我正在尝试将4个表与INNER JOIN链接在一起

Question

我的代码看起来像这样

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;

希望尝试获得艺术家的姓名，流派和歌曲以进行匹配和展示。 可是我明白了

Unknown column 'Artists.ArtistID' in 'on clause'

对SQL和INNER JOINS来说，我是新手，任何帮助和解释都很棒！

Answer 1

尝试这个

SELECT Artists.name, Genre.genreName, Songs.name
FROM Songs
INNER JOIN Albums ON Songs.Albums_albumID=Albums.albumsID
INNER JOIN Artists ON Albums.Artists_ArtistID=Artists.ArtistID
INNER JOIN Genre ON Artists.ArtistID=Genre.genreID;

检查联接中表的顺序和出现情况

Answer 2

您会在第一行连接中提到Songs表。 但是您只在最后一行中加入ON函数表，就像您的代码一样，您应该尝试以下方法：

SELECT 
Artists.name, Genre.genreName, Songs.name
FROM Songs
INNER JOIN Albums 
ON Songs.Albums_albumID=Albums.albumsID
INNER JOIN Artists 
ON Albums.Artists_ArtistID=Artists.ArtistID
INNER JOIN Genre 
ON Artists.ArtistID=Genre.genreID;