【发布时间】:2014-04-02 05:45:40
【问题描述】:
我的代码是这样的
CREATE TABLE Genre (
genreID INT NOT NULL DEFAULT 0,
genreName VARCHAR(20) NULL,
PRIMARY KEY (genreID));
CREATE TABLE Artists (
ArtistID INT NOT NULL DEFAULT 0,
name VARCHAR(45) NULL,
Genre_genreID INT NOT NULL,
PRIMARY KEY (ArtistID),
FOREIGN KEY (Genre_genreID)
REFERENCES Genre(genreID));
CREATE TABLE Albums (
albumsID INT NOT NULL DEFAULT 0,
name VARCHAR(45) NULL,
Artists_ArtistID INT NOT NULL,
PRIMARY KEY (albumsID),
FOREIGN KEY (Artists_ArtistID)
REFERENCES Artists(ArtistID));
CREATE TABLE Songs (
songID INT NOT NULL,
name VARCHAR(45) NULL,
length TIME NULL,
Albums_albumsID INT NOT NULL DEFAULT 0,
PRIMARY KEY (songID),
FOREIGN KEY (Albums_albumsID)
REFERENCES Albums (albumsID));
SELECT Artists.name, Genre.genreName, Songs.name
FROM Songs
INNER JOIN Genre ON Artists.ArtistID=Genre.genreID
INNER JOIN Artists ON Albums.Artists_ArtistID=Artists.ArtistID
INNER JOIN Albums ON Songs.Albums_albumID=Albums.albumsID;
希望尝试将艺术家、流派和歌曲的名称进行匹配和展示。然而我得到了
Unknown column 'Artists.ArtistID' in 'on clause'
我是 SQL 和 INNER JOINS 的新手,任何帮助和解释都会很棒!
【问题讨论】:
标签: mysql sql inner-join on-clause