Workbenchでテーブルを作成できません、errno 121

Question

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 `mydb` DEFAULT CHARACTER SET latin1 COLLATE latin1_swedish_ci ; 
USE `mydb` ; 

-- ----------------------------------------------------- 
-- Table `mydb`.`restaurants` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`restaurants` (
    `id` INT NOT NULL AUTO_INCREMENT , 
    `name` VARCHAR(128) NOT NULL , 
    `description` VARCHAR(1024) NOT NULL , 
    `address` VARCHAR(1024) NOT NULL , 
    `phone` VARCHAR(16) NOT NULL , 
    `url` VARCHAR(128) NOT NULL , 
    `min_order` INT NOT NULL , 
    `food_types` SET('pizza', 'sushi', 'osetian_pie') NOT NULL , 
    PRIMARY KEY (`id`) , 
    UNIQUE INDEX `name_UNIQUE` (`name` ASC) , 
    UNIQUE INDEX `id_UNIQUE` (`id` ASC)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `mydb`.`regions` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`regions` (
    `id` INT NOT NULL AUTO_INCREMENT , 
    `restaurant` INT NOT NULL , 
    `name` VARCHAR(128) NOT NULL , 
    PRIMARY KEY (`id`) , 
    INDEX `restaurant_idx` (`restaurant` ASC) , 
    UNIQUE INDEX `id_UNIQUE` (`id` ASC) , 
    CONSTRAINT `restaurant` 
    FOREIGN KEY (`restaurant`) 
    REFERENCES `mydb`.`restaurants` (`id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `mydb`.`food` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`food` (
    `id` INT NOT NULL , 
    `type` ENUM('pizza', 'sushi', 'osetian_pie') NOT NULL , 
    `name` VARCHAR(45) NOT NULL , 
    `ingredients` VARCHAR(256) NULL , 
    `image` VARCHAR(256) NOT NULL , 
    PRIMARY KEY (`id`) , 
    UNIQUE INDEX `id_UNIQUE` (`id` ASC)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `mydb`.`food_variant` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`food_variant` (
    `id` INT NOT NULL AUTO_INCREMENT , 
    `size` VARCHAR(16) NOT NULL , 
    `weight` VARCHAR(16) NOT NULL , 
    `price` INT NOT NULL , 
    `food` INT NOT NULL , 
    `restaurant` INT NOT NULL , 
    PRIMARY KEY (`id`) , 
    UNIQUE INDEX `id_UNIQUE` (`id` ASC) , 
    INDEX `food_idx` (`food` ASC) , 
    INDEX `restaurant_idx` (`restaurant` ASC) , 
    CONSTRAINT `food` 
    FOREIGN KEY (`food`) 
    REFERENCES `mydb`.`food` (`id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION, 
    CONSTRAINT `restaurant` 
    FOREIGN KEY (`restaurant`) 
    REFERENCES `mydb`.`restaurants` (`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; 

Error is: 
    Executing SQL script in server 
    ERROR: Error 1005: Can't create table 'mydb.food_variant' (errno: 121) 

重複する制約はありません。それはどこにある？

Answer 1

あなたはカラムと同じ識別子を持つ少なくとも一つの制約を命名しているので、これはそうである：

CONSTRAINT `fk_restaurant` 
    FOREIGN KEY (`restaurant`) 
    REFERENCES `mydb`.`restaurants` (`id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 

：

/* You already have a column named `restaurant` in this table, 
    but are naming the FK CONSTRAINT `restaurant` also... */ 
CONSTRAINT `restaurant` 
    FOREIGN KEY (`restaurant`) 
    REFERENCES `mydb`.`restaurants` (`id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 

は、のようにfk_restaurantのような制約のために異なる識別子を使用すべきfoodテーブルの同じもの：

/* Name it fk_food */ 
    CONSTRAINT `fk_food` 
    FOREIGN KEY (`food`) 
    REFERENCES `mydb`.`food` (`id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION, 
    /* Name it fk_restaurant */ 
    CONSTRAINT `fk_restaurant` 
    FOREIGN KEY (`restaurant`) 
    REFERENCES `mydb`.`restaurants` (`id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 

これは私が見る唯一の3人ですが、私が見逃した人がいるかもしれません。