Question

我试图将以下命令提供给MySQL CLI ，这似乎很好，但是当我再添加一个表时，它抱怨语法错误。

以下是命令

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，它表示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),
);

由于新添加的命令中没有抱怨错误，因此不确定哪里出错了。

- UPDATE -

那是固定的，但还有一个问题。添加以下表格抛出

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？ THX

Answer 1

检查代码的最后一行

FOREIGN KEY(accept_uid) REFERENCES User(uid),

删除末尾的逗号。

Answer 2

删除,行尾

FOREIGN KEY(accept_uid) REFERENCES User(uid),
--------------------------------------------^

Answer 3

第三张桌子上有一个额外的逗号。

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), <---- Extra comma
);

Answer 4

@Daniel，您不能将名称 CONTAINS 用于表，因为它是一个SQL关键字。请使用其他名称创建表格。

Answer 5

幸运的是我想出了第二个问题，外键不正确，我需要使用

        FOREIGN KEY(uid, wid) REFERENCES Wishlist(uid, wid),

引用愿望清单中的主键组合

Answer 6

FOREIGN KEY(accept_uid) REFERENCES User(uid),<-- Comma should be removed

试试这个：

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)
);

SQL语法创建表中的错误

6 个答案: