所以我有这个片段:
SELECT Recording_artist.artist_name,Musical_genre.musical_genre,
COUNT(Musical_genre.musical_genre) AS Songs
FROM Recording_artist
FULL OUTER JOIN Album
ON Recording_artist.recording_artist_id = Album.recording_artist_id
RIGHT OUTER JOIN Song
ON Album.album_id = Song.album_id
INNER JOIN Musical_genre
ON Album.musical_genre_id = Musical_genre.musical_genre_id
GROUP BY Recording_artist.artist_name,Musical_genre.musical_genre;
我需要放一个WHERE musical_genre = 'Rock'
,但似乎甲骨文不希望我把它放在任何地方......
答案 0 :(得分:0)
如果我理解你的问题,这就是你想要的?:
SELECT Recording_artist.artist_name,Musical_genre.musical_genre,
COUNT(Musical_genre.musical_genre) AS Songs
FROM Recording_artist
FULL OUTER JOIN Album
ON Recording_artist.recording_artist_id = Album.recording_artist_id
RIGHT OUTER JOIN Song
ON Album.album_id = Song.album_id
INNER JOIN Musical_genre
ON Album.musical_genre_id = Musical_genre.musical_genre_id
WHERE Musical_genre.musical_genre = 'Rock'
GROUP BY Recording_artist.artist_name,Musical_genre.musical_genre;
并且,这个问题有mysql标签。但似乎是甲骨文的问题。