Question

该程序是费用跟踪程序，需要无错误。为了实现该目标，我需要从第11行开始重新启动所有操作。

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int Size;
    int order;
    System.out.println("Put in the amount of expenses you have");
    Size = sc.nextInt();
    System.out.println("put in all your expenses");
    int userInput[] = new int[Size];
    for (int i = 0; i < userInput.length; i++)
        userInput[i] = sc.nextInt();
    System.out
            .println("do you want it ascending or descending order. If you want it in ascending press 1 or if you want descending press 2");
    order = sc.nextInt();
    System.out.print("expenses not sorted : ");
    printExpenses(userInput);
    if (order == 1) {
        expensesAscending(userInput);
    } else if (order == 2) {
        expensedescending(userInput);
    }else if (order>2){
        //How do i make it so that if they press three or above the program restarts
    }
}

Answer 1

您可以使用永恒循环，直到用户输入有效的条目：

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    int Size;
    int order;
    int userInput[];
    do { // loop starts here
        System.out.println("Put in the amount of expenses you have");
        Size = sc.nextInt();
        System.out.println("put in all your expenses");
        userInput = new int[Size];
        for (int i = 0; i < userInput.length; i++)
            userInput[i] = sc.nextInt();
        System.out.println(
                "do you want it ascending or descending order. If you want it in ascending press 1 or if you want descending press 2");
        order = sc.nextInt();
    } while (order < 1 || order > 2); // if the input is not 1 or 2, it goes back
    System.out.print("expenses not sorted : ");
    printExpenses(userInput);
    if (order == 1) {
        expensesAscending(userInput);
    } else {
        expensedescending(userInput);
    }
}

但是，如果用户输入错误，我不建议您重新启动整个程序。最好再次询问他输入或他是否要重新启动整个程序。

Answer 2

在不向您提供代码的情况下，您需要的称为 while循环。 while循环将继续“执行操作”（在这种情况下，尝试使用户输入正确的输入），直到满足条件为止（在这种情况下，order的值为1或2）。

例如：

int order = null; 
while (order != 1 && order != 2){
    System.out.println("do you want it ascending or descending order."
        + "If you want it in ascending press 1 or if you want descending press 2");
    order = sc.nextInt(); 
}

但是，我也认为您可能需要重新考虑界面设计。错误和不正确的输入处理很重要，但是这使用户可以轻松地在第一次尝试时就正确地做事。例如，用户可能更容易记住如何通过输入“ a”代表升序，输入“ d”代表降序来获得所需的结果。

如果输入了无效的选项，我该如何循环代码

2 个答案: