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Java 无法解析userInput_Java_User Input - Fatal编程技术网

Java 无法解析userInput

Java 无法解析userInput,java,user-input,Java,User Input,在我的代码中 import java.util.Scanner; public class JavaLEssonFour { static Scanner userinput= new Scanner(System.in); public static void main (String[] args) { String contYorN = "Y"; int h=1; while (contYorN.equalsIgn

在我的代码中

import java.util.Scanner;
public class JavaLEssonFour 
{
    static Scanner userinput= new Scanner(System.in);

    public static void main (String[] args)
    {
        String contYorN = "Y";
        int h=1;
        while (contYorN.equalsIgnoreCase("y"))
        {
            System.out.println(h);
            System.out.print("Continue y or n? ");
            contYorN = userInput.nextline();
            h++;


        }
    }}
在第13行:
contYorN=userInput.nextline()

我得到一个错误:

无法解析userInput


我使用的是Java JDK 1.7和Eclipse indigo。

你先写了“userinput”都很小,然后用大写字母I写了“userinput”。Java认为这是两个独立的对象,因此无法解析“userinput”

你先写了“userinput”都很小,然后用大写字母I写了“userinput”。Java认为这是两个单独的对象,因此无法解析'userInput'

相同的'nextline()'应该是'nextline()'相同的'nextline()'应该是'nextline()'
userInput
userInput
不一样
userinput
userinput
不一样!!