获得表格中的元素偏移量 [英] Getting offsetTop of element in a table

问题描述

我似乎无法弄清楚如何获得表格中元素的偏移量。它适用于表格外的元素，但表格中的所有元素都返回相同的结果，并且通常位于页面的顶部。我在Firefox和Chrome中尝试了这一点。如何获得表中元素的offsetTop？

解决方案

offsetTop 返回一个相对于 offsetParent 的值;您需要通过所有父母递归添加 offsetParent.offsetTop ，直到 offsetParent 为 null 。考虑使用 jQuery 的 offset （）方法。

编辑：如果你不想使用jQuery，你可以编写一个像这样的方法（未经测试）：

 函数偏移量（elem）{
 if（！elem）elem = this; 
 
 var x = elem.offsetLeft; 
 var y = elem.offsetTop; 
 
 while（elem = elem.offsetParent）{
 x + = elem.offsetLeft; 
 y + = elem.offsetTop; 
} 
 
 return {left：x，top：y}; 
} 
  

I can't seem to figure out how to get the offsetTop of an element within a table. It works fine on elements outside tables, but all of the elements within a table return the same result, and it's usually at the top of the page. I tried this in Firefox and Chrome. How do I get the offsetTop of an element in a table?

解决方案

offsetTop returns a value relative to offsetParent; you need to recursively add offsetParent.offsetTop through all of the parents until offsetParent is null. Consider using jQuery's offset() method.

EDIT: If you don't want to use jQuery, you can write a method like this (untested):

function offset(elem) {
    if(!elem) elem = this;

    var x = elem.offsetLeft;
    var y = elem.offsetTop;

    while (elem = elem.offsetParent) {
        x += elem.offsetLeft;
        y += elem.offsetTop;
    }

    return { left: x, top: y };
}

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