Question

我一直在尝试将这些密钥添加到我的表中，但是我收到了错误

ORA-01735：ALTER TABLE选项无效

我的代码：

ALTER TABLE Room
ADD FOREIGN KEY (RoomType_ID) REFERENCES RoomType(RoomType_ID), 
ADD FOREIGN KEY (Reservation_ID) REFERENCES Reservation(Reservation_ID), 
ADD FOREIGN KEY (Gust_ID) REFERENCES Gust(Gust_ID);

表

CREATE TABLE Gust ( Gust_ID INT NOT NULL PRIMARY KEY, First_Name VARCHAR(50), Last_Name VARCHAR(50), Email VARCHAR(20), phone_number INT(10), Address VARCHAR(30) )

CREATE TABLE Reservation ( Reservation_ID INT NOT NULL PRIMARY KEY, Start_Date Date, End_Date Date )

CREATE TABLE Room ( Room_ID INT NOT NULL PRIMARY KEY, Price INT )

CREATE TABLE RoomType ( RoomType_ID INT NOT NULL PRIMARY KEY, Class VARCHAR(10), ExtraPrice INT )

Answer 1

多个ALTER的

FOREIGN KEYS语句不起作用。

每个CONSTRAINT都应单独添加：

CREATE TABLE Gust ( Gust_ID INT PRIMARY KEY, First_Name VARCHAR(50), Last_Name VARCHAR(50), Email VARCHAR(20), phone_number INT, Address VARCHAR(30) );
CREATE TABLE Reservation ( Reservation_ID INT PRIMARY KEY, Start_Date Date, End_Date Date );
CREATE TABLE RoomType ( RoomType_ID INT NOT NULL PRIMARY KEY, Class VARCHAR(10), ExtraPrice INT );

CREATE TABLE Room(Room_ID int PRIMARY KEY, Price INT, Reservation_ID int,Gust_ID int );

ALTER TABLE Room ADD FOREIGN KEY (Room_ID) REFERENCES RoomType(RoomType_ID);

ALTER TABLE Room ADD FOREIGN KEY (Reservation_ID) REFERENCES Reservation(Reservation_ID); 
-- the table Room is assumed to have a column Reservation_ID 

ALTER TABLE Room ADD FOREIGN KEY (Gust_ID) REFERENCES Gust(Gust_ID);
    -- the table Room is assumed to have a column Gust_ID

如果不想要系统生成的约束名，那么使用这些;

ALTER TABLE Room ADD CONSTRAINT fk_RoomType_ID FOREIGN KEY (Room_ID) 
                                               REFERENCES RoomType(RoomType_ID);

ALTER TABLE Room ADD CONSTRAINT fk_Reservation_ID FOREIGN KEY (Reservation_ID) 
                                                REFERENCES Reservation(Reservation_ID); 

ALTER TABLE Room ADD CONSTRAINT fk_Gust_ID FOREIGN KEY (Gust_ID) 
                                           REFERENCES Gust(Gust_ID);

可能是首选。

P.S。：

不要将NOT NULL与PRIMARY KEY，PRIMARY KEY一起使用包括NOT NULL。
无法为INT列定义长度。

将外键添加到现有表

1 个答案: