另一个答案很好，如果您使用的是SQLite3.25或更新版本（或者我认为是MySQL8），那么可以使用 dense_rank() 窗口功能：

WITH rankings(Album, Artist, TitleCount, ranking) AS (
  SELECT al.title, ar.name, count(al.title)
       , dense_rank() OVER (ORDER BY count(al.title) DESC)
  FROM tracks AS t
  INNER JOIN albums AS al ON al.albumid = t.albumid
  INNER JOIN artists AS ar ON ar.artistid = al.artistid
  GROUP BY al.albumid)
SELECT Album, Artist, TitleCount
FROM rankings
WHERE ranking <= 5
ORDER BY ranking;

使用以下代码

SELECT
  albums.title AS Album,
  artists.name AS Artist,
  COUNT(*) as TitleCount
FROM tracks
INNER JOIN albums ON albums.albumid = tracks.albumid
INNER JOIN artists ON artists.artistid = albums.artistid
GROUP BY albums.albumid 
HAVING COUNT(*) IN
(
  SELECT DISTINCT COUNT(*)
  FROM tracks
  GROUP BY albumid
  ORDER BY count(*) DESC

)
ORDER BY TitleCount DESC LIMIT 5;

mysql和sqlite LIMIT 两个条款都缺少一个 WITH TIES 选项，这正是你需要的。所以改用子查询：选择五个最大数量的轨迹（这是一个罕见的实际合并的情况） GROUP BY 与 DISTINCT - GROUP BY 要获得每张专辑的数量， DISTINCT 要获得五个最高点（不同的计数），请选择具有尽可能多的曲目的专辑。因为这是一个聚合结果，所以它属于 HAVING 条款：

SELECT
  albums.title AS Album,
  artists.name AS Artist,
  COUNT(*) as TitleCount
FROM tracks
INNER JOIN albums ON albums.albumid = tracks.albumid
INNER JOIN artists ON artists.artistid = albums.artistid
GROUP BY albums.albumid 
HAVING COUNT(*) IN
(
  SELECT DISTINCT COUNT(*)
  FROM tracks
  GROUP BY albumid
  ORDER BY count(*) DESC
  LIMIT 5
)
ORDER BY TitleCount DESC;