Tic Tac Toe win条件未正确激活

Question

我刚才问过一个问题。经过大量的代码重写，我的井字脚趾看起来很好，除了一个明显的问题。当我尝试玩游戏时，赢的条件似乎是随机激活的--我会在一行中有X-O-X，并说“玩家x赢了！”，但当我有O-O-O时，它不会检测到赢。不过画画还行。

赢取条件代码：

 // win conditions. if true, set win==true; else set win==false
            if (square[0].getText().equals(square[1].getText())
                    && square[1].getText().equals(square[2].getText())
                    != square[0].getText().isEmpty()) {
                    win = true;}

            if (square[3].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[5].getText())
                    != square[3].getText().isEmpty()) {
                    win = true;}

            if (square[6].getText().equals(square[7].getText())
                    && square[7].getText().equals(square[8].getText())
                    != square[6].getText().isEmpty()) {
                    win = true;}

            if (square[0].getText().equals(square[3].getText())
                    && square[3].getText().equals(square[6].getText())
                    != square[0].getText().isEmpty()) {
                    win = true;}

            if (square[1].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[7].getText())
                    != square[1].getText().isEmpty()) {
                    win = true;}

            if (square[2].getText().equals(square[5].getText())
                    && square[5].getText().equals(square[8].getText())
                    != square[2].getText().isEmpty()) {
                    win = true;}

            if (square[0].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[8].getText())
                    != square[0].getText().isEmpty()) {
                    win = true;}

            if (square[6].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[2].getText())
                    != square[6].getText().isEmpty()) {
                    win = true;}

            else{win = false;
            }

和动作代码：

  public void actionPerformed (ActionEvent e) {
            //one more move has gone by, calculate player turn + player letter
            move++;
            if (move % 2 == 0) {
                    player = 1; letter = "X";
            }else{
                    player = 2; letter = "O";
            }

            playergo.setText("It is player " + player + "'s go!");

            //set square letter to player's letter, disable square so no further moves can be made there
            for (int i=0; i<=8; i++){
                    if (e.getSource() == square[i]){
                            square[i].setText(letter);
                            square[i].setEnabled(false);
                    }
            }

这里有完整的pastebin代码。提前谢谢大家！！！

Answer 1

因为在win条件中没有使用if/else，所以您将始终检查最后一个if，如果该false，则将设置win=false