【发布时间】:2015-03-25 05:24:15
【问题描述】:
我正在尝试为健身房管理系统创建数据库,但我无法弄清楚为什么会出现此错误。我试图在这里搜索答案,但找不到。
ERROR 1215 (HY000): Cannot add foreign key constraint
CREATE TABLE sales(
saleId int(100) NOT NULL AUTO_INCREMENT,
accountNo int(100) NOT NULL,
payName VARCHAR(100) NOT NULL,
nextPayment DATE,
supplementName VARCHAR(250),
qty int(11),
workoutName VARCHAR(100),
sDate datetime NOT NULL DEFAULT NOW(),
totalAmount DECIMAL(11,2) NOT NULL,
CONSTRAINT PRIMARY KEY(saleId, accountNo, payName),
CONSTRAINT FOREIGN KEY(accountNo) REFERENCES accounts(accountNo) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT FOREIGN KEY(payName) REFERENCES paymentFor(payName) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT FOREIGN KEY(supplementName) REFERENCES supplements(supplementName) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT FOREIGN KEY(workoutName) REFERENCES workouts(workoutName) ON DELETE CASCADE ON UPDATE CASCADE
);
ALTER TABLE sales AUTO_INCREMENT = 2001;
这是父表。
CREATE TABLE accounts(
accountNo int(100) NOT NULL AUTO_INCREMENT,
accountType VARCHAR(100) NOT NULL,
firstName VARCHAR(50) NOT NULL,
lastName VARCHAR(60) NOT NULL,
birthdate DATE NOT NULL,
gender VARCHAR(7),
city VARCHAR(50) NOT NULL,
street VARCHAR(50),
cellPhone VARCHAR(10),
emergencyPhone VARCHAR(10),
email VARCHAR(150) NOT NULL,
description VARCHAR(350),
occupation VARCHAR(50),
createdOn datetime NOT NULL DEFAULT NOW(),
CONSTRAINT PRIMARY KEY(accountNo)
);
ALTER TABLE accounts AUTO_INCREMENT = 1001;
CREATE TABLE supplements(
supplementId int(100) NOT NULL AUTO_INCREMENT,
supplementName VARCHAR(250) NOT NULL,
manufacture VARCHAR(100),
description VARCHAR(150),
qtyOnHand INT(5),
unitPrice DECIMAL(11,2),
manufactureDate DATE,
expirationDate DATE,
CONSTRAINT PRIMARY KEY(supplementId, supplementName)
);
ALTER TABLE supplements AUTO_INCREMENT = 3001;
CREATE TABLE workouts(
workoutId int(100) NOT NULL AUTO_INCREMENT,
workoutName VARCHAR(100) NOT NULL,
description VARCHAR(7500) NOT NULL,
duration VARCHAR(30),
CONSTRAINT PRIMARY KEY(workoutId, workoutName)
);
ALTER TABLE workouts AUTO_INCREMENT = 4001;
CREATE TABLE paymentFor(
payId int(100) NOT NULL AUTO_INCREMENT,
payName VARCHAR(100) NOT NULL,
amount DECIMAL(11,2),
CONSTRAINT PRIMARY KEY(payId, payName)
);
ALTER TABLE paymentFor AUTO_INCREMENT = 5001;
你们能帮我解决这个问题吗?谢谢。
【问题讨论】:
-
不确定是不是这样,但在
sales中,supplementName和workoutName都不是像它们在workouts表和supplement表中引用的列一样。我相信与外键相关的列必须完全匹配的理论。
标签: mysql foreign-keys