Javascript / jquery，获取所有divs位置在（x，y）。转发接触？

编程入门行业动态更新时间:2024-10-25 22:34:28

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可能重复：如何获取位于点击点的所有元素的列表？

我知道我可以通过使用 document.elementFromPoint（x， y）。

问题是我需要获取包含触摸事件位置的每个div。

我如何将触摸传播到下面的元素？ hacky解决方案，显示/隐藏元素，而重新生成事件，或指针事件样式与css，但我不能使用这些，他们可能会导致闪烁...

下图说明了我需要做什么：

如果紫色，绿色和蓝色框代表div元素，红点是触摸位置，我需要一个函数，返回div3，div2 ，div1。

解决方案

此代码不会闪烁：

$（body）click（function（e）{ var x = e.pageX，y = e.pageY; var res = [] var ele = document.elementFromPoint（x，y）; while（ele& ele.tagName！=BODY& ele.tagName！=HTML ）{ res.push（ele）; ele.style.display =none; ele = document.elementFromPoint（x，y）; } for（var i = 0; i

Possible Duplicate: How to get a list of all elements that resides at the clicked point?

I know I can get the element with the highest z-index by using document.elementFromPoint(x,y).

The problem is I need to get every div that contains the touch event's location.

How can I propagate touches to elements underneath?

I have seen some hacky solutions that show/hide elements while re-generating the event, or pointer-events style with css, however I cannot use these and they may cause flickering...

The following diagram illustrates what I need to do:

If the purple, green, and blue boxes represent the div elements, and the red dot is the touch location, I need a function that would return "div3, div2, div1".

解决方案

No flickering with this code:

$("body").click(function(e){ var x = e.pageX, y = e.pageY; var res = []; var ele = document.elementFromPoint(x,y); while(ele && ele.tagName != "BODY" && ele.tagName != "HTML"){ res.push(ele); ele.style.display = "none"; ele = document.elementFromPoint(x,y); } for(var i = 0; i < res.length; i++){ res[i].style.display = ""; } console.log(res); });