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我正在與MySQL的錯誤，我不明白爲什麼：＃1005 - 無法創建表 'feedback.answer'（錯誤：150）
SET @[email protected]@UNIQUE_CHECKS, UNIQUE_CHECKS=0;
SET @[email protected]@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0;
SET @[email protected]@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 [email protected]_SQL_MODE;
SET [email protected]_FOREIGN_KEY_CHECKS;
SET [email protected]_UNIQUE_CHECKS;
錯誤：
#1005 - Can't create table 'feedback.answer' (errno: 150)
我從這個創造我的表作爲模板：
我的思維過程添加answer_i d到答案表是我希望用戶能夠多次填寫相同的調查問卷。
爲什麼答案表會拋出錯誤？
編輯： 服務器版本：5.5.29-0ubuntu0.12.04.2 我導入此使用phpMyAdmin
也許這個問題是有幫助的：http://stackoverflow.com/questions/1457305/mysql-creating-tables-with-foreign-keys-giving-errno-150但我看不到你的任何要求失蹤了。 – Barmar
我在沒有創建反饋數據庫的情況下在sqlfiddle中寫道，它可以工作。我不確定這個問題會出現什麼問題。 –
我也嘗試在sqlfiddle中重新創建，並且不能。也許這與改變默認字符集或整理有關，因爲sqlfiddle不會讓我擺弄這些。 – Barmar