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我正在构建一个纸牌游戏。当用户点击卡片时，卡片会淡出。但是，目前，有一个严重的错误。用户可以多次单击一张卡，将多个卡号添加到数组中（本质上是选择很多卡）。

我该如何解决这个问题？到目前为止，这是我的代码：

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

任何帮助，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();
    }

}
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2 回答 2

2

尝试这个

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

如果事件被 inline 调用onclick：

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

演示

于 2012-12-15T21:48:30.807 回答
0

您应该首先删除元素上的单击事件，然后使用回调函数执行淡出效果，该回调函数将重新附加单击事件。

$('#clicked_id').off('click').fadeOut('fast', function () {
    $('#clicked_id').on('click');
});

这是完全未经测试的，只是一个理论，但它可能会奏效！

于 2012-12-15T21:55:08.930 回答