Java输入不起作用（初学者）

小编典典

Java输入不起作用（初学者）

java

由于某些原因，我的代码将不接受最后一行“您想订购的商品：”的输入

谁能告诉我我的错误在这里吗？它正在正确地编译一切。我只是一个初学者，所以请简单地告诉我。

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");
                    }

阅读 582

2020-11-26

共1个答案

小编典典

您需要在调用in.nextLine()行的后面立即调用in.nextInt()
，原因是仅要求下一个整数不会占用输入中的整个行，因此您需要通过调用来跳至输入中的下一个新行字符in.nextLine()

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

每次您需要在调用不消耗整行的方法后获取新行时，都必须执行此操作。考虑改用BufferedReader对象！

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

Scanner.nextInt()如果无法将输入解析为整数，则会引发相同的错误。

关于您对错误的评论，有以下一种：

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

}

相反，应如下所示：

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

另外，这也不是绝对必要的，但是我建议您在实例化列表时确实声明ArrayList的泛型类型，这样就不必将对Menu.get（）的进一步调用强制转换为String了。

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"应该改为choice2.equals("yes")。使用equalsIgnoreCase代替代替equals将忽略大小写差异，这在这种情况下可能有用。

2020-11-26