エラー：150/MySQL 'テーブルを作成できません'

Question

MySQL Workbenchのモデル作成者からデータベースモデルを作成しました。しかし、SQLファイルに変換しようとすると、エラーコードが出てきます。エラーはここ

Can't create table project2i2w . customer (error: 150 "Foreign key constraint is incorrectly formed") 0.032 sec.

を言う私のスクリプトです：

-- MySQL Workbench Forward Engineering 

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'; 

-- ----------------------------------------------------- 
-- Schema project2i2w 
-- ----------------------------------------------------- 
DROP SCHEMA IF EXISTS `project2i2w` ; 

-- ----------------------------------------------------- 
-- Schema project2i2w 
-- ----------------------------------------------------- 
CREATE SCHEMA IF NOT EXISTS `project2i2w` DEFAULT CHARACTER SET utf8 ; 
USE `project2i2w` ; 

-- ----------------------------------------------------- 
-- Table `project2i2w`.`services` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `project2i2w`.`services` (
    `service_id` INT NOT NULL AUTO_INCREMENT, 
    `service_name` VARCHAR(50) NOT NULL, 
    `service_desc` VARCHAR(250) NOT NULL, 
    `os_id` INT NOT NULL, 
    `customer_id` INT NOT NULL, 
    PRIMARY KEY (`service_id`)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `project2i2w`.`customer` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `project2i2w`.`customer` (
    `customer_id` INT NOT NULL AUTO_INCREMENT, 
    `first_name` VARCHAR(45) NOT NULL, 
    `last_name` VARCHAR(45) NOT NULL, 
    `city` VARCHAR(45) NOT NULL, 
    `street` VARCHAR(20) NOT NULL, 
    `province` VARCHAR(30) NOT NULL, 
    `postal` CHAR(7) NOT NULL, 
    PRIMARY KEY (`customer_id`), 
    CONSTRAINT `fk_customer_services` 
    FOREIGN KEY (`customer_id`) 
    REFERENCES `project2i2w`.`services` (`customer_id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `project2i2w`.`op_sys` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `project2i2w`.`op_sys` (
    `os_id` INT NOT NULL AUTO_INCREMENT, 
    `os_name` VARCHAR(25) NOT NULL, 
    `os_desc` VARCHAR(250) NULL, 
    `os_ver` VARCHAR(10) NOT NULL, 
    PRIMARY KEY (`os_id`), 
    CONSTRAINT `fk_op_sys_services1` 
    FOREIGN KEY (`os_id`) 
    REFERENCES `project2i2w`.`services` (`os_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; 

-- ----------------------------------------------------- 
-- Data for table `project2i2w`.`services` 
-- ----------------------------------------------------- 
START TRANSACTION; 
USE `project2i2w`; 
INSERT INTO `project2i2w`.`services` (`service_id`, `service_name`, `service_desc`, `os_id`, `customer_id`) VALUES (1, 'Virus Removal', 'Removing of viruses.', 1, 1); 
INSERT INTO `project2i2w`.`services` (`service_id`, `service_name`, `service_desc`, `os_id`, `customer_id`) VALUES (2, 'Tune-up', 'Tuning system performance.', 2, 2); 
INSERT INTO `project2i2w`.`services` (`service_id`, `service_name`, `service_desc`, `os_id`, `customer_id`) VALUES (3, 'Rooting', 'Accessing super-user status.', 3, 3); 
INSERT INTO `project2i2w`.`services` (`service_id`, `service_name`, `service_desc`, `os_id`, `customer_id`) VALUES (4, 'Restore', 'Restoring to stock factory condition.', 4, 4); 
INSERT INTO `project2i2w`.`services` (`service_id`, `service_name`, `service_desc`, `os_id`, `customer_id`) VALUES (5, 'Fix Permissions', 'Fixing file permissions.', 5, 5); 

COMMIT; 


-- ----------------------------------------------------- 
-- Data for table `project2i2w`.`customer` 
-- ----------------------------------------------------- 
START TRANSACTION; 
USE `project2i2w`; 
INSERT INTO `project2i2w`.`customer` (`customer_id`, `first_name`, `last_name`, `city`, `street`, `province`, `postal`) VALUES (1, 'Kathryn', 'Trollinger', 'Kanata', '2337 Merivale Road', 'ON', 'K2K 1L9'); 
INSERT INTO `project2i2w`.`customer` (`customer_id`, `first_name`, `last_name`, `city`, `street`, `province`, `postal`) VALUES (2, 'Lori ', 'Frederick', 'Toronto', '799 Yonge Street', 'ON', 'M4W 1J7'); 
INSERT INTO `project2i2w`.`customer` (`customer_id`, `first_name`, `last_name`, `city`, `street`, `province`, `postal`) VALUES (3, 'Maxine ', 'Coley', 'Ingersoll', '1305 Albert Street', 'ON', 'N5C 1S2'); 
INSERT INTO `project2i2w`.`customer` (`customer_id`, `first_name`, `last_name`, `city`, `street`, `province`, `postal`) VALUES (4, 'June ', 'Harrison', 'Windsor', '1089 Goyeau Ave', 'ON', 'N9A 1H9'); 
INSERT INTO `project2i2w`.`customer` (`customer_id`, `first_name`, `last_name`, `city`, `street`, `province`, `postal`) VALUES (5, 'Joan ', 'Dunn', 'Oshawa', '615 Toy Avenue', 'ON', 'L1H 7M3'); 

COMMIT; 


-- ----------------------------------------------------- 
-- Data for table `project2i2w`.`op_sys` 
-- ----------------------------------------------------- 
START TRANSACTION; 
USE `project2i2w`; 
INSERT INTO `project2i2w`.`op_sys` (`os_id`, `os_name`, `os_desc`, `os_ver`) VALUES (1, 'Windows', 'OS by Microsoft', '10'); 
INSERT INTO `project2i2w`.`op_sys` (`os_id`, `os_name`, `os_desc`, `os_ver`) VALUES (2, 'Mac OSX', 'OS by Apple', 'El Capitan'); 
INSERT INTO `project2i2w`.`op_sys` (`os_id`, `os_name`, `os_desc`, `os_ver`) VALUES (3, 'Android', 'OS by Google', 'Marshmellow'); 
INSERT INTO `project2i2w`.`op_sys` (`os_id`, `os_name`, `os_desc`, `os_ver`) VALUES (4, 'iOS', 'OS by Apple', '10'); 
INSERT INTO `project2i2w`.`op_sys` (`os_id`, `os_name`, `os_desc`, `os_ver`) VALUES (5, 'Linux', 'Open-source OS', 'Ubuntu'); 

COMMIT; 

誰かが私のためにこれを修正し、それが働いていない理由を教えてもらえば、私は非常にそれをapreciateだろう。

ニック

Answer 1

あなたはcustomer_idのために何らかの種類のキーが必要です。それはそのトリックを行います。

CREATE TABLE IF NOT EXISTS `project2i2w`.`services` (
`service_id` INT NOT NULL AUTO_INCREMENT, 
`service_name` VARCHAR(50) NOT NULL, 
`service_desc` VARCHAR(250) NOT NULL, 
`os_id` INT NOT NULL, 
`customer_id` INT NOT NULL, 
INDEX `customerid` (`customer_id`), 
PRIMARY KEY (`service_id`)) 
ENGINE = InnoDB; 

---

- 表project2i2w。 customer

---

CREATE TABLE IF NOT EXISTS `project2i2w`.`customer` (
`customer_id` INT NOT NULL AUTO_INCREMENT, 
`first_name` VARCHAR(45) NOT NULL, 
`last_name` VARCHAR(45) NOT NULL, 
`city` VARCHAR(45) NOT NULL, 
`street` VARCHAR(20) NOT NULL, 
`province` VARCHAR(30) NOT NULL, 
`postal` CHAR(7) NOT NULL, 
PRIMARY KEY (`customer_id`), 
CONSTRAINT `fk_customer_services` 
FOREIGN KEY (`customer_id`) 
REFERENCES `project2i2w`.`services` (`customer_id`) 
ON DELETE NO ACTION 
ON UPDATE NO ACTION) 
    ENGINE = InnoDB; 

この

Answer 2

はこの試してみてください。外部キーは、参照テーブルの主キーでなければなりません

CREATE TABLE IF NOT EXISTS `project2i2w`.`services` (
    `service_id` INT NOT NULL AUTO_INCREMENT, 
    `service_name` VARCHAR(50) NOT NULL, 
    `service_desc` VARCHAR(250) NOT NULL, 
    `os_id` INT NOT NULL, 
    PRIMARY KEY (`service_id`)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 

CREATE TABLE IF NOT EXISTS `project2i2w`.`customer` (
    `customer_id` INT NOT NULL AUTO_INCREMENT, 
    `services_id` INT NOT NULL, 
    `first_name` VARCHAR(45) NOT NULL, 
    `last_name` VARCHAR(45) NOT NULL, 
    `city` VARCHAR(45) NOT NULL, 
    `street` VARCHAR(20) NOT NULL, 
    `province` VARCHAR(30) NOT NULL, 
    `postal` CHAR(7) NOT NULL, 
    PRIMARY KEY (`customer_id`), 
    CONSTRAINT `fk_customer_services` 
    FOREIGN KEY (`services_id`) 
    REFERENCES `project2i2w`.`services` (`service_id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 

を。