获取每个元素的位置

Question

$(function(){

    var $animatedEls = $(".marked");

    $(window).scroll(function(e) {

            var offset = 0;

            $.each($animatedEls, function(i, item) {

                offset = $(item).offset().top;

                console.log($(item).offset());

            });

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>
<div>...</div>
<h2 class="marked">This sucks.</h2>

我试图在滚动时获取一些匹配元素的位置。然而，每个元素的输出都是相同的数字。

输出：

Object {top: 2480, left: 0}
Object {top: 2480, left: 0}
Object {top: 2480, left: 0}

为什么每个元素的偏移量都相同？当我滚动时，这些值也在改变。

编辑：好的。该代码段适用于此处，但不适用于我的网站。非常烦人。

Answer 1

问题出在.each的使用上

应该这样使用：

var $animatedEls = $('.marked');

        $(window).scroll(function(e) {

            $.each($animatedEls, function(index, item) {
                console.log($(item).offset());
            }

        }

获取每个元素的位置

Get position of each element

javascript

each

jquery

offset