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.)

这是对奥哈德·纳冯的回答的一点补充。

如果元素的中心属于另一个元素，我们就找不到它。

为了确保元素的其中一个点是可见的

function isElementVisible(elem) {
    if (!(elem instanceof Element)) throw Error('DomUtil: elem is not an element.');
    const style = getComputedStyle(elem);
    if (style.display === 'none') return false;
    if (style.visibility !== 'visible') return false;
    if (style.opacity === 0) return false;
    if (elem.offsetWidth + elem.offsetHeight + elem.getBoundingClientRect().height +
        elem.getBoundingClientRect().width === 0) {
        return false;
    }
    var elementPoints = {
        'center': {
            x: elem.getBoundingClientRect().left + elem.offsetWidth / 2,
            y: elem.getBoundingClientRect().top + elem.offsetHeight / 2
        },
        'top-left': {
            x: elem.getBoundingClientRect().left,
            y: elem.getBoundingClientRect().top
        },
        'top-right': {
            x: elem.getBoundingClientRect().right,
            y: elem.getBoundingClientRect().top
        },
        'bottom-left': {
            x: elem.getBoundingClientRect().left,
            y: elem.getBoundingClientRect().bottom
        },
        'bottom-right': {
            x: elem.getBoundingClientRect().right,
            y: elem.getBoundingClientRect().bottom
        }
    }

    for(index in elementPoints) {
        var point = elementPoints[index];
        if (point.x < 0) return false;
        if (point.x > (document.documentElement.clientWidth || window.innerWidth)) return false;
        if (point.y < 0) return false;
        if (point.y > (document.documentElement.clientHeight || window.innerHeight)) return false;
        let pointContainer = document.elementFromPoint(point.x, point.y);
        if (pointContainer !== null) {
            do {
                if (pointContainer === elem) return true;
            } while (pointContainer = pointContainer.parentNode);
        }
    }
    return false;
}

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

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');
}

仅供参考，应该注意getBoundingClientRect()在某些情况下可以工作。

例如，使用display: none简单检查元素是否被隐藏，可能看起来像这样:

var box = element.getBoundingClientRect();
var visible = box.width && box.height;

这也很方便，因为它还涵盖了零宽度、零高度和位置:固定的情况。但是，它不应该报告使用opacity: 0或visibility: hidden隐藏的元素(但是也不会报告offsetParent)。

对我来说，所有其他的解决方案在某些情况下都失效了。

获胜的答案如下:

http://plnkr.co/edit/6CSCA2fe4Gqt4jCBP2wu?p=preview

最终，我认为最好的解决方案是$(elem).is(':visible')——然而，这不是纯javascript。它是jquery..

所以我偷看了他们的来源，找到了我想要的

jQuery.expr.filters.visible = function( elem ) {
    return !!( elem.offsetWidth || elem.offsetHeight || elem.getClientRects().length );
};

这是来源:https://github.com/jquery/jquery/blob/master/src/css/hiddenVisibleSelectors.js