在 int 中输入小数时打印错误

Question

我正在编写一个名为 Revenue 的程序，我已经满足了除一个之外的所有要求。

  System.out.print("\t Enter quantity(ies): ");
    quantity = keyboard.nextInt();
    if (quantity < 0 || quantity > 150 || quantity == '.'){

        System.out.println();
        System.out.println("\t Invalid item price.");
        System.out.println("\t Please run the program again");
        System.out.println();
        System.out.println("Thank you for using \"Temple\" store");
        System.exit(0);

我们必须向用户询问他们购买的商品的数量，并且不能有小数点或“。”在里面。例如，当要求用户输入他们想要购买的物品数量时，他们将输入数字。如果他们输入一个带小数的数字，它应该打印出来

     Invalid item price.
     Please run the program again

Thank you for using "Temple" store

Answer 1

keyboard.nextInt() 不允许使用小数。如果您键入类似“3.14.15”的内容，它将引发异常。对于错误处理，您需要捕获此异常并打印出比默认情况下看到的堆栈跟踪更好的错误消息。

Answer 2

使用 try-catch 块来完成你想要的：

  System.out.print("\t Enter quantity(ies): ");
  try{
    quantity = keyboard.nextInt();
    if (quantity < 0 || quantity > 150)
     throw new IllegalArgumentException();
  } catch (IllegalArgumentException e) {
     System.out.println();
     System.out.println("\t Invalid item price.");
     System.out.println("\t Please run the program again");
     System.out.println();
     System.out.println("Thank you for using \"Temple\" store");
     System.exit(-1); //-1 signs an error to the application that launched the program
  }

Answer 3

使用下面的代码：

  String quantity = keyboard.next();  // use keyboard.next() instead of keyboard.nextInt() 
                 // because it will throw Exception if value is other 
                                       //than int   

   boolean flag = quantity.matches("[0-9]+");
    if (flag) {
        if (Integer.parseInt(quantity) < 0
                || Integer.parseInt(quantity) > 150) {
            flag = false;
        }
    }

    if (!flag) {
        System.out.println();
        System.out.println("\t Invalid item price.");
        System.out.println("\t Please run the program again");
        System.out.println();
        System.out.println("Thank you for using \"Temple\" store");
        System.exit(0);
    }