检查元素在DOM中是否可见

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

结合上面的几个答案:

function isVisible (ele) {
    var style = window.getComputedStyle(ele);
    return  style.width !== "0" &&
    style.height !== "0" &&
    style.opacity !== "0" &&
    style.display!=='none' &&
    style.visibility!== 'hidden';
}

就像AlexZ说的，这可能会比你的一些其他选择更慢，如果你更具体地知道你在寻找什么，但这应该抓住所有隐藏元素的主要方式。

但是，这也取决于你认为什么是可见的。例如，一个div的高度可以设置为0px，但内容仍然可见，这取决于溢出属性。或者，可以将div的内容设置为与背景相同的颜色，这样用户就不会看到它，但仍然可以在页面上显示。或者一个div可以移出屏幕或隐藏在其他div后面，或者它的内容可以是不可见的，但边界仍然可见。在一定程度上，“可见”是一个主观术语。

2015-07-09 14:25:01

其他回答

这可能会有帮助: 将元素隐藏在最左边的位置，然后检查offsetLeft属性。如果你想使用jQuery，你可以简单地检查:visible选择器并获得元素的可见状态。

HTML:

<div id="myDiv">Hello</div>

CSS:

<!-- for javaScript-->
#myDiv{
   position:absolute;
   left : -2000px;
}

<!-- for jQuery -->
#myDiv{
    visibility:hidden;
}

javaScript:

var myStyle = document.getElementById("myDiv").offsetLeft;

if(myStyle < 0){
     alert("Div is hidden!!");
}

jQuery:

if(  $("#MyElement").is(":visible") == true )
{  
     alert("Div is visible!!");        
}

js小提琴

2013-10-29 21:53:30

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>

2018-05-27 09:18:36

如果我们只是收集检测能见度的基本方法，让我不要忘记:

opacity > 0.01; // probably more like .1 to actually be visible, but YMMV

至于如何获取属性:

element.getAttribute(attributename);

所以，在你的例子中:

document.getElementById('snDealsPanel').getAttribute('visibility');

But wha? It doesn't work here. Look closer and you'll find that visibility is being updated not as an attribute on the element, but using the style property. This is one of many problems with trying to do what you're doing. Among others: you can't guarantee that there's actually something to see in an element, just because its visibility, display, and opacity all have the correct values. It still might lack content, or it might lack a height and width. Another object might obscure it. For more detail, a quick Google search reveals this, and even includes a library to try solving the problem. (YMMV)

看看下面的问题，它们可能是这个问题的副本，有很好的答案，包括来自强大的约翰·雷西格的一些见解。但是，您的特定用例与标准用例略有不同，因此我将避免标记:

如何判断一个DOM元素是否在当前视口中可见? 如何检查一个元素是否真的可见javascript?

(EDIT: OP SAYS HE'S SCRAPING PAGES, NOT CREATING THEM, SO BELOW ISN'T APPLICABLE) A better option? Bind the visibility of elements to model properties and always make visibility contingent on that model, much as Angular does with ng-show. You can do that using any tool you want: Angular, plain JS, whatever. Better still, you can change the DOM implementation over time, but you'll always be able to read state from the model, instead of the DOM. Reading your truth from the DOM is Bad. And slow. Much better to check the model, and trust in your implementation to ensure that the DOM state reflects the model. (And use automated testing to confirm that assumption.)

2013-10-29 22:04:49

为了详细说明大家的精彩回答，下面是Mozilla Fathom项目中使用的实现:

/**
 * Yield an element and each of its ancestors.
 */
export function *ancestors(element) {
    yield element;
    let parent;
    while ((parent = element.parentNode) !== null && parent.nodeType === parent.ELEMENT_NODE) {
        yield parent;
        element = parent;
    }
}

/**
 * Return whether an element is practically visible, considering things like 0
 * size or opacity, ``visibility: hidden`` and ``overflow: hidden``.
 *
 * Merely being scrolled off the page in either horizontally or vertically
 * doesn't count as invisible; the result of this function is meant to be
 * independent of viewport size.
 *
 * @throws {Error} The element (or perhaps one of its ancestors) is not in a
 *     window, so we can't find the `getComputedStyle()` routine to call. That
 *     routine is the source of most of the information we use, so you should
 *     pick a different strategy for non-window contexts.
 */
export function isVisible(fnodeOrElement) {
    // This could be 5x more efficient if https://github.com/w3c/csswg-drafts/issues/4122 happens.
    const element = toDomElement(fnodeOrElement);
    const elementWindow = windowForElement(element);
    const elementRect = element.getBoundingClientRect();
    const elementStyle = elementWindow.getComputedStyle(element);
    // Alternative to reading ``display: none`` due to Bug 1381071.
    if (elementRect.width === 0 && elementRect.height === 0 && elementStyle.overflow !== 'hidden') {
        return false;
    }
    if (elementStyle.visibility === 'hidden') {
        return false;
    }
    // Check if the element is irrevocably off-screen:
    if (elementRect.x + elementRect.width < 0 ||
        elementRect.y + elementRect.height < 0
    ) {
        return false;
    }
    for (const ancestor of ancestors(element)) {
        const isElement = ancestor === element;
        const style = isElement ? elementStyle : elementWindow.getComputedStyle(ancestor);
        if (style.opacity === '0') {
            return false;
        }
        if (style.display === 'contents') {
            // ``display: contents`` elements have no box themselves, but children are
            // still rendered.
            continue;
        }
        const rect = isElement ? elementRect : ancestor.getBoundingClientRect();
        if ((rect.width === 0 || rect.height === 0) && elementStyle.overflow === 'hidden') {
            // Zero-sized ancestors don’t make descendants hidden unless the descendant
            // has ``overflow: hidden``.
            return false;
        }
    }
    return true;
}

它检查每个父元素的不透明度、显示和矩形。

2020-12-28 06:58:46

所以我找到了最可行的方法:

function visible(elm) {
  if(!elm.offsetHeight && !elm.offsetWidth) { return false; }
  if(getComputedStyle(elm).visibility === 'hidden') { return false; }
  return true;
}

这是基于以下事实:

显示:所有元素(即使是嵌套的元素)都没有宽度和高度。可见性即使对于嵌套的元素也是隐藏的。

因此不需要测试offsetParent或在DOM树中循环来测试哪个父对象具有可见性:hidden。这应该可以在ie9中工作。

你可能会说，如果透明度:0和折叠的元素(有宽度但没有高度-反之亦然)也不是真正可见的。但话说回来，它们并不是隐藏的。

2016-04-20 08:19:52

检查元素在DOM中是否可见

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