Question

所以我已经在这段代码上工作了一段时间了，这让我有些失落。请记住，我是Java的新手，所以我对Java的使用有点慢。我创建了一些形状类，这些类实现了与getArea，getPerimeter和getDescription方法的接口。有多种形状，但这并不是问题所在。当我尝试实现开关盒以允许用户选择他或她想要添加的形状时，问题就来了。由于形状数组将允许我尝试添加的形状的地址，我收到相同的消息的次数很多。我意识到我犯的错误很可能是初学者，非常感谢您的帮助，谢谢。另外，如果您能给我一个关于如何按形状的面积对形状进行分类的线索，将不胜感激。

public class ShapeApp2 {

/**
 * @param args the command line arguments
 */

public static void main(String[] args) {
     Shape[] test = new Shape[10];
     System.out.println("Choose a shape or type stop to break away?");
     Scanner sc = new Scanner(System.in);
     String Shape = sc.nextLine();

     for (int i=0; i<test.length; i++) {

     switch (Shape) {
         case "Rectangle":
             System.out.println("You have chosen a Rectangle");
             test[i] = new Rectangle();
             System.out.println("Enter another one now");
             break;
         case "Square":
             System.out.println("You have chosen a Square");
             test[i] = new Square();
             System.out.println("Enter another one now");
             break;         
         case "Equilateral Triangle":
             System.out.println("You have chosen an Equilateral Triangle");
             test[i] = new Equilateral_Triangle();
             System.out.println("Enter another one now");
             break;
         case "Right Triangle":
             System.out.println("You have chosen a Right Triangle");
             test[i] = new Right_Triangle();
             System.out.println("Enter another one now");
             break;
         case "Isosceles Triangle":
             System.out.println("You have chosen an Isosceles Triangle");
             test[i] = new Isosceles_Triangle();
             System.out.println("Enter another one now");
             break;
         case "Scalene Triangle":
             System.out.println("You have chosen a Scalene Triangle");
             test[i] = new Scalene_Triangle();
             System.out.println("Enter another one now");
             break;
         case "Stop":
             break;
      }
         System.out.println(test[i]);
     }

}

}

这里还有一些用于上下文的Shape类。

package shapeapp2;

/**
 *
 * @author my-pc
 */
public class Rectangle implements Shape {
    private double length;
    private double width;
    private String shapeName;
    public Rectangle(){
        length = 4.0;
        width = 5.0;
        shapeName = "Rectangle";
    }
     public double getArea(){
         double Area;
         Area = length * width;
         return Area;
     }
     public double getPerimeter() {
         double Perimeter;
         Perimeter = (2*length) + (2*width);
         return Perimeter;
     }
     public String getDescription() {
         return shapeName;
     }
}



 package shapeapp2;

/**
 *
 * @author my-pc
 */
public class Square implements Shape {
    private double length;
    private double width;
    private String shapeName;
    public Square(){
        length = 8.0;
        width = 8.0;
        shapeName = "Square";
    }
     public double getArea(){
         double Area;
         Area = length * width;
         return Area;
     }
     public double getPerimeter() {
         double Perimeter;
         Perimeter = (2*length) + (2*width);
         return Perimeter;
     }
     public String getDescription() {
         return shapeName;
     }
}

Answer 1

在循环之前，您仅从用户读取一次形状。您想在循环中阅读它。那是唯一有意义的事情。

String Shape = sc.nextLine();
for (int i=0; i<test.length; i++) {

应该是

for (int i=0; i<test.length; i++) {
    String Shape = sc.nextLine();

此外，您应该重命名该变量。 Shape看起来像一个类名。

如何在Java中更正此切换情况？

1 个答案: