使用jQuery滚动后获取元素的新x/y位置 [英] get new x/y positions of element after scroll using jQuery
问题描述
让我们说我有一个<a>标记，如下所示:
Lets say I have an <a> tag as follows:
<body>
<div class="wrapper">
<a href="#" class="a1">Click Me</a>
</div>
</body>
和我的CSS是:
body{ padding:10px;}
.wrapper { height:1000px; width:500px;}
目前，我正在使用Jquery的.offset()来获取<a>标签的X/Y位置.
Currently I am using .offset() of Jquery to get the X/Y positions of <a> tag.
var offset = $(".a1").offset();
var top = offset.top;
var left = offset.left;
现在，当我滚动页面并检查<a>标签的x，y坐标时，它们保持不变，即页面滚动无效.
滚动与屏幕相关的页面后，我想获得<a>标记的新X，Y位置.
如果此<a>标记在向下滚动后被隐藏，我希望其位置为负值.
Now when I scroll the page and check of <a> tag's x,y co-ordinates, they remain the same, i.e. ineffective of page scrolling.
I want to get the new X,Y positions of <a> tag after scrolling the page related to the screen.
If this <a> tag gets hidden after scrolling down, I want its position in negative values.
检查此小提琴: http://jsfiddle.net/xQh5J/9/
请帮助.
推荐答案
链接元素的位置w.r.t.滚动时，文档不会更改.要获取元素的位置w.r.t. 窗口的左上角，您可以将其取为 offset()并减去窗口的scrollTop形式:
The link element's position w.r.t. the document does not change when you scroll. To get the position of the element w.r.t. the window's top-left, you can take it's offset() and subtract the window's scrollTop form what you get:
var offset = $(".a1").offset();
var w = $(window);
alert("(x,y): ("+(offset.left-w.scrollLeft())+","+(offset.top-w.scrollTop())+")");
演示: http://jsfiddle.net/xQh5J/10/
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