标签:mysql workbench mysql-workbench
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)
我看不到重复的约束.它在哪里?
解决方法:
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)
应该使用不同的标识符作为约束,如fk_restaurant,如:
CONSTRAINT `fk_restaurant`
FOREIGN KEY (`restaurant` )
REFERENCES `mydb`.`restaurants` (`id` )
ON DELETE NO ACTION
ON UPDATE NO ACTION)
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)
这是我看到的唯一三个,但可能还有其他我错过了.
标签:mysql,workbench,mysql-workbench 来源: https://codeday.me/bug/20191002/1843211.html
本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享; 2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关; 3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关; 4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除; 5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。