Question

当我执行此脚本来创建2个表时，CUSTOMER表中的STORE列引用USERS表中的ID列（两列都是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;

我收到此错误：

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InnoDB，我检查了外键列的数据类型，并确保我在引用的表格中使用主键。

有什么想法吗？

Answer 1

父表中的列可能是INT UNSIGNED？

两个表中的数据类型必须完全相同。

Foregin Key Constaints

您可能会遇到外键约束错误的原因：

您正尝试在目标表上引用不存在的键。确保它是另一张桌子上的钥匙（可以是primary or unique key），
列的类型不相同（例外是列在参考表上可以是nullable）。

Answer 2

列定义必须在父表和子表中都匹配。

家长

CREATE TABLE IF NOT EXISTS `part_finder`.`users` (

  `id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT, -- <----- this is unsigned

  ...
  PRIMARY KEY (`id`))

儿童

CREATE TABLE IF NOT EXISTS `part_finder`.`customer` (
...

`store` INT(10) NOT NULL, -- <----------- this is signed

...
  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)

更改：

`store` INT(10) NOT NULL, -- <----------- this is signed

致：

`store` INT(10) UNSIGNED NOT NULL, -- <---- now this too is unsigned

请参阅：MySQL: Foregin Key Constaints

MySQL Workbench正向工程师错误1215：无法添加外键约束

2 个答案: