2013-11-23 48 views
-2

所以我有這樣的片段:SQL WHERE插入

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; 

我需要把WHERE musical_genre = 'Rock',但似乎預言不想要我把在任何地方......

回答

0

如果我明白你的問題正確,這是你想要什麼?:

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; 

和,這個問題有mysql標記。但似乎像甲骨文的問題。

+0

在這裏等一下,你的意思是告訴我縮進事宜嗎?我把你的縮進版本,它似乎工作。哇OK – user3015045

+0

@ user3015045 oop!縮進不是問題。我在GROUP BY之前添加了WHERE。 ;-) –

+0

但我之前做過,我得到一個錯誤?是的,我忘了標記爲Oracle。 – user3015045