我正在尝试使用下面的代码为JButton编写actoinPerformed代码。如果我在＆＃34; count ++＆＃34;中切断代码，那么一切都可以正常工作。但是，当我添加部件以检查获胜者时，单击按钮时没有任何反应。当我在程序的命令行版本中使用它们时，我正在使用的方法（即placeXToken，placeOToken和checkWinner）都可以正常工作，所以我不认为问题在那里。我认为它可能与尝试在actionPerformed方法中打开JOptionPanes有关，所以（如你所见）我将它们更改为命令行打印，但我仍然遇到同样的问题。当我尝试添加checkWinner部分时，我完全迷失了它为什么停止工作。任何帮助将不胜感激。
@Override
public void actionPerformed(ActionEvent e) {
String action = e.getActionCommand();
colInput = Integer.parseInt(action);
if (count % 2 == 0) {
instructionslabel.setText("<html>It is " + names[otherIndex] + "'s turn."
+ "<br/>Please select where you would like to place your token.</html");
placeXToken(colInput);
}
else {
instructionslabel.setText("<html>It is " + names[randIndex] + "'s turn."
+ "<br/>Please select where you would like to place your token.</html");
placeOToken(colInput);
}
count++;
//Checks to see if there is a winner or a tie each turn
if (checkWinner() != null) {
//what to do in case of a tie
if (checkWinner() == notoken) {
System.out.println("tie");
/*int tie = JOptionPane.showConfirmDialog(
getParent(), "<html>The game ended in a tie.<br/>Would you like to play again?</html>",
"Tie Game",
JOptionPane.YES_NO_OPTION);
if (tie == 0)
main(null);
else {
JOptionPane.showMessageDialog(getParent(),
"Thank you for playing. Goodbye.", "Goodbye.", JOptionPane.PLAIN_MESSAGE);
close();
}*/
}
//what to do if X wins
if (checkWinner() == xtoken) {
System.out.println("xwin");
/*int xwin = JOptionPane.showConfirmDialog(getParent(),
names[randIndex] + " won the game.Would you like to play again?",
"X Win",
JOptionPane.YES_NO_OPTION);
if (xwin == 0)
main(null);
else {
JOptionPane.showMessageDialog(getParent(),
"Thank you for playing. Goodbye.", "Goodbye.", JOptionPane.PLAIN_MESSAGE);
close();
}*/
}
//what to do if O wins
if (checkWinner()== otoken) {
System.out.println("owin");
/*int owin = JOptionPane.showConfirmDialog(getParent(),
names[otherIndex] + " won the game.Would you like to play again?",
"O Win",
JOptionPane.YES_NO_OPTION);
if (owin == 0)
main(null);
else {
JOptionPane.showMessageDialog(getParent(),
"Thank you for playing. Goodbye.", "Goodbye.", JOptionPane.PLAIN_MESSAGE);
close();
}*/
}
}
}
这是checkWiner方法。也许某人可能能够指出我在将其从用于命令行转换为GUI时可能犯的错误。
public static Icon checkWinner() {
// Checks for horizontal win
for (int row = 0; row < 7; row = row++) {
for (int col = 0; col < 5; col =col++) {
if ((slotlabels[row][col].getIcon() != notoken)
&& (slotlabels[row][col+1].getIcon() != notoken)
&& (slotlabels[row][col+2].getIcon() != notoken)
&& (slotlabels[row][col+3].getIcon() != notoken)
&& ((slotlabels[row][col].getIcon() == slotlabels[row][col+1].getIcon())
&& (slotlabels[row][col+1].getIcon() == slotlabels[row][col+2].getIcon())
&& (slotlabels[row][col+2].getIcon() == slotlabels[row][col+3].getIcon())))
return slotlabels[row][col].getIcon(); // returns the token (X or O) of the winner
}
}
// Checks for vertical win
for (int col = 0; col < 8; col =col++) {
for (int row =0; row < 4; row = row++) {
if((slotlabels[row][col].getIcon() != notoken)
&& (slotlabels[row+1][col].getIcon() != notoken)
&& (slotlabels[row+2][col].getIcon() != notoken)
&& (slotlabels[row+3][col].getIcon() != notoken)
&& ((slotlabels[row][col].getIcon() == slotlabels[row+1][col].getIcon())
&& (slotlabels[row+1][col].getIcon() == slotlabels[row+2][col].getIcon())
&& (slotlabels[row+2][col].getIcon() == slotlabels[row+3][col].getIcon())))
return slotlabels[row][col].getIcon(); // returns the token (X or O) of the winner
}
}
// Checks for diagonal left-top to right-bottom win
for (int row = 0; row < 4; row = row++) {
for (int col = 0;col < 5; col = col++) {
if((slotlabels[row][col].getIcon() != notoken)
&& (slotlabels[row+1][col+1].getIcon() != notoken)
&& (slotlabels[row+2][col+2].getIcon() != notoken)
&& (slotlabels[row+3][col+3].getIcon() != notoken)
&& ((slotlabels[row][col].getIcon() == slotlabels[row+1][col+1].getIcon())
&& (slotlabels[row+1][col+1].getIcon() == slotlabels[row+2][col+2].getIcon())
&& (slotlabels[row+2][col+2].getIcon() == slotlabels[row+3][col+3].getIcon())))
return slotlabels[row][col].getIcon(); // returns the token (X or O) of the winner
}
}
// Checks for diagonal right-top to left-bottom win
for (int row = 0; row < 4; row = row++) {
for (int col = 3; col < 8; col = col++) {
if((slotlabels[row][col].getIcon() != notoken)
&& (slotlabels[row+3][col-4].getIcon() != notoken)
&& (slotlabels[row+6][col-8].getIcon() != notoken)
&& (slotlabels[row+9][col-12].getIcon() != notoken)
&& ((slotlabels[row][col].getIcon() == slotlabels[row+3][col-4].getIcon())
&& (slotlabels[row+3][col-4].getIcon() == slotlabels[row+6][col-8].getIcon())
&& (slotlabels[row+6][col-8].getIcon() == slotlabels[row+9][col-12].getIcon())))
return slotlabels[row][col].getIcon(); // returns the token (X or O) of the winner
}
}
// Checks to see if the game resulted in a tie
boolean tie = true;
for (int row = 0; row < 7; row = row++) {
for (int col = 0; col < 8; col = col++) {
if (slotlabels[row][col].getIcon() == notoken)
tie = false;
}
}
if (tie == true)
return notoken;
// Otherwise, there is no winner and no tie, so do nothing with this.
return null;
}
答案 0 :(得分：0)
在IDE的调试器中启动应用程序，并为行count++设置断点。然后，调试器允许您逐步执行代码，这样您就可以看到执行的每一行，并且还会显示所有可见变量的值。
这样，您应该能够看到代码中断的位置。
如果您没有IDE，那么便宜的解决方法是在任何地方添加System.out.println()一些有用的文字，以便您可以看到输出停止的位置。
最后，您可能希望将代码包装在带有
的大try/catch块中
catch(Exception e) { e.printStackTrace(); }
最后，您可以看到代码可能在控制台上抛出的任何异常。
答案 1 :(得分：0)
我不小心从命令行（我在网格中制作3x4单元格以使其看起来很漂亮）转换到GUI（每个插槽距离前一个只有+1），然后我结束了将for循环中的增量保留为“row = row ++”和“col = col ++”。 Java不喜欢这个。