java - 扫描仪不接受另一个字符串的输入并跳过它

Question

为什么扫描仪不接受另一个字符串的输入并跳过它？我无法理解，这是我的代码:

import java.util.Scanner;

public class demo {

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);
        String name;
        String address;
        int age;

        System.out.println("Enter your name");
        name = s.nextLine();
        System.out.println("My name is :" + name);

        System.out.println("Enter your age");
        age = s.nextInt();
        System.out.println("My age is :" + age);

        System.out.println("Enter your address");
        address = s.nextLine();
        System.out.println("My address is :" + address);
    }
}

输出:

Enter your name
dk
My name is :dk
Enter your age
22
My age is :22
Enter your address
My address is :

Answer 1

很简单。您可以在 age = s.nextInt();

之后使用 s.nextLine();

Scanner 提供了方法 nextInt() 来请求用户输入，并且不消耗最后一个换行符 (\n)。

System.out.println("Enter your age");
age = s.nextInt();
s.nextLine(); // consumes the \n character
System.out.println("My age is :" + age);