如何在鼠標上創建彈出div並在單擊時保留

Question

我正在嘗試創建彈出窗口，可以顯示當鼠標懸停在它上面並且在點擊鏈接時會停留。問題是我已經讓彈出窗口顯示當鼠標懸停在它上面但它會在鼠標離開時消失。如何我點擊時彈出窗口會保持顯示。這是我的代碼：

HTML

<div id="pop1" class="popbox">
    <h2>Job Info Search</h2>
    <h2>WRKNo : <input type="text"  /></h2>
    <h2>Result</h2>
    <p>Customer Name : <input type="text"  /></p>
    <p>Caller Number : <input type="text"  /></p>
    <p>Complosed : <input type="text"  /></p>
    <p>Cate : <input type="text"  /></p>
    <p>Det : <input type="text"  /></p>
    <p>Feedback : <input type="text"  /></p>
    <p>WRKNo : <input type="text"  /></p>
</div>




This is a popbox test.  <a href="#" class="popper" data-popbox="pop1">Hover here</a> to see how it works.

CSS

.popbox {
    display: none;
    position: absolute;
    z-index: 99999;
    width: 400px;
    padding: 10px;
    background: #EEEFEB;
    color: #000000;
    border: 1px solid #4D4F53;
    margin: 0px;
    -webkit-box-shadow: 0px 0px 5px 0px rgba(164, 164, 164, 1);
    box-shadow: 0px 0px 5px 0px rgba(164, 164, 164, 1);
}
.popbox h2
{
    background-color: #4D4F53;
    color:  #E3E5DD;
    font-size: 14px;
    display: block;
    width: 100%;
    margin: -10px 0px 8px -10px;
    padding: 5px 10px;
}

使用Javascript

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js" type="text/javascript"></script>
<script>
$(function() {
    var moveLeft = 0;
    var moveDown = 0;
    $('a.popper').hover(function(e) {

        var target = '#' + ($(this).attr('data-popbox'));

        $(target).show();
        moveLeft = $(this).outerWidth();
        moveDown = ($(target).outerHeight() / 2);
    }, function() {
        var target = '#' + ($(this).attr('data-popbox'));
        $(target).hide();
    });

    $('a.popper').mousemove(function(e) {
        var target = '#' + ($(this).attr('data-popbox'));

        leftD = e.pageX + parseInt(moveLeft);
        maxRight = leftD + $(target).outerWidth();
        windowLeft = $(window).width() - 40;
        windowRight = 0;
        maxLeft = e.pageX - (parseInt(moveLeft) + $(target).outerWidth() + 20);

        if(maxRight > windowLeft && maxLeft > windowRight)
        {
            leftD = maxLeft;
        }

        topD = e.pageY - parseInt(moveDown);
        maxBottom = parseInt(e.pageY + parseInt(moveDown) + 20);
        windowBottom = parseInt(parseInt($(document).scrollTop()) + parseInt($(window).height()));
        maxTop = topD;
        windowTop = parseInt($(document).scrollTop());
        if(maxBottom > windowBottom)
        {
            topD = windowBottom - $(target).outerHeight() - 20;
        } else if(maxTop < windowTop){
            topD = windowTop + 20;
        }

        $(target).css('top', topD).css('left', leftD);


    });

});
</script>

我有什么想法可以做到這一點？

Answer 1

嘗試這個：

$('a.popper').hover(function (e) {    
    var target = '#' + ($(this).attr('data-popbox'));
    $(target).show();
    moveLeft = $(this).outerWidth();
    moveDown = ($(target).outerHeight() / 2);
}, function () {
    var target = '#' + ($(this).attr('data-popbox'));
    if (!($("a.popper").hasClass("show"))) {
        $(target).hide(); //dont hide popup if it is clicked
    }
});
$('a.popper').click(function (e) {
    var target = '#' + ($(this).attr('data-popbox'));
    if (!($(this).hasClass("show"))) {
        $(target).show();
    }
    $(this).toggleClass("show");
});

FIDDLE在這里。

Answer 2

在jquery中使用click方法 ，就像$('a.popper').click(); 並在具有彈出窗口的click方法中傳入參數，類似於使用懸停方法和鼠標懸停方法的方式