Java 输入不起作用（初学者）

由于某种原因，我的代码不接受最后一行的输入“您想订购什么：”

谁能告诉我我的错误是什么？它编译正确，一切顺利。我只是一个初学者，所以请告诉我基本术语。

import java.util.Scanner;
import java.util.*;

class RestaurantMain {
    public static void main(String[] args)
    {

        //Create an array list
        ArrayList menu = new ArrayList();

        //Variables//
        int choice;
        int customerChoice;
        boolean trueFalse;
        int restart = 0;
        String choice2;
        String addItems = "";
        int menuCount = 0;
        int indexCount = 0;
        String item = "";

        //Import input device
        Scanner in = new Scanner(System.in);

        ArrayList theMenu = new ArrayList();

        System.out.println("Welcome to the Cooper's restaurant system!");
        System.out.println("How can I help?");
        System.out.println("");
        System.out.println("1. Customer System");
        System.out.println("2. Management System");
        System.out.println("");
        System.out.println("");
        System.out.print("Which option do you choose: ");
        choice = in.nextInt();

            if (choice == 1) {
                System.out.println("Our menu's are as follows:");
                System.out.println("");
                System.out.println("1. Drinks");
                System.out.println("2. Starters");
                System.out.println("3. Mains");
                System.out.println("4. Desserts");
                System.out.println("");
                System.out.println("Please note - You MUST order 5 items.");
                System.out.println("");
                System.out.print("What menu would you like to follow? ");
                customerChoice = in.nextInt();

                    if (customerChoice == 1) {
                        System.out.println("Drinks Menu");
                            System.out.println("Would you like to order? ");
                            choice2 = in.nextLine();
                                if (choice2 == "yes") {
                                    System.out.println("Please enter the amount of items you want to order: ");
                                    while (indexCount <= menuCount);
                                        System.out.println("Please enter your item: ");
                                        item = in.nextLine(); {
                                        theMenu.add(item);
                                    }
                                    }

                    }
                    if (customerChoice == 2) {
                        System.out.println("Starters Menu");
                    }
                    if (customerChoice == 3) {
                        System.out.println("Mains menu");
                    }
                    if (customerChoice == 4) {
                        System.out.println("Desserts Menu");
                    }

你需要打电话in.nextLine()就在您拨打电话的线路之后in.nextInt()原因是仅询问下一个整数不会消耗输入中的整行，因此您需要通过调用向前跳到输入中的下一个换行符in.nextLine()

customerChoice = in.nextInt();
in.nextLine();

每次在调用不消耗整行的方法后需要获取新行时，几乎都必须执行此操作。考虑使用BufferedReader对象代替！

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int integer = Integer.parseInt(reader.readLine());

这将引发相同的错误Scanner.nextInt()如果输入无法解析为整数，则执行此操作。

关于您对错误的评论，有一个：

while (indexCount <= menuCount);
System.out.println("Please enter your item: ");
item = in.nextLine(); {
 theMenu.add(item);
}

}

应该像下面这样：

while(indexCount <= menuCount){
    System.out.println("Please enter your item: ");
    item = in.nextLine();
    theMenu.add(item);
}

另外，这并不是绝对必要的，但我建议您在实例化列表时声明 ArrayList 的泛型类型，以便进一步调用 theMenu.get() 不需要转换为字符串。

ArrayList<String> theMenu = new ArrayList<String>();

比较字符串时，请确保使用str.equals("string to compare with")方法，而不是等于运算符 (==）。因此，例如，choice2 == "yes"应该是choice2.equals("yes"). Using equalsIgnoreCase代替equals将忽略大小写差异，这在这种情况下可能有用。