var visible = document.getElementById("yourelementID's");
 if (visible){
          // make events
 } else
 {
   //other events
 }

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;
}

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)

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。

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)。

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.)