#1005 -无法创建表“feedback.answer”(errno: 150)

Question

我在mysql中遇到了一个错误，我不明白为什么：

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 `feedback` DEFAULT CHARACTER SET utf8 COLLATE utf8_general_ci ;
USE `feedback` ;

-- -----------------------------------------------------
-- Table `feedback`.`application`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`application` (
  `application_id` INT NOT NULL AUTO_INCREMENT,
  `app_name` VARCHAR(45) NULL,
  PRIMARY KEY (`application_id`),
  UNIQUE INDEX `app_name_UNIQUE` (`app_name` ASC))
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `feedback`.`user`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`user` (
  `user_id` INT NOT NULL AUTO_INCREMENT,
  `firstname` VARCHAR(45) NOT NULL,
  `lastname` VARCHAR(45) NULL,
  `email` VARCHAR(45) NOT NULL,
  `customer_length` VARCHAR(45) NULL,
  PRIMARY KEY (`user_id`))
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `feedback`.`users_has_application`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`users_has_application` (
  `user_id` INT NOT NULL,
  `application_id` INT NOT NULL,
  PRIMARY KEY (`user_id`, `application_id`),
  INDEX `fk_users_has_application_application1_idx` (`application_id` ASC),
  INDEX `fk_users_has_application_users_idx` (`user_id` ASC),
  CONSTRAINT `fk_users_has_application_users`
    FOREIGN KEY (`user_id`)
    REFERENCES `feedback`.`user` (`user_id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_users_has_application_application1`
    FOREIGN KEY (`application_id`)
    REFERENCES `feedback`.`application` (`application_id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `feedback`.`survey`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`survey` (
  `survey_id` INT NOT NULL AUTO_INCREMENT,
  `name` VARCHAR(45) NULL,
  `description` VARCHAR(255) NULL,
  `is_active` TINYINT(1) NULL,
  PRIMARY KEY (`survey_id`))
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `feedback`.`question`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`question` (
  `question_id` INT NOT NULL,
  `question_text` VARCHAR(255) NULL,
  PRIMARY KEY (`question_id`))
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `feedback`.`option`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`option` (
  `option_id` INT NOT NULL AUTO_INCREMENT,
  `question_id` INT NOT NULL,
  `option_number` INT NOT NULL,
  `option_text` TEXT NULL,
  INDEX `fk_option_question1_idx` (`question_id` ASC),
  PRIMARY KEY (`option_id`),
  UNIQUE INDEX `uk_question_option_number_key` (`question_id` ASC, `option_number` ASC),
  CONSTRAINT `fk_option_question1`
    FOREIGN KEY (`question_id`)
    REFERENCES `feedback`.`question` (`question_id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `feedback`.`answer`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`answer` (
  `answer_id` INT NOT NULL AUTO_INCREMENT,
  `user_id` INT NOT NULL,
  `option_id` INT NOT NULL,
  `date_submitted` DATETIME NOT NULL,
  PRIMARY KEY (`answer_id`),
  INDEX `fk_answer_user1_idx` (`user_id` ASC),
  INDEX `fk_answer_option1_idx` (`option_id` ASC),
  CONSTRAINT `fk_answer_user1`
    FOREIGN KEY (`user_id`)
    REFERENCES `feedback`.`user` (`user_id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_answer_option1`
    FOREIGN KEY (`option_id`)
    REFERENCES `feedback`.`option` (`option_id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `feedback`.`survey_has_question`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `feedback`.`survey_has_question` (
  `survey_id` INT NOT NULL,
  `question_id` INT NOT NULL,
  `question_number` INT NULL,
  PRIMARY KEY (`survey_id`, `question_id`),
  INDEX `fk_survey_has_question_question1_idx` (`question_id` ASC),
  INDEX `fk_survey_has_question_survey1_idx` (`survey_id` ASC),
  UNIQUE INDEX `unique_order_key` (`survey_id` ASC, `question_number` ASC),
  CONSTRAINT `fk_survey_has_question_survey1`
    FOREIGN KEY (`survey_id`)
    REFERENCES `feedback`.`survey` (`survey_id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_survey_has_question_question1`
    FOREIGN KEY (`question_id`)
    REFERENCES `feedback`.`question` (`question_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;

错误：

#1005 - Can't create table 'feedback.answer' (errno: 150)

我将根据这个模板创建我的表：

https://dba.stackexchange.com/questions/16002/survey-database-design-associate-an-answer-to-a-user/16047#16047

我在回答表中添加answer_id的想法是，我希望用户能够多次填写相同的调查。

为什么答案表会抛出错误？

编辑：服务器版本: 5.5.29-0ubuntu0.12.04.2我正在使用phpmyadmin导入它

Answer 1

您的代码在MYSQL服务器5.1上运行，没有错误。

errno: 150的一个常见原因是当您创建一个引用还不存在的PK的FK约束时。在创建“答案”表之前，请确保首先创建“用户”和“选项”表。

为了帮助调试，您可以每次删除一个FK约束，以查看哪个约束触发了问题。

如果您按照所示的顺序执行代码，我将不会看到任何FK问题。

Answer 2

尝尝这个

 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';

-- -----------------------------------------------------
-- Table `feedback`.`application`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `application` (
`application_id` INT NOT NULL AUTO_INCREMENT,
`app_name` VARCHAR(45) NULL,
PRIMARY KEY (`application_id`),
UNIQUE INDEX `app_name_UNIQUE` (`app_name` ASC))
;

你的工作代码在小提琴里