof

...

Das ist meine Frage..

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;