有什么方法,我可以检查如果一个元素是可见的纯JS(没有jQuery) ?

因此,给定一个DOM元素,我如何检查它是否可见?我试着:

window.getComputedStyle(my_element)['display']);

但这似乎并不奏效。我想知道我应该检查哪些属性。我想到了:

display !== 'none'
visibility !== 'hidden'

还有我可能漏掉的吗?


当前回答

来自http://code.jquery.com/jquery-1.11.1.js的jQuery代码有一个isHidden参数

var isHidden = function( elem, el ) {
    // isHidden might be called from jQuery#filter function;
    // in that case, element will be second argument
    elem = el || elem;
    return jQuery.css( elem, "display" ) === "none" || !jQuery.contains( elem.ownerDocument, elem );
};

因此,看起来有一个与所有者文档相关的额外检查

我想知道这是否真的适用于以下情况:

基于zIndex隐藏在其他元素后面的元素 完全透明的元素使它们不可见 位于屏幕外的元素(即左:-1000px) 具有可见性的元素:隐藏 有显示的元素:无 没有可见文本或子元素的元素 高度或宽度设置为0的元素

其他回答

let element = document.getElementById('element');
let rect = element.getBoundingClientRect();

if(rect.top == 0 && 
  rect.bottom == 0 && 
  rect.left == 0 && 
  rect.right == 0 && 
  rect.width == 0 && 
  rect.height == 0 && 
  rect.x == 0 && 
  rect.y == 0)
{
  alert('hidden');
}
else
{
  alert('visible');
}

公认的答案对我不起作用。

2020年分解。

The (elem.offsetParent !== null) method works fine in Firefox but not in Chrome. In Chrome position: fixed will also make offsetParent return null even the element if visible in the page. User Phrogz conducted a large test (2,304 divs) on elements with varying properties to demonstrate the issue. https://stackoverflow.com/a/11639664/4481831 . Run it with multiple browsers to see the differences. Demo: //different results in Chrome and Firefox console.log(document.querySelector('#hidden1').offsetParent); //null Chrome & Firefox console.log(document.querySelector('#fixed1').offsetParent); //null in Chrome, not null in Firefox <div id="hidden1" style="display:none;"></div> <div id="fixed1" style="position:fixed;"></div> The (getComputedStyle(elem).display !== 'none') does not work because the element can be invisible because one of the parents display property is set to none, getComputedStyle will not catch that. Demo: var child1 = document.querySelector('#child1'); console.log(getComputedStyle(child1).display); //child will show "block" instead of "none" <div id="parent1" style="display:none;"> <div id="child1" style="display:block"></div> </div> The (elem.clientHeight !== 0). This method is not influenced by position: fixed and it also check if element parents are not-visible. But it has problems with simple elements that do not have a css layout and inline elements, see more here Demo: console.log(document.querySelector('#inline1').clientHeight); //zero console.log(document.querySelector('#div1').clientHeight); //not zero console.log(document.querySelector('#span1').clientHeight); //zero <div id="inline1" style="display:inline">test1 inline</div> <div id="div1">test2 div</div> <span id="span1">test3 span</span> The (elem.getClientRects().length !== 0) may seem to solve the problems of the previous 3 methods. However it has problems with elements that use CSS tricks (other then display: none) to hide in the page. Demo console.log(document.querySelector('#notvisible1').getClientRects().length); console.log(document.querySelector('#notvisible1').clientHeight); console.log(document.querySelector('#notvisible2').getClientRects().length); console.log(document.querySelector('#notvisible2').clientHeight); console.log(document.querySelector('#notvisible3').getClientRects().length); console.log(document.querySelector('#notvisible3').clientHeight); <div id="notvisible1" style="height:0; overflow:hidden; background-color:red;">not visible 1</div> <div id="notvisible2" style="visibility:hidden; background-color:yellow;">not visible 2</div> <div id="notvisible3" style="opacity:0; background-color:blue;">not visible 3</div>

结论。

所以我向你们展示的是没有什么方法是完美的。要进行适当的可见性检查,必须结合使用后3种方法。

有许多情况下,这将不一定工作,但在我的情况下,我正在使用这个,它为我所需要的工作。所以,如果你正在寻找一个基本的解决方案(不包括所有的可能性),如果这个简单的解决方案适合你的特殊需求,它“可能”对你有帮助。

var element= document.getElementById('elementId');

if (element.style.display == "block"){

<!-- element is visible -->

} else {

<!-- element is hidden-->

}

如果元素是常规可见的(display:block和visibility:visible),但有些父容器是隐藏的,那么我们可以使用clientWidth和clienttheight来检查。

function isVisible (ele) {
  return  ele.clientWidth !== 0 &&
    ele.clientHeight !== 0 &&
    (ele.style.opacity !== '' ? parseFloat(ele.style.opacity) > 0 : true);
}

活塞(点击这里)

这就是我所做的:

HTML和CSS:默认情况下使元素隐藏

<html>
<body>

<button onclick="myFunction()">Click Me</button>

<p id="demo" style ="visibility: hidden;">Hello World</p> 

</body>
</html> 

JavaScript:增加了一个代码来检查可见性是否被隐藏:

<script>
function myFunction() {
   if ( document.getElementById("demo").style.visibility === "hidden"){
   document.getElementById("demo").style.visibility = "visible";
   }
   else document.getElementById("demo").style.visibility = "hidden";
}
</script>