使用扫描仪输入分配类

时间:2015-11-06 03:56:34

标签: java

    import java.util.Scanner;

public class Library {
    public static void main(String [] args) {
    /*
     * Patron class: name + 3 books
     * Book class: author + title
     */
    Book TwelveYears = new Book ("12 Years a Slave", "Solomon Northup");
    Book Diary = new Book ("The Diary of John Smith", "John Smith");
    Book Fahrenheit451 = new Book ("Fahrenheit 451", "Ray Bradbury");
    Book Gatsby = new Book ("The Great Gatsby", "Fitzgerald");
    Book Antigone = new Book ("12 Years a Slave", "Sophocles");




    System.out.println("Welcome to library account management! Here is a list of books to check out: " +TwelveYears.getBookTitle()+", "+Diary.getBookTitle()+", "+Fahrenheit451.getBookTitle()+", "+Gatsby.getBookTitle()+", "+Antigone.getBookTitle()+".");
    Scanner sc=new Scanner(System.in);
    System.out.println("Enter your name!");
    String enteredName = sc.nextLine();
    System.out.println("Enter your first book! Enter it exactly as seen above.");
    String book1 = sc.nextLine();
    System.out.println("Enter your second book!");
    String book2 = sc.nextLine();
    System.out.println("Enter your third book!");
    String book3 = sc.nextLine();
    Patron patron1 = new Patron(enteredName, book1, book2, book3);
    book1 = //line I need help on
    System.out.println("Here are the books you are checking out: " +book1+ ", " +book2+ ", " +book3+ ".");





    }

}

在这里,我正在尝试使用book1变量,并将其与正确的类成员进行协调。我认为解释我在这里尝试做什么的最好方法是举个例子:如果用户输入的book1是“12年奴隶”,那么book1的值将是“TwelveYears”,其中一个类定义在开头附近代码,所以在将来,我可以使用像book1.getAuthor这样的访问器/ mutator方法。

我只是在学习java,所以我没有办法正确解释这个问题,所以我道歉,但是我会非常感激任何帮助,如果我想要的话,我很乐意回答任何问题。在这里,但我希望我很清楚。

谢谢!

2 个答案:

答案 0 :(得分:2)

如果必须根据字符串决定要创建哪个对象,则只需对书名进行一系列检查,并确定要创建的对象。假设TwelveYearsGreatGatsby都从Book继承,那么您可以这样做:

Book unknownBook;
switch(bookTitleScanned)
{
    case("Twelve Years a Slave")
        unknownBook = new TwelveYears();
    case("Great Gatsby")
        unknownBook = new GreatGatsby();
    ...
}

由于polymorphism,您的Book引用可以指向TwelveYearsGreatGatsby

但是这可能是做你想做的更好的方法:
假设所有用于您目的的对象都非常相似(在您的示例中,bookTwelveYearsGreatGatsby),如果您只编写一个类可能会更容易,比如Book,让该类进行所有处理,而不是创建单独的类。

然后,您只需创建Book类的实例,将书名称作为属性分配给这些实例,并将其收集到某种Collections对象中,这将使您的代码更加灵活和可重用(嘿,如果你把它们放在Book中,你可以循环遍历ArrayList

具体而言,您的Book类可以有一个属性bookName,以及

等方法
public void getBookName()
{
    return bookName;
}

甚至:

public void handleParticularBook()
{
   if (bookName.equals("GreatGatsby"))
      System.out.println("My favorite book!");
   else if ...
}

答案 1 :(得分:1)

将所有图书放在Function GrafMet(Rng As Range) As String Dim celda As Range Dim currentcell As Range Dim nextcell As Range Set currentcell = ActiveCell For Each celda In Rng valor1 = celda.Value Set nextcell = currentcell.Offset(0, valor1) nextcell.Interior.ColorIndex = 3 Set currentcell = nextcell.Offset(0, valor1 * -1) Next celda ' this is just for returning something Dim ppp As String ppp = "OK" GrafMet = ppp End Function 中。如果与ArrayList<Book>

匹配,请重复ArrayList并使用book1.equals(a1.get(i).getBookName)
相关问题