Homework 5

第五次作业

a. 阅读 Asg_RH 文档，按用例构建领域模型。

b. 数据库建模(E-R 模型)

Mysql 物理数据库的脚本

-- MySQL Script generated by MySQL Workbench
-- Sun Apr 29 18:03:54 2018
-- Model: New Model    Version: 1.0
-- MySQL Workbench Forward Engineering

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';

-- -----------------------------------------------------
-- Schema mydb
-- -----------------------------------------------------

-- -----------------------------------------------------
-- Schema mydb
-- -----------------------------------------------------
CREATE SCHEMA IF NOT EXISTS `mydb` DEFAULT CHARACTER SET utf8 ;
USE `mydb` ;

-- -----------------------------------------------------
-- Table `mydb`.`Customer`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Customer` (
  `id` INT NOT NULL,
  `name` VARCHAR(45) NULL,
  `e-mail` VARCHAR(45) NULL,
  PRIMARY KEY (`id`))
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `mydb`.`Hotel`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Hotel` (
  `id` INT NOT NULL,
  `name` VARCHAR(45) NULL,
  `location` VARCHAR(45) NULL,
  `star-rating` INT NULL,
  PRIMARY KEY (`id`))
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `mydb`.`Reservation`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Reservation` (
  `id` INT NOT NULL,
  `checkInDate` DATETIME NULL,
  `checkOutDate` DATETIME NULL,
  `roomInfo` VARCHAR(50) NULL,
  `expense` DOUBLE NULL,
  `Customer_id` INT NOT NULL,
  `Hotel_id` INT NOT NULL,
  PRIMARY KEY (`id`, `Customer_id`, `Hotel_id`),
  INDEX `fk_Reservation_Customer1_idx` (`Customer_id` ASC),
  INDEX `fk_Reservation_Hotel1_idx` (`Hotel_id` ASC),
  CONSTRAINT `fk_Reservation_Customer1`
    FOREIGN KEY (`Customer_id`)
    REFERENCES `mydb`.`Customer` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_Reservation_Hotel1`
    FOREIGN KEY (`Hotel_id`)
    REFERENCES `mydb`.`Hotel` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;


-- -----------------------------------------------------
-- Table `mydb`.`Room`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Room` (
  `id` INT NOT NULL,
  `type` VARCHAR(45) NULL,
  `pricePerDay` DOUBLE NULL,
  `isReserved` TINYINT NULL,
  `Reservation_id` INT NOT NULL,
  `Reservation_Customer_id` INT NOT NULL,
  `Hotel_id` INT NOT NULL,
  PRIMARY KEY (`id`, `Reservation_id`, `Reservation_Customer_id`, `Hotel_id`),
  INDEX `fk_Room_Reservation1_idx` (`Reservation_id` ASC, `Reservation_Customer_id` ASC),
  INDEX `fk_Room_Hotel1_idx` (`Hotel_id` ASC),
  CONSTRAINT `fk_Room_Reservation1`
    FOREIGN KEY (`Reservation_id` , `Reservation_Customer_id`)
    REFERENCES `mydb`.`Reservation` (`id` , `Customer_id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_Room_Hotel1`
    FOREIGN KEY (`Hotel_id`)
    REFERENCES `mydb`.`Hotel` (`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;

简单叙说 数据库逻辑模型 与 领域模型 的异同

另外数据库逻辑模型也更具体，需要给出字段类型

• 同：
  都是在定义领域概念，并且建立关系

上一篇    

原文:大专栏  Homework 5