
本文旨在帮助开发者解决在JavaScript井字棋游戏中,当棋盘填满时,平局判断与胜负判断冲突的问题。通过修改`checkWin()`和`draw()`函数,确保在棋盘填满且没有玩家获胜时,正确地判定为平局,避免胜负判断的错误覆盖。
在开发井字棋游戏时,一个常见的挑战是如何正确地判断平局,尤其是在棋盘填满的同时,还要避免与胜负判断产生冲突。以下提供一种解决方案,通过修改 checkWin() 和 draw() 函数,确保逻辑的正确性。
原代码中,draw() 函数仅简单地检查棋盘是否已满,并直接输出 "tie"。然而,如果最后一步导致某位玩家获胜,checkWin() 函数会先判断出胜者,但 draw() 函数随后仍然会执行,导致最终输出 "tie",覆盖了正确的胜负结果。
核心思路是让 checkWin() 函数返回胜者(如果有),否则返回空字符串。draw() 函数返回一个布尔值,表示是否平局。在主循环中,首先检查是否有胜者,如果没有,再检查是否平局。
将 checkWin() 函数修改为返回胜者,如果没有胜者则返回 null。
function checkWin(){
if (squares[0].textContent == 'X' && squares[1].textContent == 'X' && squares[2].textContent == 'X') return 'X Wins';
if (squares[0].textContent == 'O' && squares[1].textContent == 'O' && squares[2].textContent == 'O') return 'O Wins';
if (squares[3].textContent == 'X' && squares[4].textContent == 'X' && squares[5].textContent == 'X') return 'X Wins';
if (squares[3].textContent == 'O' && squares[4].textContent == 'O' && squares[5].textContent == 'O') return 'O Wins';
if (squares[6].textContent == 'X' && squares[7].textContent == 'X' && squares[8].textContent == 'X') return 'X Wins';
if (squares[6].textContent == 'O' && squares[7].textContent == 'O' && squares[8].textContent == 'O') return 'O Wins';
if (squares[0].textContent == 'X' && squares[3].textContent == 'X' && squares[6].textContent == 'X') return 'X Wins';
if (squares[0].textContent == 'O' && squares[3].textContent == 'O' && squares[6].textContent == 'O') return 'O Wins';
if (squares[1].textContent == 'X' && squares[4].textContent == 'X' && squares[7].textContent == 'X') return 'X Wins';
if (squares[1].textContent == 'O' && squares[4].textContent == 'O' && squares[7].textContent == 'O') return 'O Wins';
if (squares[2].textContent == 'X' && squares[5].textContent == 'X' && squares[8].textContent == 'X') return 'X Wins';
if (squares[2].textContent == 'O' && squares[5].textContent == 'O' && squares[8].textContent == 'O') return 'O Wins';
if (squares[0].textContent == 'X' && squares[4].textContent == 'X' && squares[8].textContent == 'X') return 'X Wins';
if (squares[0].textContent == 'O' && squares[4].textContent == 'O' && squares[8].textContent == 'O') return 'O Wins';
if (squares[2].textContent == 'X' && squares[4].textContent == 'X' && squares[6].textContent == 'X') return 'X Wins';
if (squares[2].textContent == 'O' && squares[4].textContent == 'O' && squares[6].textContent == 'O') return 'O Wins';
return null;
}修改 draw() 函数,使其返回一个布尔值,表示是否平局。只有当所有格子都被填充且没有玩家获胜时,才返回 true。
function draw(){
return squares.every(square => square.textContent !== '');
}在主循环中,先调用 checkWin() 函数,如果返回了胜者,则显示胜者信息。否则,调用 draw() 函数,如果返回 true,则显示平局信息。
squares.forEach(function(e, i){
squares[i].onclick = ()=>{
squares[i].textContent = player = player === 'O' ? 'X' : 'O';
squares[i].textContent === 'X' ? message.textContent = "It's O's turn" :
message.textContent = "It's X's turn";
let winner = checkWin();
if (winner) {
message.textContent = winner;
} else if (draw()) {
message.textContent = 'Draw';
console.log('Tie Game');
}
}
});function checkWin(){
if (squares[0].textContent == 'X' && squares[1].textContent == 'X' && squares[2].textContent == 'X') return 'X Wins';
if (squares[0].textContent == 'O' && squares[1].textContent == 'O' && squares[2].textContent == 'O') return 'O Wins';
if (squares[3].textContent == 'X' && squares[4].textContent == 'X' && squares[5].textContent == 'X') return 'X Wins';
if (squares[3].textContent == 'O' && squares[4].textContent == 'O' && squares[5].textContent == 'O') return 'O Wins';
if (squares[6].textContent == 'X' && squares[7].textContent == 'X' && squares[8].textContent == 'X') return 'X Wins';
if (squares[6].textContent == 'O' && squares[7].textContent == 'O' && squares[8].textContent == 'O') return 'O Wins';
if (squares[0].textContent == 'X' && squares[3].textContent == 'X' && squares[6].textContent == 'X') return 'X Wins';
if (squares[0].textContent == 'O' && squares[3].textContent == 'O' && squares[6].textContent == 'O') return 'O Wins';
if (squares[1].textContent == 'X' && squares[4].textContent == 'X' && squares[7].textContent == 'X') return 'X Wins';
if (squares[1].textContent == 'O' && squares[4].textContent == 'O' && squares[7].textContent == 'O') return 'O Wins';
if (squares[2].textContent == 'X' && squares[5].textContent == 'X' && squares[8].textContent == 'X') return 'X Wins';
if (squares[2].textContent == 'O' && squares[5].textContent == 'O' && squares[8].textContent == 'O') return 'O Wins';
if (squares[0].textContent == 'X' && squares[4].textContent == 'X' && squares[8].textContent == 'X') return 'X Wins';
if (squares[0].textContent == 'O' && squares[4].textContent == 'O' && squares[8].textContent == 'O') return 'O Wins';
if (squares[2].textContent == 'X' && squares[4].textContent == 'X' && squares[6].textContent == 'X') return 'X Wins';
if (squares[2].textContent == 'O' && squares[4].textContent == 'O' && squares[6].textContent == 'O') return 'O Wins';
return null;
}
function draw(){
return squares.every(square => square.textContent !== '');
}
squares.forEach(function(e, i){
squares[i].onclick = ()=>{
squares[i].textContent = player = player === 'O' ? 'X' : 'O';
squares[i].textContent === 'X' ? message.textContent = "It's O's turn" :
message.textContent = "It's X's turn";
let winner = checkWin();
if (winner) {
message.textContent = winner;
} else if (draw()) {
message.textContent = 'Draw';
console.log('Tie Game');
}
}
});通过修改 checkWin() 和 draw() 函数的返回值,并在主循环中先检查胜者,再检查平局,可以有效地解决井字棋游戏中平局判断与胜负判断冲突的问题。这种方法确保了在任何情况下,游戏都能正确地判断胜负或平局。记住,清晰的逻辑和适当的函数设计是解决问题的关键。
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