进行时:
DELETE FROM `jobs` WHERE `job_id` =1 LIMIT 1
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))
这是我的桌子:
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`);
照原样,必须先删除Advertisers表中的行,然后才能删除它引用的Jobs表中的行。这个:
ALTER TABLE `advertisers` ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) REFERENCES `jobs` (`advertiser_id`);
…实际上与应有的相反。实际上,这意味着您必须在作业表中有一条记录,然后才是广告商。因此,您需要使用:
ALTER TABLE `jobs` ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) REFERENCES `advertisers` (`advertiser_id`);
纠正外键关系后,您的delete语句将起作用。