Question

我正在尝试扫描一个单词，然后按下输入后，将使用for循环将其放置在ArrayList的第n个位置。我在哪里放置（i）以使其有效？我不能把它放进去像topScan.next（i）;？

for ( int i = 0; i>topsNo; i++); {
    System.out.println("Enter next topping :");
    Scanner topScan = new Scanner(System.in);
    b.currentTops = topScan.next();
}

编辑：

我说不清楚因此我将添加其余的课程：

public static void main(String [] Args) {
PizzaBase a = new PizzaBase();
Pizza p = new Pizza();
PizzaToppings b = new PizzaToppings();


//SCAN FOR KEYWORD INGRIDIENTS TO SEE WHAT IS OK


Scanner s = new Scanner(System.in);
String base;
System.out.println("Enter base: ");

base = s.next();
a.setBase(base);

int topsNo = 0;

System.out.println("Enter number of desired toppings: ");
Scanner nt = new Scanner(System.in);

if(nt.hasNextInt()){
       topsNo = nt.nextInt();
    }else{
       System.out.println("Try again (re-enter number of toppings 1-9)");
    }

编辑：根据要求，PizzaToppings类：

public class PizzaToppings {


List<String> tops = new ArrayList<String>();
List<Double> prices = new ArrayList<Double>();
List<String> currentTops = new ArrayList<String>();
double topPrice;
public void pizzaTop() {

    currentTops.add("mushrooms");
    currentTops.add("cheese");
    currentTops.add("ham");
    currentTops.add("chicken");

    for(int i = 0; i<currentTops.size(); i++){

    if(currentTops.get(i).equalsIgnoreCase("cheese")){ 
           topPrice+=(1.0);
       } else if(currentTops.get(i).equalsIgnoreCase("sweetcorn")){
           topPrice+=(2.0);
       } else if(currentTops.get(i).equalsIgnoreCase("mushrooms")){
           topPrice+=(1.2);
       } else if(currentTops.get(i).equalsIgnoreCase("chicken")){
           topPrice+=(1.25);
       }
       else{
           System.out.println("Sorry but topping "+ currentTops.get(i)
                + " cannot be offered.");
           break;
       }
     }


}

Answer 1

我假设你试图遍历for-loop，直到用户输入了添加的浇头数量。这是一些代码来执行此操作。此外，每次使用时都不需要创建新的扫描仪，所以我改变了它。

    Scanner theKeyboard = new Scanner(System.in);
    String base;
    System.out.println("Enter base: ");

    base = theKeyboard.next();
    a.setBase(base);



    System.out.println("Enter number of desired toppings: ");
    int topsNo = theKeyboard.nextInt();

    //Alternately use a while loop here incrementing 'i' each time a topping is entered
    for (int i = 0; i < topsNo; i++)
    {
        System.out.println("Enter next topping :");
        b.currentTops = theKeyboard.next();
    }

扫描到ArrayList Java

1 个答案: