mysql - 从 SQL DUMP 导入数据库模式

Question

我有以下数据库模式，其中有一堆表和外键，当我尝试导入 sql 转储时，我不断收到以下错误。

Can't create table errno 150

我知道它正在尝试创建具有尚未创建的表的依赖关系的表，但我不明白如何在不清除所有外键的情况下导入模式，然后根据给定的答案重新创建它们在 Stack 和谷歌上。

必须有一个更简单的方法，大公司有数百张 table 做什么？

我有下面的 sql 语句，如有任何建议，我们将不胜感激。谢谢

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DROP TABLE IF EXISTS `contact_interest`;
DROP TABLE IF EXISTS `contact_seeking`;
DROP TABLE IF EXISTS `interests`;
DROP TABLE IF EXISTS `job_current`;
DROP TABLE IF EXISTS `job_desired`;
DROP TABLE IF EXISTS `job_listings`;
DROP TABLE IF EXISTS `my_contacts`;
DROP TABLE IF EXISTS `profession`;
DROP TABLE IF EXISTS `seeking`;
DROP TABLE IF EXISTS `status`;
DROP TABLE IF EXISTS `zip_code`;


CREATE TABLE `contact_interest` (
  `contact_id` int(10) unsigned NOT NULL,
  `interest_id` int(10) unsigned NOT NULL,
  KEY `mycontacts_contactinterest_fk` (`contact_id`),
  KEY `interests_contactinterest_fk` (`interest_id`),
  CONSTRAINT `mycontacts_contactinterest_fk` FOREIGN KEY (`contact_id`) REFERENCES `my_contacts` (`contact_id`),
  CONSTRAINT `interests_contactinterest_fk` FOREIGN KEY (`interest_id`) REFERENCES `interests` (`interest_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;


CREATE TABLE `contact_seeking` (
  `contact_id` int(10) unsigned NOT NULL,
  `seeking_id` int(10) unsigned NOT NULL,
  KEY `contactid_contactseeking_fk` (`contact_id`),
  KEY `seeking_contactseeking_fk` (`seeking_id`),
  CONSTRAINT `contactid_contactseeking_fk` FOREIGN KEY (`contact_id`) REFERENCES `my_contacts` (`contact_id`),
  CONSTRAINT `seeking_contactseeking_fk` FOREIGN KEY (`seeking_id`) REFERENCES `seeking` (`seeking_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;


CREATE TABLE `interests` (
  `interest_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `interest` varchar(50) DEFAULT NULL,
  PRIMARY KEY (`interest_id`)
) ENGINE=InnoDB AUTO_INCREMENT=25 DEFAULT CHARSET=latin1;


CREATE TABLE `job_current` (
  `contact_id` int(10) unsigned NOT NULL,
  `title` varchar(20) DEFAULT NULL,
  `salary` decimal(8,2) DEFAULT NULL,
  `start_date` date DEFAULT NULL,
  KEY `mycontacts_jobcurrent_fk` (`contact_id`),
  CONSTRAINT `mycontacts_jobcurrent_fk` FOREIGN KEY (`contact_id`) REFERENCES `my_contacts` (`contact_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;


CREATE TABLE `job_desired` (
  `contact_id` int(10) unsigned NOT NULL,
  `title` varchar(20) DEFAULT NULL,
  `salary_low` decimal(8,2) DEFAULT NULL,
  `salary_high` decimal(8,2) DEFAULT NULL,
  `available` date DEFAULT NULL,
  `years_exp` int(11) DEFAULT NULL,
  KEY `mycontacts_jobdesired_fk` (`contact_id`),
  CONSTRAINT `mycontacts_jobdesired_fk` FOREIGN KEY (`contact_id`) REFERENCES `my_contacts` (`contact_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;


CREATE TABLE `job_listings` (
  `job_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `title` varchar(25) DEFAULT NULL,
  `salary` decimal(8,2) DEFAULT NULL,
  `zip_code` char(5) DEFAULT NULL,
  `description` varchar(50) DEFAULT NULL,
  PRIMARY KEY (`job_id`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=latin1;


CREATE TABLE `my_contacts` (
  `contact_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `last_name` varchar(30) DEFAULT NULL,
  `first_name` varchar(20) DEFAULT NULL,
  `phone` char(10) DEFAULT NULL,
  `email` varchar(50) DEFAULT NULL,
  `gender` char(1) DEFAULT NULL,
  `birthday` date DEFAULT NULL,
  `prof_id` int(11) unsigned NOT NULL,
  `status_id` int(10) unsigned NOT NULL,
  `zip_code` char(5) DEFAULT NULL,
  PRIMARY KEY (`contact_id`),
  KEY `profession_mycontacts_fk` (`prof_id`),
  KEY `zipcode_mycontacts_fk` (`zip_code`),
  KEY `status_my_contacts_fk` (`status_id`),
  CONSTRAINT `profession_mycontacts_fk` FOREIGN KEY (`prof_id`) REFERENCES `profession` (`prof_id`),
  CONSTRAINT `status_my_contacts_fk` FOREIGN KEY (`status_id`) REFERENCES `status` (`status_id`),
  CONSTRAINT `zipcode_mycontacts_fk` FOREIGN KEY (`zip_code`) REFERENCES `zip_code` (`zip_code`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=latin1;


CREATE TABLE `profession` (
  `prof_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `profession` varchar(30) DEFAULT NULL,
  PRIMARY KEY (`prof_id`)
) ENGINE=InnoDB AUTO_INCREMENT=16 DEFAULT CHARSET=latin1;


CREATE TABLE `seeking` (
  `seeking_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `seeking` varchar(40) DEFAULT NULL,
  PRIMARY KEY (`seeking_id`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=latin1;


CREATE TABLE `status` (
  `status_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `status` varchar(30) DEFAULT NULL,
  PRIMARY KEY (`status_id`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=latin1;


CREATE TABLE `zip_code` (
  `zip_code` char(5) NOT NULL DEFAULT '',
  `city` varchar(20) DEFAULT NULL,
  `state` char(2) DEFAULT NULL,
  PRIMARY KEY (`zip_code`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

Answer 1

我发现我只需要两行就可以解决我的问题，我在顶部添加了一个 0，在底部添加了一个 1，这很好。抱歉浪费您的时间...

SET FOREIGN_KEY_CHECKS = 0;

SET FOREIGN_KEY_CHECKS = 1;