Java简单的PizzaOrder

Question

我被困住了，这里有一个错误。我正在制作一个简单的披萨订购程序。该程序只是输入披萨的大小和配料，然后它会显示价格和所有内容。我想做的是，当我在顶部选择中按 Enter 键而不输入任何内容时，它应该返回零。这是代码

    <pre>
      public static void main(String[] args)
    {
      Scanner keyboard = new Scanner(System.in);
      System.out.println("Enter The Size of Pizza you " + "want: (S/M/L/C)");
    String option = keyboard.nextLine().trim()
    .toUpperCase();

    if(!option.isEmpty())
        option = option.substring(0, 1);


    double pizzaPrice;
    double pizzaSize;


    if(option.equals(PIZZA_MEDIUM))
    {
        pizzaPrice = PRICE_MEDIUM;
        pizzaSize = MEDIUM_DIAMETER;
    }
    else if (option.equals(PIZZA_LARGE))
    {
        pizzaPrice = PRICE_LARGE;
        pizzaSize = LARGE_DIAMETER;
    }
    else if (option.equals(PIZZA_COLOSSAL))
    {
        pizzaPrice = PRICE_COLOSSAL;
        pizzaSize = COLOSSAL_DIAMETER;
    }
    else
    {
        option = PIZZA_SMALL;           
        pizzaPrice = PRICE_SMALL;
        pizzaSize = SMALL_DIAMETER;
    }


    System.out.println("Pizza Size: " + option);
    System.out.println("Enter The Number of Toppings" +
    "you want:(0-8)");
    int pizzaTopping=0;


    if(keyboard.hasNextInt())
        pizzaTopping = keyboard.nextInt();      


    if(pizzaTopping < MIN_TOPPINGS)
    {
        pizzaTopping = MIN_TOPPINGS;
    }
    else if(pizzaTopping > MAX_TOPPINGS)
    {
        pizzaTopping = MAX_TOPPINGS;
    }
    else    
    {
        pizzaTopping = pizzaTopping;
    }

我已经使用披萨尺寸的字符串来完成它

   if(!option.isEmpty())

我认为它不适用于整数。我该怎么办？

谢谢

Answer 1

如果没有值(或任何其他无效值)，请尝试以下代码将其设置为零 (0):

String toppingsString = keyboard.nextLine().trim();
int toppings = 0;
try
{
     toppings = Integer.parseInt(toppingsString);
}
catch(NumberFormatException nfe)
{
    toppings = 0;
}