我试图在Javascript中编写一个视频扑克游戏,作为一种获得它的基础知识的方式,我遇到了一个问题,jQuery点击事件处理程序被多次触发。
它们附着在按钮上,用于下注,在游戏中可以很好地进行第一次下注(只发射一次);但在第二只手的投注中,每次按下赌注或下注按钮时,它都会触发单击事件两次(因此每次按下赌注的正确金额为两次)。总的来说,当按下打赌按钮一次时,点击事件触发的次数遵循这种模式——其中序列的i项是用于从游戏开始的i手的投注:1、2、4、7、11、16、22、29、37、46,这似乎是n(n+1)/2 +1,无论价值如何——我不够聪明,我使用了OEIS。:)
这是一个点击事件处理程序的函数;希望这很容易理解(如果不容易,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
}
});
}
如果你有任何想法或建议,或者我的问题的解决方案与这里的另一个问题的解决方案相似,请告诉我(我看过许多类似标题的帖子,但没有幸运地找到一个适合我的解决方案)。