Question

所以我刚完成了这个项目，我正在为了好玩而做。我对GUI有一个基本的了解，但并不是全部，并且实际上没有把它放到我的一个项目中。我不确定如何在我的代码中实现这一切，但人们可以给我建议或者至少指出我正确的方向吗？

游戏是HangMan。

代码：公共课HangMan {

public static void main(String[] args) {
    Scanner sc = new Scanner (System.in);
    String mysteryGuess = "hello";
    String userGuess = "";
    int wrongGuesses = 0;
    boolean[] guessed = new boolean[mysteryGuess.length()];

    System.out.println("Hello and welcome to Hang Man!");
    loop:
    for(;;){
        String[] wordAsArray = convertToStringArray(mysteryGuess);

        for (int i = 0; i<wordAsArray.length;i++)
            if(wordAsArray[i].equals(userGuess))
                guessed[i]=true;
        System.out.println("Word so far:" + visibleWord(wordAsArray,guessed));
        System.out.println("What is your guess?");
        userGuess = sc.next();
        boolean guessResult = guess(userGuess,wordAsArray,guessed);
        if (guessResult==(true))
            System.out.println("Correct");
        else{
            System.out.println("Incorrect");
            wrongGuesses++;
        }
        if (didWin(guessed)==true)
            break loop;
    }
    System.out.println("Good Job! The word was " + mysteryGuess);
    System.out.println("You only got " + wrongGuesses + " wrong!");
}


//This method creates an array version of the parameter word
//For example, if word contained the data "hello", then this method
//would return {"h", "e", "l", "l", "o"}
//Parameters:   word - a single word
//Returns:      an array containing each letter in word
public static String[] convertToStringArray(String word) {
    String [] pWord = new String [word.length()];
    for (int i = 0; i<pWord.length; i++){
        pWord[i] = word.substring(i,i+1);
    }
    return pWord;

}


//This method determines whether the player has won the game of HangMan
//Parameters:   guessed - array of boolean values
//Returns:      true - if every value in guessed is true
//              false - if at least one value in guessed is false
public static boolean didWin(boolean[] guessed) {
    boolean bGuess = true;
    loop:
    for (int i = 0; i<guessed.length;i++){
        if(guessed[i]==false){
            bGuess = false;
            break loop;
        }


    }
        return bGuess;
}


//This method determines what portion of the hidden word is visible
//For example, if the parameters are as follows:
//     wordAsArray: {"h", "e", "l", "l", "o"}
//     guessed: {true, false, false, false, true}
//Then the method should return "h???o"
//Parameters:   wordAsArray - the individual letters to be guessed
//              guessed - array of boolean values; a true value means the corresponding letter has been guessed
//Returns:      A string representing how much of the word has been guessed (unguessed letters are represented by ?'s)
public static String visibleWord(String[] wordAsArray, boolean[] guessed) {
    String visibleWord="";
    String[] holder = new String [wordAsArray.length];
    for (int i = 0; i<holder.length;i++)
        holder[i]=wordAsArray[i];
    for(int i = 0; i<holder.length;i++){
        if (guessed[i] == true)
            holder[i]=holder[i];
        if (guessed[i] == false)
            holder[i]="?";
    }
    for(int i = 0; i<holder.length;i++){
        visibleWord=visibleWord+holder[i];
    }
    return visibleWord;
}


//This method checks to see if a player has made a successful guess in the game of Hang Man
//For example, if the parameters are as follows:
//     letter: "e"
//     wordAsArray: {"h", "e", "l", "l", "o"}
//     guessed: {true, false, false, false, true}
//Then the guessed array would be changed to:
//      guessed: {true, true, false, false, true}
//And the method would return false
//Parameters:   letter - the letter that the user has just guessed
//              wordAsArray - an array of individual letters that are to be guessed
//              guessed - array of boolean values; a true value means the corresponding letter has been guessed
//Returns:  true - if letter matches an unguessed letter in wordAsArray
//          false - otherwise
public static boolean guess(String letter, String[] wordAsArray, boolean[] guessed) {
    boolean appearsAtLeastOnce=false;
    for(int i = 0; i<wordAsArray.length;i++)
        if(letter.equalsIgnoreCase(wordAsArray[i])){
            guessed[i] = true;
            appearsAtLeastOnce=true;
        }
    return appearsAtLeastOnce;



}

谢谢你的时间！

Answer 1

如果您以前从未制作过GUI程序，那么尝试将控制台程序转移到GUI程序并不是一件容易的事。您需要了解事件驱动的编程。我建议您查看Swing tutorials

但有些提示。你想要一个“半gui”程序。您只需使用JOptionPane s作为输入。假设你想得到一个数字输入。你会做这样的事情

String numberString = JOptionPane.showInputDialog(null, "Enter a Number");
int number = Integer.parseInteger(numberString);

执行第一行后，会自动弹出输入窗格。要求输入。结果是一个String，所以你有来解析它以得到一个数字。

此外，如果您只想 diplay 一条消息，请使用

JOptionPane.showMessageDialog(null, message);

你可以这样做，显示一些结果。在上述情况下，当您只想显示消息时，您不需要使其等于任何内容。因此，您可以使用System.out.println()而不是JOPtionpane.showMesageDialog()代替scan.next()，而不是使用JOptionPane.showInputDialog()

如何在我的代码中实现GUI？

1 个答案: