我正在尝试编写一些代码来询问用户他们想要完成什么功能(例如查找一个点在图形上的哪个象限,等等)。但我也希望代码要求用户重新输入一个数字,如果它不是介于1和6之间(包括1和6)。我试图通过创建一个do-while循环来做到这一点,但由于某些原因,它甚至不会循环。任何让它变得更短/更整洁的技巧都会得到很好的评价。
下面是我遇到的问题:
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));
发布于 2019-04-16 09:04:36
只有当用户输入有效的数字而不是输入无效的数字时,才会出现循环循环。在做这类事情时,我倾向于使用集合,所以布尔逻辑很容易,即使在2天没有睡觉和10壶冰咖啡之后也是如此:)
Set<Integer> validInputs = new Set();
validInputs.add(1);
validInputs.add(2);
do {
// your stuff here
} while (!validInputs.contains(inputMethod));
如果您曾经远离基于控制台的用户交互,那么拥有一组有效的值在swing和JavaFX中都会有所帮助。
发布于 2019-04-16 09:26:08
添加whichMethod = scan.nextInt()作为默认值中的最后一行。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。
https://stackoverflow.com/questions/55699078
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