下面是一个(纯纯的JS)函数，它执行大量的检查，确保给定的元素对用户可见:

function isVisible(element) {
    // Check if the element is null or undefined
    if (!element) return false;

    // Get the element's bounding client rect
    const boundingRect = element.getBoundingClientRect();

    // Check if the element has a positive width and height
    if (boundingRect.width <= 0 || boundingRect.height <= 0) return false;

    // Check if the element's top and left values are within the viewport
    const top = boundingRect.top;
    const left = boundingRect.left;
    const viewportWidth = window.innerWidth || document.documentElement.clientWidth;
    const viewportHeight = window.innerHeight || document.documentElement.clientHeight;
    if (top > viewportHeight || left > viewportWidth) return false;

    // Check if the element's right and bottom values are within the viewport
    const right = boundingRect.right;
    const bottom = boundingRect.bottom;
    if (right < 0 || bottom < 0) return false;

    // Check if the element is hidden by the overflow property
    const parentNode = element.parentNode;
    if (parentNode && getComputedStyle(parentNode).overflow === 'hidden') {
        const parentRect = parentNode.getBoundingClientRect();
        if (top < parentRect.top || bottom > parentRect.bottom || left < parentRect.left || right > parentRect.right) {
            return false;
        }
    }

    const elementComputedStyle = getComputedStyle(element);

    // Check if the element has a z-index of less than 0
    const zIndex = elementComputedStyle.zIndex;
    if (zIndex < 0) return false;

    // Check if the element has a display value of 'none' or an opacity of 0
    const display = elementComputedStyle.display;
    const opacity = elementComputedStyle.opacity;
    if (display === 'none' || opacity === '0') return false;

    // Check if the element is hidden by an ancestor element with a display value of 'none' or an opacity of 0
    let ancestorElement = element.parentElement;
    while (ancestorElement) {
        const ancestorComputedStyle = getComputedStyle(ancestorElement);
        const ancestorDisplay = ancestorComputedStyle.display;
        const ancestorOpacity = ancestorComputedStyle.opacity;
        if (ancestorDisplay === 'none' || ancestorOpacity === '0') return false;
        ancestorElement = ancestorElement.parentElement;
    }

    // Initialize a variable to keep track of whether the element is obscured by another element
    let obscured = false;

    // Check if the element is obscured by another element according to its position
    if (elementComputedStyle.position === 'absolute' || elementComputedStyle.position === 'fixed' ||
        elementComputedStyle.position === 'relative' || elementComputedStyle.position === 'sticky' ||
        elementComputedStyle.position === 'static') {
        let siblingElement = element.nextElementSibling;
        while (siblingElement) {
            if (siblingElement.getBoundingClientRect().top > boundingRect.bottom || siblingElement.getBoundingClientRect().left > boundingRect.right) {
                break;
            }
            if (siblingElement.getBoundingClientRect().bottom > boundingRect.top && siblingElement.getBoundingClientRect().right > boundingRect.left) {
                obscured = true;
                break;
            }
            siblingElement = siblingElement.nextElementSibling;
        }
        if (obscured) return false;
    }

    // If all checks have passed, the element is visible
    return true;
}

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

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.)

如果你正在抓取网站，一个非常低效的方法对我来说是突出显示任何元素，然后截图，然后检查截图是否发生了变化。

//Screenshot

function makeSelected(element){
    let range = new Range()
    range.selectNode(element)
    let selection = window.getSelection()
    selection.removeAllRanges()
    selection.addRange(range)
}
// screenshot again and check for diff

Chrome 105(以及Edge和Opera)和Firefox 106引入了element . checkvisibility()，如果元素是可见的，则返回true，否则返回false。

该函数检查了使元素不可见的各种因素，包括display:none、可见性、内容可见性和不透明度:

let element = document.getElementById("myIcon");
let isVisible = element.checkVisibility({
    checkOpacity: true,      // Check CSS opacity property too
    checkVisibilityCSS: true // Check CSS visibility property too
});

旁注:checkVisibility()以前被称为isVisible()。看这个GitHub问题。 参见这里的checkVisibility()规范草案。

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。