这个Java GUI代码有什么问题？

Question

我刚刚开始学习Java GUI并在练习事件处理时遇到了这个问题. 这是初始窗口

当我在文本字段中输入一个数字时,应该说出猜测的数字是更高,更低还是匹配.如果不匹配则会提示输入另一个号码.但是窗户就挂了. 输入数据后

我猜它落在一个无限循环中.这是代码.帮我找出问题所在.谢谢.

import java.awt.*;
import java.awt.event.*;
import javax.swing.*;

public class RandomNumGame extends JFrame {
    private JLabel promptLabel, resultLabel, answerLabel;
    private int tries=1, randomNum, guessNum;
    private JButton button;
    private JTextField txt; 
    private boolean guessed;

    public RandomNumGame() {
        setLayout(new FlowLayout());

        promptLabel = new JLabel("Guess a number(1-1000): ");
        add(promptLabel);

        txt = new JTextField(7);
        add(txt);

        button = new JButton("Guess!");
        add(button);

        resultLabel = new JLabel("");
        add(resultLabel);

        /*answerLabel = new JLabel("");
        add(answerLabel);
        */

        Event e = new Event();
        button.addActionListener(e);
    }

    private class Event implements ActionListener{
        public void actionPerformed(ActionEvent e){
            randomNum = (int )(Math.random() * 1000 + 1);
            guessed=false;
            do{
                try{
                    guessNum = (int)(Double.parseDouble(txt.getText()));
                    if(guessNum>randomNum){
                        resultLabel.setText("Your number is higher. Try Again");
                    }
                    else if(guessNum<randomNum){
                        resultLabel.setText("Your number is lower. Try Again");
                    }
                    else{
                        resultLabel.setText("Your number matched!");
                        guessed=true;
                    }
                }
                catch(Exception ee){
                    resultLabel.setText("Enter a legit number. What are you stupid?");
                }
            }while(!guessed);

        }
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        RandomNumGame ran = new RandomNumGame();
        ran.setDefaultCloseOperation(EXIT_ON_CLOSE);
        ran.setSize(300, 120);
        //ran.pack();
        ran.setVisible(true);
        ran.setTitle("Random Number Game");
    }

}

Answer 1

您的方法中不需要 xc2xb4t 循环actionPerformed。只要让它执行一次，然后检查它是否被猜对了。当用户输入错误的号码时，您基本上就陷入了循环。为了使其更平滑，仅在guessedis时创建随机数true，并仅执行一次猜测和条件。

\n\n

// Change constructor\npublic RandomNumGame() {\n    ...\n    guessed = true; // initialize to true to create a new number when you click\n}\nprivate class Event implements ActionListener{\n    public void actionPerformed(ActionEvent e){\n        if(guessed) { // If the number was guessed, on the next click you get a new random number\n            randomNum = (int )(Math.random() * 1000 + 1);\n            guessed = false;\n        }\n        try{\n            guessNum = (int)(Double.parseDouble(txt.getText()));\n            if(guessNum>randomNum){\n               resultLabel.setText("Your number is higher. Try Again");\n            }\n            else if(guessNum<randomNum){\n               resultLabel.setText("Your number is lower. Try Again");\n            }\n            else{\n               resultLabel.setText("Your number matched! Click again for a new Number");\n               guessed=true;\n            }\n        }\n        catch(Exception ee){\n            resultLabel.setText("Enter a legit number. What are you stupid?");\n        }\n    }\n}\n

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