MySQL外键:无法添加或更新子行:外键约束失败

时间:2015-05-15 03:57:23

标签: mysql database foreign-keys constraints

我创建了几个表,一个是父表和三个子表。

CREATE TABLE `members` (
 `memberID` int(11) NOT NULL AUTO_INCREMENT,
 `username` varchar(255) NOT NULL,
 `password` varchar(60) NOT NULL,
 PRIMARY KEY (`memberID`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1

CREATE TABLE `pets` (
 `petsID` int(11) NOT NULL AUTO_INCREMENT,
 `Gender` varchar(25) NOT NULL,
 `Age` varchar(25) NOT NULL,
 `Size` varchar(25) NOT NULL,
 `memberID` int(11) NOT NULL,
 PRIMARY KEY (`petsID`),
 KEY `fk_pets` (`memberID`),
 CONSTRAINT `pets_ibfk_1` FOREIGN KEY (`memberID`) REFERENCES `members` (`memberID`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1

CREATE TABLE `owner` (
 `ownerID` int(11) NOT NULL AUTO_INCREMENT,
 `nameOwner` varchar(25) NOT NULL,
 `ageOwner` varchar(25) NOT NULL,
 `bioOwner` varchar(800) NOT NULL,
 `memberID` int(11) NOT NULL,
 PRIMARY KEY (`ownerID`),
 KEY `fk_owner` (`memberID`),
 CONSTRAINT `owner_ibfk_1` FOREIGN KEY (`memberID`) REFERENCES `members` (`memberID`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1

CREATE TABLE `service` (
 `serviceID` int(11) NOT NULL AUTO_INCREMENT,
 `nameService` varchar(25) NOT NULL,
 `webService` varchar(25) NOT NULL,
 `memberID` int(11) NOT NULL,
 PRIMARY KEY (`serviceID`),
 KEY `fk_service` (`memberID`),
 CONSTRAINT `service_ibfk_1` FOREIGN KEY (`memberID`) REFERENCES `members` (`memberID`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1

测试完我的桌子后,我发现了这个错误:

INSERT INTO owner (nameOwner, ageOwner, bioOwner) VALUES ('', '', '')
Cannot add or update a child row: a foreign key constraint fails (`DB`.`owner`, CONSTRAINT `owner_ibfk_1` FOREIGN KEY (`memberID`) REFERENCES `members` (`memberID`) ON DELETE CASCADE ON UPDATE CASCADE)

我已经尝试截断我的父表,但它仍然会出现此错误。我是数据库设计的初学者,所以请像我一样解释。

1 个答案:

答案 0 :(得分:0)

CONSTRAINT `owner_ibfk_1` FOREIGN KEY (`memberID`) REFERENCES `members` (`memberID`)

对owner表的这个约束意味着该表中的每一行都需要有一个有效的memberID。如果它存在于memberID表的members列中,则仅视为有效

您的insert声明:

INSERT INTO owner (nameOwner, ageOwner, bioOwner) VALUES ('', '', '')

提供要插入的memberID因此无论使用什么 1 显然不在members表中。因此违反约束。

最简单的解决方法是在memberID语句中提供有效的insertmembers中已存在该语句。

举个例子,假设你在42中没有members,这会有效:

insert into members (memberID, username, password) values (
    42,
    'paxdiablo',
    'never_you_mind');
insert into owner (nameOwner, ageOwner, bioOwner, memberID) values (
    'Pax Diablo',
    'too damn old',
    'no idea what this should be',
    42);

如果您知道会员ID应该是什么,您可以使用以下内容:

insert into owner (nameOwner, ageOwner, bioOwner, memberID) 
    select 'Pax Diablo',
           'too damn old',
           'no idea what this should be',
           memberID
      from members where username = 'paxdiablo';

虽然您可能希望members.memberID是唯一的。

1 这里实际发生的事情取决于您使用的MySQL版本以及它是否以严格的SQL模式运行。我认为,从内存中,它将使insert失败或使用零作为默认值。