正在跳过扫描仪nextLine [英] Scanner nextLine is being skipped

问题描述

所以这是我正在使用的代码：

So this is the code I am using:

System.out.println("Create a name.");
name = input.nextLine();
System.out.println("Create a password.");
password = input.nextLine();

但是当它达到这一点时，它只是说创建一个名字。和创建密码。两个在同一时间然后我必须输入一些东西。所以它基本上是跳过我需要键入字符串的Scanner部分。在创建一个名字之后。和创建密码。是打印出来然后输入，名称和密码都改为我输入的内容。如何解决这个问题？

But when it reaches this point it just says "Create a name." and "Create a password." both at the same time and then I have to type something. So it's basically skipping the Scanner parts where I need to type a String. After "Create a name." and "Create a password." is outprinted and I type then, both name and password are changing to what I typed in. How do I fix this?

这是完整的课程。我只是测试所以它实际上不是一个程序：

This is the full class. I am just testing so it isn't actually going to be a program:

package just.testing;

import java.util.Scanner;
public class TestingJava
{
    static int age;
    static String name;
    static String password;
    static boolean makeid = true;
    static boolean id = true;

    public static void main(String[] args){
        makeid(null);
        if(makeid == true){
            System.out.println("Yay.");
        }else{

        }
    }
    public static void makeid(String[] args){
        System.out.println("Create your account.");
        Scanner input = new Scanner(System.in);
        System.out.println("What is your age?");
        int age = input.nextInt();
        if(age<12){
            System.out.println("You are too young to make an account.");
            makeid = false;
            return;
        }
        System.out.println("Create a name.");
        name = input.nextLine();
        System.out.println("Create a password.");
        password = input.nextLine();
        return;
    }
}

抱歉我的语法不好。我不是英文，所以我很难解释这一点。

And sorry for my bad grammar. I am not English so it is kinda hard for me to explain this.

推荐答案

nextInt（）吃了输入数字但没有EOLN：

The nextInt() ate the input number but not the EOLN:

Create your account.
What is your age?
123 <- Here's actually another '\n'

所以第一次打电话到创建名称后的nextLine（）接受它作为输入。

so the first call to nextLine() after create name accept that as an input.

System.out.println("Create a name.");
name = input.nextLine(); <- This just gives you "\n"

User Integer.parseInt（input.nextLine（ ））或在读取数字后添加另一个input.nextLine（）将解决此问题：

User Integer.parseInt(input.nextLine()) or add another input.nextLine() after reading the number will solve this:

int age = Integer.parseInt(input.nextLine());

或

int age = input.nextInt();
input.nextLine()

另见此处。

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