查找元素是否可见（JavaScript） [英] Finding if element is visible (JavaScript )

问题描述

我有一个JavaScript函数，试图确定一个div是否可见，并使用该变量执行各种进程。我能成功地通过改变在none和block之间的显示来交换元素的可见性;但我不能存储这个值...

我已经尝试获取元素显示属性值，并查找元素ID是否可见，但两者都没有工作。当我尝试.getAttribute它总是返回null;我不知道为什么，因为我知道id被定义并且具有显示属性。

这是我尝试过的两种不同方法的代码：

  var myvar = $（＃mydivID）。is（：visible）; 
 var myvar = document.getElementById（mydivID）。getAttribute（display）; 
  

非常感谢任何指导或帮助。

解决方案

显示不是属性，它是样式属性中的CSS属性。

您可能会查找

  var myvar = document.getElementById（mydivID）。style.display ; 
  

或

  var myvar = $（＃mydivID）。css（'display'）; 
  

I have a javascript function that tries to determine whether a div is visible and does various processes with that variable. I am successfully able to swap an elements visibility by changing it's display between none and block; but I cannot store this value...

I have tried getting the elements display attribute value and finding if the the element ID is visible but neither has worked. When I try .getAttribute it always returns null; I am not sure why because I know that id is defined and it has a display attribute.

Here is the code of the two different methods I have tried:

var myvar = $("#mydivID").is(":visible");
var myvar = document.getElementById("mydivID").getAttribute("display");

Any guidance or assistance would be greatly appreciated.

解决方案

Display is not an attribute, it's a CSS property inside the style attribute.

You may be lookig for

var myvar = document.getElementById("mydivID").style.display;

or

var myvar = $("#mydivID").css('display');

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