MySQL Workbench 正向工程师错误 1215 : Cannot add foreign key constraint

Question

当我执行此脚本创建 2 个表时，其中 CUSTOMER 表中的 STORE 列引用 USERS 表中的 ID 列(两列均为 INT):

SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0;
SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0;
SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='TRADITIONAL,ALLOW_INVALID_DATES';

CREATE SCHEMA IF NOT EXISTS `part_finder` DEFAULT CHARACTER SET utf8 ;
USE `part_finder` ;

-- -----------------------------------------------------
-- Table `part_finder`.`users`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `part_finder`.`users` (
  `id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
  `account` MEDIUMINT(7) UNSIGNED NOT NULL COMMENT 'Account organisation user belongs to',
  `name` VARCHAR(32) NULL DEFAULT NULL,
  `passenc` VARCHAR(32) NULL DEFAULT NULL,
  `email` VARCHAR(55) NULL DEFAULT NULL,
  `rank` DECIMAL(1,0) NULL DEFAULT '0',
  `ip_reg` VARCHAR(15) NULL DEFAULT NULL,
  `ip_visit` VARCHAR(15) NULL DEFAULT NULL,
  `dtreg` INT(11) NOT NULL,
  `dtvisit` INT(11) NOT NULL,
  `visits` SMALLINT(5) UNSIGNED NULL DEFAULT '0',
  `pass` VARCHAR(25) NULL DEFAULT NULL,
  `make_filter_on` TINYINT(1) NULL DEFAULT FALSE,
  `brand_filter_on` TINYINT(1) NULL DEFAULT FALSE,
  PRIMARY KEY (`id`))
ENGINE = InnoDB
AUTO_INCREMENT = 3;


-- -----------------------------------------------------
-- Table `part_finder`.`customer`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `part_finder`.`customer` (
  `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
  `store` INT(10) NOT NULL,
  `name` VARCHAR(32) NULL,
  `address` VARCHAR(45) NULL,
  `address_2` VARCHAR(45) NULL,
  `city` VARCHAR(15) NULL,
  `state` TINYINT UNSIGNED NOT NULL,
  `zip` CHAR(5) NULL,
  `phone` VARCHAR(15) NULL,
  `website` VARCHAR(45) NULL,
  `email` VARCHAR(55) NULL,
  PRIMARY KEY (`id`),
  INDEX `fk_customer_users_idx` (`store` ASC),
  CONSTRAINT `fk_customer_users`
    FOREIGN KEY (`store`)
    REFERENCES `part_finder`.`users` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;


SET SQL_MODE=@OLD_SQL_MODE;
SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS;
SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS;

我收到这个错误:

ERROR: Error 1215: Cannot add foreign key constraint

-- -----------------------------------------------------
-- Table `part_finder`.`customer`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `part_finder`.`customer` (
  `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
  `store` INT(10) NOT NULL,
  `name` VARCHAR(32) NULL,
  `address` VARCHAR(45) NULL,
  `address_2` VARCHAR(45) NULL,
  `city` VARCHAR(15) NULL,
  `state` TINYINT UNSIGNED NOT NULL,
  `zip` CHAR(5) NULL,
  `phone` VARCHAR(15) NULL,
  `website` VARCHAR(45) NULL,
  `email` VARCHAR(55) NULL,
  PRIMARY KEY (`id`),
  INDEX `fk_customer_users_idx` (`store` ASC),
  CONSTRAINT `fk_customer_users`
    FOREIGN KEY (`store`)
    REFERENCES `part_finder`.`users` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB

SQL script execution finished: statements: 6 succeeded, 1 failed

Fetching back view definitions in final form.
Nothing to fetch

我使用的是 InnoDB，我已经检查了我的外键列的数据类型，并确保我使用的是引用表中的主键。

有什么想法吗？

Answer 1

也许父表中的列是 INT UNSIGNED？

它们在两个表中的数据类型必须完全相同。

Foreign Key Constraints

可能出现外键约束错误的原因:

1. 您正在尝试引用目标表中不存在的键。确保它是另一个表上的键(它可以是主键或唯一键)，

2. 列的类型不一样(除了列在引用表上可以是 nullable)。