使用 Scanner 类的问题

Question

该代码可用于输入妻子的姓名和年龄，但无法打印儿子的姓名，尽管它正确显示了他们的年龄。

import java.util.Scanner;

public class Check2
{
    public static void main(String[] args)
    {
    String wife;
    String son1;
    String son2;
    int wifeAge;
    int son1Age;
    int son2Age;

    Scanner keyboard = new Scanner(System.in);

    System.out.println("Wife's name? ");
    wife = keyboard.nextLine();
    System.out.println("Her age? ");
    wifeAge = keyboard.nextInt();
    System.out.println();

    System.out.println("First son's name? ");
    son1 = keyboard.nextLine();
    keyboard.nextLine();
    System.out.println("His age? ");
    son1Age = keyboard.nextInt();
    System.out.println();

    System.out.println("Second son's name? ");
    son2 = keyboard.nextLine();
    keyboard.nextLine();
    System.out.println("His age? ");
    son2Age = keyboard.nextInt();
    System.out.println();

    keyboard.nextLine();

    System.out.println("My wife's name is " + wife + ". She is " +
                       wifeAge + " years old.\nOur first son is " +
                       son1 + ". He is " + son1Age + ".\nOur " +
                       "second son is " + son2 + ". He is " +
                       son2Age + ".");
    }
}

Answer 1

您的代码中有多余的keyboard.nextLine();。

son1 = keyboard.nextLine();
keyboard.nextLine(); // you don't need this here.

son2 = keyboard.nextLine();
keyboard.nextLine(); // nor here

在每个keyboard.nextInt(); 之后保留额外的keyboard.nextLine(); 行，您的程序应该可以正常运行。

wifeAge = keyboard.nextInt();
keyboard.nextLine(); // put it here
...
son1Age = keyboard.nextInt();
keyboard.nextLine(); // and here

nextInt() 只读取一个整数值，而不是一个新行。如果您需要读取新行，则每次从键盘读取整数值时都需要输入keyboard.nextLine();。

希望这会有所帮助！

Answer 2

import java.util.Scanner;

public class Check2
{
    public static void main(String[] args)
    {
String wife;
String son1;
String son2;
int wifeAge;
int son1Age;
int son2Age;

Scanner keyboard = new Scanner(System.in);

System.out.println("Wife's name? ");
wife = keyboard.nextLine();
System.out.println("Her age? ");
wifeAge = keyboard.nextInt();
System.out.println();

System.out.println("First son's name? ");
keyboard.nextLine();
son1 = keyboard.nextLine();
System.out.println("His age? ");
son1Age = keyboard.nextInt();
System.out.println();

System.out.println("Second son's name? ");
keyboard.nextLine();    
son2 = keyboard.nextLine();
System.out.println("His age? ");
son2Age = keyboard.nextInt();
System.out.println();

keyboard.nextLine();

System.out.println("My wife's name is " + wife + ". She is " +
                   wifeAge + " years old.\nOur first son is " +
                   son1 + ". He is " + son1Age + ".\nOur " +
                   "second son is " + son2 + ". He is " +
                   son2Age + ".");
    }
}

Answer 3

nextLine() 不是您想要的方法。 您想改用next()，如以下更正和简化的代码所示。您真的不想管理换行符，只需让扫描仪将所有输入视为由空格分隔的标记字符串，并按顺序读取它们。输入字符串使用next()方法，输入整数使用nextInt()方法...

import java.util.Scanner;

public class Check2
{
    public static void main(String[] args)
    {
        String wife;
        String son1;
        String son2;
        int wifeAge;
        int son1Age;
        int son2Age;

        Scanner keyboard = new Scanner(System.in);

        System.out.println("Wife's name? ");
        wife = keyboard.next();
        System.out.println("Her age? ");
        wifeAge = keyboard.nextInt();
        System.out.println();

        System.out.println("First son's name? ");
        son1 = keyboard.next();
        System.out.println("His age? ");
        son1Age = keyboard.nextInt();
        System.out.println();

        System.out.println("Second son's name? ");
        son2 = keyboard.next();
        System.out.println("His age? ");
        son2Age = keyboard.nextInt();
        System.out.println();

        System.out.println("My wife's name is " + wife + ". She is " +
                   wifeAge + " years old.\nOur first son is " +
                   son1 + ". He is " + son1Age + ".\nOur " +
                   "second son is " + son2 + ". He is " +
                   son2Age + ".");
        }
   }