java中的连续错误

Question

可能重复:
如何比较Java中的字符串？

在我的Java应用程序中.要接受挑战,请按y,但程序不会继续.谁能帮我这个？

这是我的代码:

import java.util.Scanner;


    public class MainApp 
    {


        public static void main(String[] args)
        {
            Scanner scanner = new Scanner(System.in);
            System.out.println("Hello User");
            System.out.println("Please Enter your first name");
            String name;
            name =scanner.next();

        System.out.println("\n" + " Hello " + name + "how are you today?");
        String y="";
        y=scanner.next();


        System.out.println("\n" + " Well " +  name + " I am " + y + " Too." +"\n" + " I have a riddle for you " + name + ", wll you Attempt it?" + "\n" + " Type y for yes , or n for no");
        String v;
        v=scanner.next();


        if(v == "y")
            {
            System.out.println("\n" + "How much wood could a wood chuck chuck if a wood chuck could chuck wood? :) " );
            }

        else if(v == "n")
            {
            System.out.println("Ok then " + name + " suit youself, Goodbye! :) ");
            }
        else 
            { 
            System.out.println("Please pick y, or n , and make sure it is in lower case");
            }

        }

    }

Answer 1

if(v == "y")

您应该String.equals()用来比较字符串.

所以条件的代码应该是这样的:

if ("y".equals(v)) //checking "y".equals() prevents null access if v is null

说明:
将operator==被检查身份 -如果两个操作数是完全相同的对象.这不是你想要的,因为很少有字符串的情况.

equals()另一方面,该方法 - 如果两个对象彼此相等,则检查相等性.