java - 对象库的方法存在问题

Question

我正在尝试编写一个程序来创建一个包含不同书籍的图书馆，我为图书馆中的每个项目设置了多个副本，每次我 checkout 一个项目时，我希望它仅从该项目中扣除一份副本我检查了特定的对象，但它从所有对象中获取了一个副本。不知道如何解决这个问题。

public abstract class Item{
  private int identify;
  private String title;
  private int copies;

  public Item(){
    identify=0;
    title="N/A";
    copies=0;
  }
  public Item(int id, int copy, String t){
    identify=id;
    copies=copy;
    title=t;
  }
  public void setIdentificationNumber(int id){
    identify = id;
  }
  public void setTitle(String t){
    title=t;
  }
  public void setNumberCopies(int num){
    copies=num;
  }
  public int getIdentificationNumber(){
    return identify;
  }
  public String getTitle(){
    return title;
  }
  public int getNumberCopies(){
    return copies;
  }
  public void checkOut(){
    if(copies>0){
      copies-=1;
      System.out.println("You have checked out "+title+". Thank You");
    }
    else{
      System.out.println("All copies of "+title+" are checked out!");
    }
  }
  public void checkIn(){
    copies+=1;
  }
}

问题也可能出在我的客户端方法中，我已在下面发布了该方法的代码。

import java.util.Arrays;
import java.util.Scanner;
public class Library{
  static String title;
  static String author;
  static int id;
  static int copies;
  static String date;
  static Book[] database = new Book[100];
  static int count=0;

  public static void main(String[] args){
    int i;
    Scanner s = new Scanner(System.in);
    do{
      addBook();
      System.out.println("would you like to add another book?");
      i=s.nextInt();
    }while(i == 0);
    database[0].viewDetails();
    database[1].viewDetails();
    checkingOut();
    database[0].viewDetails();
    database[1].viewDetails();
  }
  public static void addBook(){
    Scanner s = new Scanner(System.in);
    System.out.println("Enter the title of the book you want to add to the collection");
    title=s.nextLine();
    System.out.println("Enter the author of the book you want to add to the collection");
    author=s.nextLine();
    System.out.println("Enter the publishing date of the book you want to add to the collection");
    date=s.nextLine();
    System.out.println("Enter the ID number of the book you want to add to the collection");
    id=s.nextInt();
    System.out.println("Enter the the number of copies that will be added into the collection");
    copies=s.nextInt();

    Book Book1 = new Book(date, author, copies, id, title);
    database[count] = Book1;
    count++;
  }
  public static void checkingOut(){
    boolean found=false;
    int idSearch;
    int i=0;
    Scanner s = new Scanner(System.in);
    System.out.println("Enter the ID number of the book you want to check out");
    idSearch=s.nextInt();
    while(i<database.length && found!=true){
      if(database[i].getIdentificationNumber() == idSearch){
        found = true;
      }
      i++;
    }
    if(found==true){
      database[i].checkOut();
      System.out.println("There are "+database[i].getNumberCopies()+" copies left");
    }
    else{System.out.println("There is no book with that ID number!");}
  }
}

在我的 addBook 方法中，每次制作一本新书时，我都会创建一个名为 book1 的新对象，因此我认为每次添加一本书时，它可能会更改所有书籍对象。我不太确定编写该方法的更好方法。

这也是我的书籍方法

public class Book extends WrittenItem{
  public Book(){
    super();
  }
  public Book(String date, String a, int copy, int id, String t){
    super(a, date, copy, id, t);

  }
  public void viewDetails(){
    System.out.println("ID: "+getIdentificationNumber()+"\nTitle: "+getTitle()+"\nAuthor: "+getAuthor()+" Date written: "+getDate()+"\nCopies available: "+getNumberCopies());
  }
}

任何帮助将不胜感激。

Answer 1

我无法从您的代码中 100% 判断，但我假设 Book 扩展了 Item。在这种情况下，请为 Book 构造函数尝试类似的操作

public class Book extends Item {
    String author;
    String date;

    public Book(String date, String author, int copies, int id, String title) {
        super(id, copies, title); // Item constructor matches this super() call
                                  // public Item(int id, int copy, String t)
        this.author = author;
        this.date = date;
    }
}

您希望Book 在Item 上具有相同的副本。因此，当您 checkOut() 时，Item 编号等于您从 Book 构造函数输入的值。如果您不将 super() 放入构造函数中，您的 Item copies 将保持为 0，并且您将始终获得 System .out.println(““+title+”的所有副本均已 check out !”); 因为 copies 永远不会 > 0。