Question

我只是希望扫描程序将新行读为空字符串，然后如果用户按Enter键继续进行下一步处理。所以有效输入必须是y，n，enter。知道怎么做吗？

这是我的代码：

String gender = "", employed = "";
Scanner in = new Scanner(System.in);
System.out.print("Gender M/F, press enter to skip... ");
while(!in.hasNext("[mfMF]$")){
    System.out.print("Invalid, please choose m/f only... ");
    in.nextLine();
}
if(in.hasNextLine()){
    gender = in.nextLine();
}
System.out.print("Employed? y/n, press enter to skip... ");
    while(!in.hasNext("[ynYN]$|")){
        System.out.print("Invalid, please choose y/n only... ");
        in.nextLine();
    }
if(in.hasNextLine()){
    employed = in.nextLine();
}
System.out.println(gender + " : " + employed);

Answer 1

因此，要检查用户是否按下了enter，您必须使用 isEmpty（）方法。这样做的方法如下所示：

String enter = in.nextLine();
if (enter.isEmpty()) {
   // do what is needed
}

Answer 2

试试这个：

import java.util.*;

class Scanner1
{
   public static void main(String args[])
   {
     Scanner sc=new Scanner(System.in);
     String s;
     do
     {
        s=sc.nextLine();
           if(!(s.equalsIgnoreCase("y")||s.equalsIgnoreCase("n")||s.equalsIgnoreCase("")))
        {
            System.out.println("Please Enter valid input");
        }

    }while(!(s.equalsIgnoreCase("y")||s.equalsIgnoreCase("n")||s.equalsIgnoreCase("")));

    }
}

如何接受输入作为Scanner.nextLine（）的有效输入？

2 个答案: