如何在jQuery中相对于父级获取offset()? [英] How to get offset() relative to parent in jQuery?

问题描述

这使我可以从主窗口的左边缘找到某些元素的位置:

This gives me the position of some element from the left edge of the main window:

$('#bar').offset().left;

如果该元素位于其他元素内，并且我想要#bar相对于#foo(它是父级)的位置，那我该怎么办呢?

If that element is situated inside some other element and I wanted the position of #bar relative to #foo (it's parent), how can I get that?

<style>
#foo { width: 200px; margin: 0 auto; }
#foo #bar { width: 50px; margin: 0 auto; }
</style>

<div id="foo">
   <span id="bar"></span>
</div>

我看到有一个名为offsetParent()的函数，但是在控制台登录时，该函数似乎没有任何名为left或x的属性.因此，不确定是否可以用来获取我需要的东西.

I saw that there is a function called offsetParent() but when console logged it doesn't seem like this function has any properties called left or x. So not sure if that can be used to get what I need.

因此在上面的示例中，偏移量应距父对象边缘约125像素，而不是距主窗口边缘约数千个像素.

So in my example above the offset should be something around 125px from the parent's edge rather than some thousands of pixels from the main windows edge.

推荐答案

使用position()方法.

Use the position() method.

$('#bar').position().left;

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