我有一个更有效的解决方案相比AlexZ的getComputedStyle()解决方案时，有位置“固定”元素，如果一个愿意忽略一些边缘情况(检查评论):

function isVisible(el) {
    /* offsetParent would be null if display 'none' is set.
       However Chrome, IE and MS Edge returns offsetParent as null for elements
       with CSS position 'fixed'. So check whether the dimensions are zero.

       This check would be inaccurate if position is 'fixed' AND dimensions were
       intentionally set to zero. But..it is good enough for most cases.*/
    return Boolean(el.offsetParent || el.offsetWidth || el.offsetHeight);
}

附注:严格来说，“可见性”首先需要定义。在我的情况下，我正在考虑一个元素可见，只要我可以运行所有DOM方法/属性上没有问题(即使不透明度为0或CSS可见性属性是“隐藏”等)。

根据MDN文档，元素的offsetParent属性将在它或它的任何父元素通过display style属性被隐藏时返回null。只要确保元素不是固定的。一个脚本来检查这个，如果你没有位置:fixed;页面上的元素可能是这样的:

// Where el is the DOM element you'd like to test for visibility
function isHidden(el) {
    return (el.offsetParent === null)
}

另一方面，如果您确实有位置固定的元素可能会在此搜索中被捕获，那么您将不得不遗憾地(并且缓慢地)使用window.getComputedStyle()。这种情况下的函数可能是:

// Where el is the DOM element you'd like to test for visibility
function isHidden(el) {
    var style = window.getComputedStyle(el);
    return (style.display === 'none')
}

选项2可能更简单一点，因为它考虑了更多的边缘情况，但我打赌它也会慢很多，所以如果你不得不多次重复这个操作，最好避免它。

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)

const isVisible = (selector) => { let selectedElement let topElement let selectedData selectedElement = document.querySelector(selector) if (!selectedElement) { return false } selectedData = selectedElement.getBoundingClientRect() if (!selectedData || !Object.keys(selectedData)) { return false } if (!(selectedData.width > 0) || !(selectedData.height > 0)) { return false } topElement = document.elementFromPoint(selectedData.top, selectedData.left) if (selectedElement !== topElement) { return false } return true } const output = document.querySelector('.text') output.innerHTML = '.x element is visible: ' + isVisible('.x') .block { width: 100px; height: 100px; background: black; } .y { background: red; margin-top: -100px; } <div class="text"></div> <div class="x block"></div> <div class="y block"></div>

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。

我有一个更有效的解决方案相比AlexZ的getComputedStyle()解决方案时，有位置“固定”元素，如果一个愿意忽略一些边缘情况(检查评论):

function isVisible(el) {
    /* offsetParent would be null if display 'none' is set.
       However Chrome, IE and MS Edge returns offsetParent as null for elements
       with CSS position 'fixed'. So check whether the dimensions are zero.

       This check would be inaccurate if position is 'fixed' AND dimensions were
       intentionally set to zero. But..it is good enough for most cases.*/
    return Boolean(el.offsetParent || el.offsetWidth || el.offsetHeight);
}

附注:严格来说，“可见性”首先需要定义。在我的情况下，我正在考虑一个元素可见，只要我可以运行所有DOM方法/属性上没有问题(即使不透明度为0或CSS可见性属性是“隐藏”等)。