Question

这个link将显示我的架构和所涉及的所有表的内容。

我的目标是，使用单个Select语句，显示每个艺术家的名称，并按照他们所拥有的“摇滚”歌曲的数量排序，即使对于没有任何艺术家的艺术家，也按照最少的顺序排列。这是我尝试过的，显然没有用。

SELECT 
    Musical_genre.musical_genre_id, 
    COUNT(Musical_genre.musical_genre_id) AS nr_rocksongs 
FROM 
    Musical_genre
JOIN 
    Album ON Album.musical_genre_id = Musical_genre.musical_genre_id
JOIN 
    Recording_artist ON Album.recording_artist_id = Recording_artist.recording_artist_id
GROUP BY 
    Album.recording_artist_id, Recording_artist.artist_name, Musical_genre.musical_genre_id
ORDER BY 
    nr_rocksongs ASC

我想要的任何想法都能得到我想要的结果吗？非常感谢所有帮助。

Answer 1

从group by中删除Musical_genre.musical_genre_id并在select

中添加artist_name

SELECT Musical_genre.musical_genre_id, Recording_artist.artist_name, 
      COUNT(Musical_genre.musical_genre_id) AS nr_rocksongs 
FROM Musical_genre
JOIN Album ON Album.musical_genre_id = Musical_genre.musical_genre_id
JOIN Recording_artist ON Album.recording_artist_id = Recording_artist.recording_artist_id
GROUP BY  Musical_genre.musical_genre_id, Recording_artist.artist_name
ORDER BY nr_rocksongs ASC

Answer 2

你需要从艺术家开始，然后加入专辑和流派。

SELECT
    ra.artist_name,
    mg.musical_genre_id,
    mg.musical_genre,
    COUNT(mg.musical_genre_id) AS nr_rocksongs
FROM
    Recording_artist ra
    LEFT JOIN Album a ON
        a.recording_artist_id = ra.recording_artist_id
    LEFT JOIN Musical_genre mg ON
        a.musical_genre_id = mg.musical_genre_id
        AND mg.musical_genre = 'Rock' 
        --use musical_genre_id = 201 if you like, but then you could remove the last left join entirely.
    LEFT JOIN Song s ON
        a.album_id = s.album_id
GROUP BY
    mg.musical_genre_id,
    mg.musical_genre,
    ra.artist_name
ORDER BY
    nr_rocksongs ASC

聚合SQL语句的问题

2 个答案: