Question

我正在尝试编写一个代码，要求用户输入他们的名字，将其存储在List中，然后要求他们输入他们的年龄，将其存储在另一个列表中。然后它询问用户是否想再试一次[是/否]。在测试代码时，我输入“Y”并期望循环让我输入另一个名字。相反，它会跳过名称输入并跳转到年龄输入。我无法弄清楚为什么。这是代码

import java.util.ArrayList;
import java.util.Scanner;

public class PatternDemoPlus {
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        ArrayList<String> names = new ArrayList<String>();
        ArrayList<Integer> ages = new ArrayList<Integer>();
        String repeat = "Y";
        while(repeat.equalsIgnoreCase("Y")){
            System.out.print("Enter the name: ");
            names.add(scan.nextLine());
            System.out.print("Enter the age: ");
            ages.add(scan.nextInt());

            System.out.print("Would you like to try again? [Y/N]");
            repeat = scan.next();
            //Notice here that if I use "repeat = scan.nextLine(); instead, the code does not allow me to input anything and it would get stuck at "Would you like to try again? [Y/N]
            System.out.println(repeat);

            //Why is it that after the first iteration, it skips names.add and jumps right to ages.add?
        }
    }
}

感谢您的帮助。谢谢。

Answer 1

使用next（）只会返回空格之前的内容。返回当前行后，nextLine（）会自动将扫描仪向下移动。

尝试更改您的代码，如下所示。

    public class PatternDemoPlus {
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        ArrayList<String> names = new ArrayList<String>();
        ArrayList<Integer> ages = new ArrayList<Integer>();
        String repeat = "Y";
        while(repeat.equalsIgnoreCase("Y")){
            System.out.print("Enter the name: ");
            String s =scan.nextLine();
            names.add(s);
            System.out.print("Enter the age: ");
            ages.add(scan.nextInt());
            scan.nextLine();
            System.out.print("Would you like to try again? [Y/N]");
            repeat = scan.nextLine();

            System.out.println(repeat);


        }
    }
}

为什么此循环的第二次迭代会跳过第一次扫描？

1 个答案: