java 正在跳过扫描仪 nextLine
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Scanner nextLine is being skipped
提问by Reflex Donutz
So this is the code I am using:
所以这是我正在使用的代码：
System.out.println("Create a name.");
name = input.nextLine();
System.out.println("Create a password.");
password = input.nextLine();
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?
但是当它到达这一点时，它只会说“创建一个名称”。和“创建密码”。两者同时进行，然后我必须输入一些内容。所以它基本上跳过了我需要输入字符串的扫描器部分。在“创建名称”之后。和“创建密码”。打印出来然后我输入，名称和密码都更改为我输入的内容。我该如何解决这个问题？
This is the full class. I am just testing so it isn't actually going to be a program:
这是全班。我只是在测试，所以它实际上不会成为一个程序：
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;
}
}
And sorry for my bad grammar. I am not English so it is kinda hard for me to explain this.
对不起，我的语法不好。我不是英语，所以我很难解释这一点。
回答by zw324
The nextInt() ate the input number but not the EOLN:
nextInt() 吃掉了输入数字，但没有吃掉 EOLN：
Create your account.
What is your age?
123 <- Here's actually another '\n'
so the first call to nextLine() after create name accept that as an input.
所以在 create name 之后对 nextLine() 的第一次调用接受它作为输入。
System.out.println("Create a name.");
name = input.nextLine(); <- This just gives you "\n"
User Integer.parseInt(input.nextLine()) or add another input.nextLine() after reading the number will solve this:
用户 Integer.parseInt(input.nextLine()) 或在读取数字后添加另一个 input.nextLine() 将解决此问题：
int age = Integer.parseInt(input.nextLine());
or
或者
int age = input.nextInt();
input.nextLine()
Also see herefor a duplicated question.
另请参阅此处的重复问题。
回答by Sean Landsman
This doesnt actually tell you why the lines are being skipped, but as you're capturing names and passwords you could use Console:
这实际上并没有告诉您为什么跳过这些行，但是当您捕获名称和密码时，您可以使用 Console：
Console console = System.console();
String name = console.readLine("Create a name.");
char[] password = console.readPassword("Create a password.");
System.out.println(name + ":" + new String(password));
回答by RajputAdya
You can also use next() instead of nextLine(). I have tested it in eclipse. its working fine.
您也可以使用 next() 代替 nextLine()。我已经在 Eclipse 中对其进行了测试。它的工作正常。
回答by Priyank Yadav
next()will work but it will not read the string after space.
next()会工作，但它不会读取空格后的字符串。
