Question

我正在尝试编写一些代码，询问用户他们想要完成什么功能（例如，找到一个点在图形上的哪个象限，等等）。但是我还希望代码要求用户重新输入一个数字（如果该数字不在1到6之间）。我试图通过创建一个do-while循环来做到这一点，但是由于某种原因，它甚至不会循环。使其更短/更清洁的任何技巧都受到赞赏。

以下是我遇到的问题：

  int whichMethod;
   do{ 
     whichMethod = scan.nextInt();
     switch(whichMethod){
       case 1:
         System.out.println("Enter x and y values:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point p = new Point(x, y);
         System.out.println("Quadrant:"+ p.quadrant());
         break;
       case 2:
         System.out.println("Enter x and y values:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case2p = new Point(x, y);
         case2p.flip();
         System.out.println("Flipped Coordinates" + case2p);
         break;
       case 3:
         System.out.println("Enter x and y values:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case3p = new Point(x, y);

         System.out.println("Enter x and y values for the 2nd Point: ");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case3p2 = new Point(x, y);
         System.out.println("Manhattan Distance:"+ 
         case3p.manhattanDistance(case3p2));
         break;
       case 4:
         System.out.println("Enter x and y values:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case4p = new Point(x, y);

         System.out.println("Enter x and y values for the 2nd Point: ");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case4p2 = new Point(x, y);
         System.out.println("Are they Vertical?: " + case4p.isVertical(case4p2));
         break;
       case 5:
         System.out.println("Enter x and y values:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case5p = new Point(x, y);

         System.out.println("Enter x and y values for the 2nd Point:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case5p2 = new Point(x, y);
         System.out.println("Slope is: " + case5p.slope(case5p2));
         break;
       case 6:
         System.out.println("Enter x and y values:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case6p = new Point(x, y);

         System.out.println("Enter x and y values for the 2nd Point:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case6p2 = new Point(x, y);

         System.out.println("Enter x and y values for the 3rd Point:");
         x = scan.nextDouble();
         y = scan.nextDouble();
         Point case6p3 = new Point(x, y);
         System.out.println("Are they Collinear?: "+ case6p.isCollinear(case6p2, case6p3));
         break;
       default:
         System.out.println("This isn't one of the methods available.");
         System.out.println("Please enter a number between 1 and 6");      
      } 

} while((whichMethod >= 1) && (whichMethod <= 6));

Answer 1

仅当用户输入有效号码而不是用户输入无效号码时，您才可以循环循环。在做这种事情时，我倾向于使用集合，这样布尔逻辑很容易，即使在两天不睡觉和十壶冷咖啡的情况下：）

Set<Integer> validInputs = new Set();
validInputs.add(1);
validInputs.add(2);


do {
// your stuff here

} while (!validInputs.contains(inputMethod));

如果您远离基于控制台的用户交互，那么拥有一组有效值将有助于swing和JavaFX。

Answer 2

在默认值的最后一行添加whichMethod = scan.nextInt（）。如果用户输入了错误的号码，这将允许用户输入有效的号码。

在switch-case语句之后，do-while循环不会循环

2 个答案: