JQuery:当元素在视图中时触发动作

Question

在我网站的页脚中,我使用counUp.js(链接:http://inorganik.github.io/countUp.js/ )来计算三个数字.我在网站底部添加了此代码:

   <script type="text/javascript">
          var c1 = new countUp("upnum1", 1, 27, 0, 4);
          var c2 = new countUp("upnum2", 1, 10, 0, 4);
          var c3 = new countUp("upnum3", 1, 27, 0, 4);
          var c4 = new countUp("upnum4", 1, 1000, 0, 4);
          window.onload = function() {
            c1.start();
            c2.start();
            c3.start();
            c4.start();
          }
    </script>

这很好用,但当然加载页面后开始计数.如果数字的包含div是"在视图中"而不是在加载页面时,如何设置触发此效果？尝试了几个jQuery的东西,但找不到有效的解决方案...谢谢!

Answer 1

可以通过简单的检查来查看元素是否在视口中.

您可以选择使用插件或纯JQuery.

相关HTML:

<div id="inViewport">Is this in the viewport?</div>

纯JQuery:

这是小提琴:http: //jsfiddle.net/zWtkc/1/

$.fn.isOnScreen = function(){

    var win = $(window);

    var viewport = {
        top : win.scrollTop(),
        left : win.scrollLeft()
    };
    viewport.right = viewport.left + win.width();
    viewport.bottom = viewport.top + win.height();

    var bounds = this.offset();
    bounds.right = bounds.left + this.outerWidth();
    bounds.bottom = bounds.top + this.outerHeight();

    return (!(viewport.right < bounds.left || viewport.left > bounds.right || viewport.bottom < bounds.top || viewport.top > bounds.bottom));

};

$(document).ready(function(){
    $(window).scroll(function(){
        if ($('#inViewport').isOnScreen()) {
            // The element is visible, do something
            alert("in viewport!");
        } else {
            // The element is NOT visible, do something else
        }
    });
});

插件:

你可以使用这个插件:https://github.com/teamdf/jquery-visible/

$(document).ready(function(){
    if ($('#inViewport').visible(true)) {
        // The element is visible, do something
    } else {
        // The element is NOT visible, do something else
    }
});

或者您可以使用此插件:http://www.appelsiini.net/projects/viewport,它允许视口选择器,您的代码将如下所示:

$('#inViewport:in-viewport').doSomething();