无法删除或更新父行:外键约束失败

Question

做的时候:

DELETE FROM `jobs` WHERE `job_id` =1 LIMIT 1 

它错误:

#1451 - Cannot delete or update a parent row: a foreign key constraint fails 
(paymesomething.advertisers, CONSTRAINT advertisers_ibfk_1 FOREIGN KEY 
(advertiser_id) REFERENCES jobs (advertiser_id))

这是我的表格:

CREATE TABLE IF NOT EXISTS `advertisers` (
  `advertiser_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  `password` char(32) NOT NULL,
  `email` varchar(128) NOT NULL,
  `address` varchar(255) NOT NULL,
  `phone` varchar(255) NOT NULL,
  `fax` varchar(255) NOT NULL,
  `session_token` char(30) NOT NULL,
  PRIMARY KEY (`advertiser_id`),
  UNIQUE KEY `email` (`email`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;


INSERT INTO `advertisers` (`advertiser_id`, `name`, `password`, `email`, `address`, `phone`, `fax`, `session_token`) VALUES
(1, 'TEST COMPANY', '', '', '', '', '', '');

CREATE TABLE IF NOT EXISTS `jobs` (
  `job_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `advertiser_id` int(11) unsigned NOT NULL,
  `name` varchar(255) NOT NULL,
  `shortdesc` varchar(255) NOT NULL,
  `longdesc` text NOT NULL,
  `address` varchar(255) NOT NULL,
  `time_added` int(11) NOT NULL,
  `active` tinyint(1) NOT NULL,
  `moderated` tinyint(1) NOT NULL,
  PRIMARY KEY (`job_id`),
  KEY `advertiser_id` (`advertiser_id`,`active`,`moderated`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;


INSERT INTO `jobs` (`job_id`, `advertiser_id`, `name`, `shortdesc`, `longdesc`, `address`, `active`, `moderated`) VALUES
(1, 1, 'TEST', 'TESTTEST', 'TESTTESTES', '', 0, 0);

ALTER TABLE `advertisers`
  ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) REFERENCES `jobs` (`advertiser_id`);

Answer 1

简单的方法是禁用外键检查; 进行更改然后重新启用外键检查.

SET FOREIGN_KEY_CHECKS=0; -- to disable them
SET FOREIGN_KEY_CHECKS=1; -- to re-enable them

Answer 2

按原样,您必须先删除广告商表中的行,然后才能删除它引用的作业表中的行.这个:

ALTER TABLE `advertisers`
  ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) 
      REFERENCES `jobs` (`advertiser_id`);

......实际上与应该是的相反.实际上,这意味着您必须在广告商之前在作业表中创建记录.所以你需要使用:

ALTER TABLE `jobs`
  ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) 
      REFERENCES `advertisers` (`advertiser_id`);

一旦纠正了外键关系,你的删除语句就可以了.

Answer 3

在您当前(可能有缺陷的)设计下,您必须先删除广告商表中的行,然后才能删除它引用的作业表中的行.

或者,您可以设置外键,以便父表中的删除会导致子表中的行自动删除.这称为级联删除.它看起来像这样:

ALTER TABLE `advertisers`
ADD CONSTRAINT `advertisers_ibfk_1`
FOREIGN KEY (`advertiser_id`) REFERENCES `jobs` (`advertiser_id`)
ON DELETE CASCADE;

话虽如此,正如其他人已经指出的那样,你的外键感觉它应该反过来,因为广告商表真的包含主键而job表包含外键.我会像这样重写它:

ALTER TABLE `jobs`
ADD FOREIGN KEY (`advertiser_id`) REFERENCES `advertisers` (`advertiser_id`);

并且不需要级联删除.

Answer 4

如果要删除表,则应该在一个步骤中执行以下查询

SET FOREIGN_KEY_CHECKS = 0; DROP TABLE table_name;

Answer 5

禁用外键检查并进行更改，然后重新启用外键检查。

SET FOREIGN_KEY_CHECKS=0; -- to disable them
DELETE FROM `jobs` WHERE `job_id` = 1 LIMIT 1 
SET FOREIGN_KEY_CHECKS=1; -- to re-enable them

Answer 6

我尝试了@Alino Manzi 提到的解决方案，但它在使用 wpdb 的 WordPress 相关表上对我不起作用。

然后我修改了下面的代码并且它起作用了

SET FOREIGN_KEY_CHECKS=OFF; //disabling foreign key

//run the queries which are giving foreign key errors

SET FOREIGN_KEY_CHECKS=ON; // enabling foreign key

Answer 7

我认为你的外键是倒退的.尝试:

ALTER TABLE 'jobs'
ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) REFERENCES `advertisers` (`advertiser_id`)

Answer 8

如果有多个工作具有相同的广告商 ID，那么您的外键应该是：

ALTER TABLE `jobs`
ADD CONSTRAINT `advertisers_ibfk_1` 
FOREIGN KEY (`advertiser_id`) 
REFERENCES `advertisers` (`advertiser_id`);

否则（如果您的情况相反），如果您希望在删除作业中的行时自动删除广告商中的行，请将“ON DELETE CASCADE”选项添加到外键的末尾：

ALTER TABLE `advertisers`
ADD CONSTRAINT `advertisers_ibfk_1` 
FOREIGN KEY (`advertiser_id`) 
REFERENCES `jobs` (`advertiser_id`)
ON DELETE CASCADE;

查看外键约束