我正在尝试用 Javascript 编写一个视频扑克游戏，作为了解它的基础知识的一种方式，但我遇到了一个问题，即 jQuery 单击事件处理程序多次触发。
它们附在用于下注的按钮上，在游戏期间对第一手牌进行下注时效果很好（仅触发一次）；但在为第二手牌下注时，每次按下下注或下注按钮时，它都会触发两次 click 事件（因此每次按下时下注正确金额的两倍）。总的来说，当按下下注按钮一次时，点击事件被触发的次数遵循这种模式——序列的第 i项是从游戏开始时第 i手的下注：1, 2, 4 , 7, 11, 16, 22, 29, 37, 46，这似乎是 n(n+1)/2 + 1 对于任何值得的东西——而且我不够聪明，无法弄清楚，我使用了OEIS . :)
这是带有正在执行的单击事件处理程序的函数；希望它很容易理解（如果没有，请告诉我，我也想在这方面做得更好）：
/** The following function keeps track of bet buttons that are pressed, until place button is pressed to place bet. **/
function pushingBetButtons() {
$("#money").text("Money left: $" + player.money); // displays money player has left
$(".bet").click(function() {
var amount = 0; // holds the amount of money the player bet on this click
if($(this).attr("id") == "bet1") { // the player just bet $1
amount = 1;
} else if($(this).attr("id") == "bet5") { // etc.
amount = 5;
} else if($(this).attr("id") == "bet25") {
amount = 25;
} else if($(this).attr("id") == "bet100") {
amount = 100;
} else if($(this).attr("id") == "bet500") {
amount = 500;
} else if($(this).attr("id") == "bet1000") {
amount = 1000;
}
if(player.money >= amount) { // check whether the player has this much to bet
player.bet += amount; // add what was just bet by clicking that button to the total bet on this hand
player.money -= amount; // and, of course, subtract it from player's current pot
$("#money").text("Money left: $" + player.money); // then redisplay what the player has left
} else {
alert("You don't have $" + amount + " to bet.");
}
});
$("#place").click(function() {
if(player.bet == 0) { // player didn't bet anything on this hand
alert("Please place a bet first.");
} else {
$("#card_para").css("display", "block"); // now show the cards
$(".card").bind("click", cardClicked); // and set up the event handler for the cards
$("#bet_buttons_para").css("display", "none"); // hide the bet buttons and place bet button
$("#redraw").css("display", "block"); // and reshow the button for redrawing the hand
player.bet = 0; // reset the bet for betting on the next hand
drawNewHand(); // draw the cards
}
});
}
如果您有任何想法或建议，或者我的问题的解决方案与此处另一个问题的解决方案相似，请告诉我（我查看了许多类似标题的线程，但没有找到可行的解决方案给我）。