Question

我正在建立纸牌游戏。当用户点击卡片时，卡片会淡出。然而，目前，存在严重的错误。用户可以多次单击一张卡，将多个卡号添加到阵列中（基本上可以选择大量的卡）。

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();
    }

}

Answer 1

试试这个

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

如果内联onclick调用该事件：

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

DEMO

Answer 2

首先应删除元素上的click事件，然后使用将重新附加click事件的回调函数执行淡出效果。

$('#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，但它可能有用！

在jQuery中使用.fadeOut（）时禁用单击

2 个答案: