Question

首先，我想我会尝试编写一个没有GUI的标准hangman程序。我能够成功做到这一点。但我无法进入GUI。我似乎最挣扎于GUI。

到目前为止，我将展示我的代码。也许你们可以指出我正确的方向？或者告诉我，我是否走在正确的轨道上？

驱动：

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

public class Hangman {
    public static void main (String[] args){
        JFrame frame = new JFrame ("Hangman");
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        HangmanPanel hmp = new HangmanPanel();

        frame.getContentPane().add(hmp);
        frame.pack();
        frame.setVisible(true);

    }


}

Panel Class（这个有点凌乱，因为我正在尝试实验，我猜想我会经历一个试错过程）如果它逻辑上关闭我会道歉。

import java.awt.Panel;
import java.util.Scanner;
import java.util.Random;
import javax.swing.BorderFactory;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;

public class HangmanPanel extends JPanel {

private Hangman hm;
    Scanner input = new Scanner(System.in);
    final String[] WordList = { "ADA", "COBOL", "LOGO", "BASIC", "PROLOG",
            "UBUNTU", "UHURU" };
    String ChosenWord = WordList[(int) Math.random() * WordList.length];
    StringBuilder display = new StringBuilder(ChosenWord.length());{




    for (int i = 0; i < ChosenWord.length(); i++)
        display.append("*");

    int NumberOfTries = 0;

    System.out.print("Let's Begin \n\n");
    System.out.println("Try to guess the word in 6 tries, or you're a dead man!\n\n");

    boolean correct = false;

    while (NumberOfTries < 6 && !correct) {
        String UserGuess = input.next();
        String Letter = UserGuess.substring(0, 1);

        if (ChosenWord.indexOf(Letter) < 0) {
            System.out
                    .printf("The letter %s does not appear anywhere in the word.\n",
                            Letter);
            NumberOfTries++;
        }

        else {
            if (display.indexOf(Letter) >= 0)
                System.out
                        .printf("The letter %s has already been entered as a guess.\n",
                                Letter);

            else {
                for (int p = 0; p < ChosenWord.length(); p++)
                    if (ChosenWord.charAt(p) == Letter.charAt(0))
                        display.setCharAt(p, Letter.charAt(0));
            }
        }

        correct = display.indexOf("*") < 0;
        draw(NumberOfTries);
    }

    if (correct)
        System.out
                .println("You have guessed "
                        + ChosenWord
                        + " correct and saved yourself from the gallos. Till next time that is.\n");

    else {
        System.out
                .printf("You've had %d strikes against you. Thus you've been hung. Better luck next time.\n",
                        NumberOfTries);
    }
}

public static void draw(int num) {
    JPanel p1 = new JPanel();
    p1.setBorder(BorderFactory.createEtchedBorder());
    final String[] status = { "____\n|   |\n |   \n|\n|",
            "____\n| |\n| O\n|\n|\n|", "____\n| |\n| O\n|/|\n|\n|",
            "____\n| |\n| O\n|/|\\\n|\n|", "____\n| |\n| O\n|/|\\\n|/\n|",
            "____\n| |\n| O\n|/|\\\n|/\\\n|" };
    p1.add(new JLabel("Etched Border"));

    if (num >= 0 && num < status.length) {
        System.out.println(status[num]);
    }

    else {
        System.out.println("Must be a Mistake. Out of Range.");
    }

}

}

Answer 1

Swing（和大多数GUI）是事件驱动的环境。这基本上意味着逻辑的执行通常由事件侦听器/处理程序

处理

首先熟悉如何编写一般用户界面，看看Creating a GUI with Swing

使用像while-loop这样的东西来获取用户输入在GUI环境中不起作用，你需要使用可用的控件构建你想要做的概念

如何将这个hangman程序导入GUI？

1 个答案: