Question

好吧，我想通过一些搜索找到它，但现在我有另一个问题当弹出第二个框时，我点击“否”＆＃39;第一个盒子仍然运行，如果我取消它，我会收到一个错误。我做错了什么？

  import javax.swing.*;
 import java.util.*;
import java.util.Scanner;

public class RPS {

public static void main(String[] args) {

    Scanner sc = new Scanner (System.in);
    String input;                      
    int user;                       
    int computer;                   
while (true){//here we go
        input = JOptionPane.showInputDialog("What'll it be? Rock, paper, or scissors?\n" +
                "1 for rock, 2 for paper, and 3 for scissors: ",JOptionPane.YES_NO_OPTION);


    user = Integer.parseInt(input);


    Random randomnum = new Random ();

    computer = randomnum.nextInt(3);




    if (user == 1 && computer == 0){System.out.println ("You played Rock! You have tied");
    JOptionPane.showMessageDialog(null, "Tie!");}

    else if (user == 1 && computer == 1){System.out.println ("You have played Rock! You have lost");
    JOptionPane.showMessageDialog(null, "Paper beats rock. You lose!");}

    else if (user == 1 && computer == 2){System.out.println ("You have played Rock! You have won");
    JOptionPane.showMessageDialog(null, "Rock beats scissors. You win!");}

    else if (user == 2 && computer == 0){System.out.println ("You have played Paper! You have won");
    JOptionPane.showMessageDialog(null, "Paper beats rock. You win!");}

    else if (user == 2 && computer == 1){System.out.println ("You have played Paper! You have tied");
    JOptionPane.showMessageDialog(null, "Tie!");}

    else if (user == 2 && computer == 2){System.out.println ("You have played Paper! You have lost");
    JOptionPane.showMessageDialog(null, "Scissors beats paper. You lose!");}

    else if (user == 3 && computer == 0){System.out.println ("You have played Scissors! You have lost");
    JOptionPane.showMessageDialog(null, "Rock beats Scissors. You lose!");}

    else if (user == 3 && computer == 1){System.out.println ("You have played Scissors! You have won");
    JOptionPane.showMessageDialog(null, "Scissors beats paper. You win!");}

    else if (user == 3 && computer == 2){System.out.println ("You have played Scissors! You have tied");
    JOptionPane.showMessageDialog(null, "Tie!");}



     int n = JOptionPane.showConfirmDialog(null,"Would you like to play again?", "Confirmation",JOptionPane.YES_NO_OPTION);
       if(n == JOptionPane.YES_OPTION) {
          JOptionPane.showMessageDialog(null,"Let's start");
       } else {
          JOptionPane.showMessageDialog(null,"Goodbye");
       } 

  }
 }
}

Answer 1

在return;之后加上JOptionPane.showMessageDialog(null,"Goodbye");语句。否则它不知道结束while循环。

如果您选择立即取消，当您按取消时也会发生错误。这是因为parseInt不知道如何读取结果输入。您应该检查showInputDialog的返回值以查看它的有效输出。我建议检查＆＃34; 1＆＃34;，＆＃34; 2＆＃34;或＆＃34; 3＆＃34; （或＆＃34;摇滚＆＃34;，＆＃34;纸张＆＃34;或＆＃34; scizzors＆＃34;）而不是使用parseInt，然后如果它不匹配它们中的任何一个来以及某种错误消息。

如何循环JOptionPane

1 个答案: