无法在Workbench中创建表,错误121

时间:2022-02-19 21:11:19
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)

I see no duplicate constraints. Where is it?

我看到没有重复的约束。它在哪里?

1 个解决方案

#1


24  

This is likely because you have named at least one constraint with the same identifier as a column:

这可能是因为您已命名至少一个具有与列相同的标识符的约束:

/* 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)

Should use a different identifier for the constraint like fk_restaurant as in :

应该像fk_restaurant一​​样使用不同的约束标识符,如:

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

And same thing in the food table:

在食物表中同样的事情:

  /* 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)

Those are the only three I see, but there could be others I missed.

这是我看到的唯一三个,但可能还有其他我错过了。

#1


24  

This is likely because you have named at least one constraint with the same identifier as a column:

这可能是因为您已命名至少一个具有与列相同的标识符的约束:

/* 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)

Should use a different identifier for the constraint like fk_restaurant as in :

应该像fk_restaurant一​​样使用不同的约束标识符,如:

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

And same thing in the food table:

在食物表中同样的事情:

  /* 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)

Those are the only three I see, but there could be others I missed.

这是我看到的唯一三个,但可能还有其他我错过了。