使用外键SQL将数据插入到表中

2022-03-12 00:00:00 sql backend database mysql sql-server

在为‘Item’表的一行插入下列值之后,我需要向‘ItemBook’表中插入一些数据:

Name='Clippers', itemLink='amazon.com' description='hair clippers'  

现在我还想将一些数据插入到ItemBook表中,但我不确定如何对具有外键的表执行此操作。以下是SQL代码:

CREATE TABLE Item (
Name VARCHAR(100) NOT NULL,
itemLink VARCHAR(100) NOT NULL,
description VARCHAR(1000) NOT NULL,
PRIMARY KEY (Name)
);

CREATE TABLE ItemBook (
ItemName VARCHAR(100) NOT NULL,
Publisher VARCHAR(100) NOT NULL,
PRIMARY KEY (ItemName),
FOREIGN KEY (ItemName) REFERENCES Item(Name)
);

我的尝试:

INSERT INTO itemBook (Name, Publisher)
VALUES ('Clippers', 'Bob');

错误消息:

Error Code: 1175. You are using safe update mode and you tried to update a table without a WHERE that uses a KEY column.  To disable safe mode, toggle the option in Preferences -> SQL Editor and reconnect.

其他尝试:

INSERT INTO eventbook (EventName, Publisher)
SELECT E.name
FROM event E
WHERE E.name = eventbook.EventName;

错误消息:

Error Code: 1054. Unknown column 'eventbook.EventName' in 'where clause'

解决方案

不使用整数ID有什么特殊原因吗?我将执行以下操作:

CREATE TABLE Item (
Id INTEGER NOT NULL IDENTITY PRIMARY KEY,
Name VARCHAR(100) NOT NULL,
itemLink VARCHAR(100) NOT NULL,
description VARCHAR(1000) NOT NULL,
);

CREATE TABLE ItemBook (
Id INTEGER NOT NULL IDENTITY PRIMARY KEY,
ItemName VARCHAR(100) NOT NULL,
Publisher VARCHAR(100) NOT NULL,
ItemId INTEGER NOT NULL,
FOREIGN KEY (ItemId) REFERENCES Item(Id)
);

INSERT INTO ItemBook Values(1, 'ItemName', 'Publisher', 0)

您的想法是什么?

编辑1.根据您的回应和示例,我为SQLITE生成了以下SQL(在其他数据库中应该也可以正常工作)

CREATE TABLE Item (
Name VARCHAR(100) NOT NULL,
ItemLink VARCHAR(100) NOT NULL,
Description VARCHAR(1000) NOT NULL,
PRIMARY KEY (Name)
);

CREATE TABLE ItemBook (
ItemName VARCHAR(100) NOT NULL,
Publisher VARCHAR(100) NOT NULL,
PRIMARY KEY (ItemName),
FOREIGN KEY (ItemName) REFERENCES Item(Name)
);

INSERT INTO Item (Name, ItemLink, Description) VALUES("Test Book", "http://www.testlink.com/", "This is a test book");

INSERT INTO ItemBook (ItemName, Publisher) Values("Test Book", "Test Publisher");

SELECT * FROM Item i JOIN ItemBook b on b.ItemName = i.Name

Check the result in this print

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