我正在尝试用 Javascript 编写一个视频扑克游戏,作为了解其基础知识的一种方式,但我遇到了 jQuery 单击事件处理程序多次触发的问题。
它们附在用于下注的按钮上,并且在游戏过程中第一手牌下注时效果很好(仅触发一次);但在第二手牌的投注中,每次按下投注或下注按钮时,它都会触发两次点击事件(因此每次按下时投注正确金额的两倍)。总体而言,当按下一次下注按钮时,点击事件被触发的次数就遵循这种模式——其中ith序列中的项用于下注ith从游戏一开始,手牌数为:1, 2, 4, 7, 11, 16, 22, 29, 37, 46,无论值多少,似乎都是 n(n+1)/2 + 1 ——而我不够聪明,无法弄清楚这一点,我用过OEIS http://oeis.org/. :)
这是正在运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类似,请告诉我(我已经查看了许多类似标题的线程,但没有找到可行的解决方案)为我)。