2016-04-14 103 views
1

我收到错误:Error 1217在我的MySQL代码中。我MySQL代码是MySQL代码中的错误代码1217

DROP TABLE IF EXISTS Formed; 
DROP TABLE IF EXISTS Album; 
DROP TABLE IF EXISTS Band; 
DROP TABLE IF EXISTS Customers; 
DROP TABLE IF EXISTS Track; 

CREATE TABLE Formed(
    FormedID int AUTO_INCREMENT, 
    YearFormed int, 
    CountryFormed varchar(50), 
    CityFormed varchar(50), 
    BandMembers varchar(400), 
    PRIMARY KEY(FormedID)) 
    ENGINE=InnoDB; 

CREATE TABLE Track (
    TrackID int AUTO_INCREMENT, 
    AlbumID int AUTO_INCREMENT, 
    Songs varchar (100), 
    TrackNumber varchar (20), 
    Title varchar (30), 
    TrackDuration varchar (4), 
    PRIMARY KEY (TrackID)) 
    ENGINE=InnoDB; 

CREATE TABLE Album(
    AlbumID int AUTO_INCREMENT, 
    TrackID int AUTO_INCREMENT, 
    BandID int AUTO_INCREMENT, 
    KEY(TrackID), 
    KEY(BandID), 
    Price varchar(5), 
    PublicationDate varchar(11), 
    Title varchar(30), 
    Genre varchar (36), 
    PRIMARY KEY(AlbumID)) 
    ENGINE=InnoDB; 

CREATE TABLE Band(
    BandID int AUTO_INCREMENT, 
    AlbumID int AUTO_INCREMENT, 
    KEY(AlbumID), 
    RecordLabel varchar(50), 
    PRIMARY KEY(BandID)) 
    ENGINE=InnoDB; 

CREATE TABLE Customers (
    CustomerID int AUTO_INCREMENT, 
    CName varchar (20), 
    CPhone int (11), 
    CEmail varchar (50), 
    CPPaid varchar (50), 
    CPDate date, 
    PRIMARY KEY (CustomerID)) 
    ENGINE=InnoDB; 

ALTER TABLE Track 
ADD FOREIGN KEY (AlbumID) REFERENCES Album(AlbumID)ON DELETE SET NULL ON UPDATE CASCADE; 

ALTER TABLE Album 
ADD FOREIGN KEY (TrackID) REFERENCES Track(TrackID)ON DELETE SET NULL ON UPDATE CASCADE; 

ALTER TABLE Album 
ADD FOREIGN KEY (BandID) REFERENCES Band(BandID)ON DELETE SET NULL ON UPDATE CASCADE; 

ALTER TABLE Band 
ADD FOREIGN KEY (AlbumID) REFERENCES Album(AlbumID)ON DELETE SET NULL ON UPDATE CASCADE; 

而我得到的错误DROP TABLE IF EXISTS AlbumError Code: 1217. Cannot delete or update a parent row: a foreign key constraint fails,也恰好为其他DROP TABLE IF EXISTS线。我是新手,在MySQL脚本中使用Foreign Keys,我不知道为什么我会收到这些错误。如果任何人都可以帮助我解决这个问题,将不胜感激。我已经尝试了所有可以通过以下链接Bogus foreign key constraint fail

+0

您可能必须以不同的顺序砸他们避免违反外键约束,首先删除外键,甚至禁用外键。 – tadman

+0

@tadman我刚刚将它们从代码中注释掉了,它仍然在发生 – smitthy

+0

@Siguza这不是重复的,因为我已经尝试了所有给出的答案,它仍然不起作用 – smitthy

回答

2

脚本即可找到答案,但在此之前Dropping表,尽量去除Foreign Key References

+0

因此,这将是沿线:'ALTER TABLE Track ADD FOREIGN KEY(AlbumID);'? – smitthy

+0

您使用的是phpmyadmin还是workbench?还是Consol!? –

+0

我正在使用MySQL工作台 – smitthy