mysql - SQL语法错误创建表

Question

我试图将以下命令提供给 MySQL CLI，这看起来不错，但是当我再添加一个表时，它会提示语法错误。

命令如下

CREATE TABLE IF NOT EXISTS User(
        uid INT,
        name VARCHAR(64) UNIQUE,
        birthday date,
        PRIMARY KEY(uid)
) ENGINE = InnoDB ;

CREATE TABLE IF NOT EXISTS UserEmail(
        uid INT,
        email VARCHAR(64),
        PRIMARY KEY(uid, email),
        FOREIGN KEY(uid) REFERENCES User(uid)
);

但是如果我想再添加一个表，它说在 PRIMARY KEY(uid) 所在行的 ')' 附近有语法错误。

CREATE TABLE IF NOT EXISTS User(
        uid INT,
        name VARCHAR(64) UNIQUE,
        birthday date,
        PRIMARY KEY(uid)
) ENGINE = InnoDB ;

CREATE TABLE IF NOT EXISTS UserEmail(
        uid INT,
        email VARCHAR(64),
        PRIMARY KEY(uid, email),
        FOREIGN KEY(uid) REFERENCES User(uid)
);

CREATE TABLE friendship(
        invite_uid  INT,
        accept_uid  INT,
        start_date  DATE,
        PRIMARY KEY(invite_uid, accept_uid),
        FOREIGN KEY(invite_uid) REFERENCES User(uid),
        FOREIGN KEY(accept_uid) REFERENCES User(uid),
);  

不确定哪里出了问题，因为新添加的命令中没有报错。

--更新--

好吧，那是固定的，但还有一个问题。添加下表抛出

无法创建表“estore.contains”(错误号:150)

CREATE TABLE contains(
        uid INT,
        wid INT,
        pid INT,
        PRIMARY KEY(uid, wid, pid),
        FOREIGN KEY(uid) REFERENCES User(uid),
        FOREIGN KEY(wid) REFERENCES Wishlist(wid),          
        FOREIGN KEY(pid) REFERENCES Product(pid)        
);

--更新 2--

我要添加的完整表格

CREATE TABLE IF NOT EXISTS User(
        uid INT,
        name VARCHAR(64) UNIQUE,
        birthday date,
        PRIMARY KEY(uid)
) ENGINE = InnoDB ;

CREATE TABLE IF NOT EXISTS UserEmail(
        uid INT,
        email VARCHAR(64),
        PRIMARY KEY(uid, email),
        FOREIGN KEY(uid) REFERENCES User(uid)
);

CREATE TABLE friendship(
        invite_uid  INT,
        accept_uid  INT,
        start_date  DATE,
        PRIMARY KEY(invite_uid, accept_uid),
        FOREIGN KEY(invite_uid) REFERENCES User(uid),
        FOREIGN KEY(accept_uid) REFERENCES User(uid)
);  

CREATE TABLE Seller(
        sid INT,
        name VARCHAR(64),
        PRIMARY KEY(sid)
);

CREATE TABLE Product(
        pid INT,
        sid INT,
        name VARCHAR(64),
        description TEXT,
        price DOUBLE,
        PRIMARY KEY(pid),
        FOREIGN KEY(sid) REFERENCES Seller(sid)
);

CREATE TABLE buy(
        uid INT,
        pid INT,
        time DATE,
        PRIMARY KEY(uid, pid),
        FOREIGN KEY(uid) REFERENCES User(uid),
        FOREIGN KEY(pid) REFERENCES Product(pid)
);

CREATE TABLE Wishlist(
        uid INT,
        wid INT,
        start_time DATE,
        end_time DATE,
        PRIMARY KEY(uid,wid),
        FOREIGN KEY(uid) REFERENCES User(uid)
);

CREATE TABLE conntains(
        uid INT,
        wid INT,
        pid INT,
        PRIMARY KEY(uid, wid, pid),
        FOREIGN KEY(uid) REFERENCES User(uid),
        FOREIGN KEY(wid) REFERENCES Wishlist(wid),          
        FOREIGN KEY(pid) REFERENCES Product(pid)        
)ENGINE = InnoDB;

有人可以帮忙吗？谢谢

Answer 1

检查代码的最后一行

FOREIGN KEY(accept_uid) REFERENCES User(uid),

去掉末尾的逗号。