javascript - 在 jQuery 中使用 .fadeOut() 时禁用单击

Question

我正在制作一款纸牌游戏。当用户单击该卡片时，该卡片就会淡出。然而，目前存在一个严重的错误。用户可以多次单击一张卡，将多个卡号添加到一个数组中(本质上是选择很多卡)。

我该如何解决这个问题？这是到目前为止我的代码:

$(clicked_id).fadeOut('fast');

如有任何帮助，我们将不胜感激!

编辑:整个代码:

<script>
var cardsPicked = new Array;

var suits = ["&hearts;", "&diams;", "&clubs;", "&spades;"];
var king = 0;
var numbers = ["A", "2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K"];
var cardRules = ["Waterfall", "You", "Me", "Girls", "Thumbmaster", "Guys", "Heaven", "Mate", "Rhyme", "Catagories", "Make a rule", "Question master", "Pour your drink!"];

var currentId;

function getCard(clicked_id) {

    clicked_id = "#" + clicked_id;

    $(clicked_id).fadeOut('fast');

    var newCard = Math.floor((Math.random() * 13) + 1);

    var newSuit = Math.floor((Math.random() * 4) + 1);
    var currentCard;
    var x = document.getElementById("pick");
    var rules = document.getElementById("rules");
    var kings = document.getElementById("kings");
    var currentCards = document.getElementById("currentCard");

    if (cardsPicked.indexOf(numbers[newCard - 1] + " " + suits[newSuit - 1]) == -1) {

        if (numbers[newCard - 1] == "K" && king < 4) {
            king = king + 1;
        }
        if (king == 4) {
            king = "All kings found!";
            alert("Fourth king picked. Down the jug!");
        }
        cardsPicked.push(numbers[newCard - 1] + " " + suits[newSuit - 1]);
        for (count = 0; count < cardsPicked.length; count++)
        currentRule = cardRules[newCard - 1];

        x.innerHTML = cardsPicked;
        currentCards.innerHTML = numbers[newCard - 1] + " " + suits[newSuit - 1];
        rules.innerHTML = currentRule;
        kings.innerHTML = king;

    } else {

        getCard();
    }

}

Answer 1

试试这个

$(clicked_id).off().fadeOut('fast');

<小时/>

如果该事件是通过内联 onclick 调用的:

$(clicked_id).prop("onclick", "").off().fadeOut('fast');

DEMO