Kann Tabelle in Workbench nicht erstellen, 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) 

Ich sehe keine doppelten Einschränkungen. Wo ist es?

Answer 1

Dies ist wahrscheinlich, weil Sie mindestens eine Einschränkung mit der gleichen Kennung wie eine Spalte mit dem Namen haben:

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

: wie fk_restaurant wie in eine andere Kennung für die Einschränkung

/* 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) 

Sollte verwenden Und das gleiche in der food Tabelle:

/* 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) 

Das sind die einzigen drei, die ich sehe, aber es könnte andere geben, die ich vermisst habe.