当我使用nextInt（）时，它会跳到下一行[duplicate]

Question

此问题已经在这里有了答案：

Scanner is skipping nextLine() after using next() or nextFoo()?（19个回答）

关闭8分钟前。

我正在尝试使用Java对象的基本原理，但是尝试时遇到了这个问题

我有狗班

public class Dogs {
    private String Name;
    private int Age;
    private String Color;
    private String Owner;
    public Dogs() {
        this.Name = "Rex";
        this.Age = 5;
        this.Color = "black";
        this.Owner = "John";
    }

    public Dogs(String name, int age, String color, String owner) {
        this.Name = name;
        this.Age = age;
        this.Color = color;
        this.Owner = owner;
    }

//all the getters and setters

和主班

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        Dogs my_dog = new Dogs();
        System.out.println("What is the dog's name? ");
        my_dog.setName(input.next());
        System.out.println("What is the dog's age? ");
        my_dog.setAge(input.nextInt());
        System.out.println("What is the dog's color? ");
        my_dog.setColor(input.nextLine());
        System.out.println("What is the owner's name? ");
        my_dog.setOwner(input.nextLine());
    }
}

当我运行它时，它可以很好地打印两个问题，但比下一个跳过...

这是转存：

What is the dog's name? 
pil
What is the dog's age? 
7
What is the dog's color? 
What is the owner's name?

我该如何解决？

Answer 1

请在读取年龄即整数值之后添加input.nextLine（）。

有关详细信息refer this

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        Dogs my_dog = new Dogs();
        System.out.println("What is the dog's name? ");
        my_dog.setName(input.next());
        System.out.println("What is the dog's age? ");
        my_dog.setAge(input.nextInt());
        // add this line
        input.nextLine()
        System.out.println("What is the dog's color? ");
        my_dog.setColor(input.nextLine());
        System.out.println("What is the owner's name? ");
        my_dog.setOwner(input.nextLine());
    }
}

Answer 2

使用input.next（）而不是input.nextLine（）将解决此问题。

nextLine（）方法返回介绍狗的年龄后按ENTER按钮时被nextInt（）跳过的行。

public static void main(String... args) {

    Scanner input = new Scanner(System.in);
    Dogs my_dog = new Dogs();
    System.out.println("What is the dog's name? ");
    my_dog.setName(input.next());
    System.out.println("What is the dog's age? ");
    my_dog.setAge(input.nextInt());
    System.out.println("What is the dog's color? ");
    my_dog.setColor(input.next());
    System.out.println("What is the owner's name? ");
    my_dog.setOwner(input.next());

}