在页面中找到位置为“绝对"的所有元素 [英] find all elements in a page with position "absolute"

问题描述

我需要获取页面中所有位置为绝对"的元素.我无法使用$('*')，因为在页面中的每个元素上进行迭代都会导致性能问题.我的页面中有将近8000个元素，每次打开上下文菜单时，我都需要计算最大的z索引.我进行了搜索，但找不到有效的解决方案.我正在寻找以下提供的解决方案，

I need to get all the elements in a page with position "absolute". I could not use $('*') because iterating on each element in the page is causing performance issue. There are nearly 8000 elements in my page and each time when i open the context menu i need to calculate the maximum z index. I have searched and i could not find an efficient solution. I am looking for solution as provided below,

$(function () {
     $("*[style*='position:absolute']").each (function () {
     alert($(this).html());
    });
});

http://jsfiddle.net/MGv9X/

但是，它仅在样式为内联时才有效.但是当从CSS指定位置时，我也需要所有具有绝对"位置的元素.

but, it only works if style is inline. But i need all the elements with "absolute" position when position is specified from CSS also.

请咨询.预先感谢.

推荐答案

您不能仅通过CSS属性>>来简单地选择元素>>，而不必遍历所有计算出的DOM元素样式，因为它们不是selectors .

You cannot that simply select an element >> by just a CSS property >> without iterating all over your computed DOM element styles, cause those are not selectors.

是的，恐怕是*全球.而且非常慢.但是为了使其更快，请首先收集您的元素，然后将方法应用于您的元素集合:

So yes, * global I'm afraid. And pretty slow. But to make it faster, collect first your elements, and afterwards apply a method to your elements collection:

jsFiddle演示

jsFiddle demo

var absElements = [];                            // Elements collector
$("*").css("position", function(i, pos) {
   if(pos==="absolute") absElements.push(this);  // Collect elements
});

// Now once the above is done...

$(absElements) // Do whatever you want with them

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