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

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。

其他回答

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;
}

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

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>

有许多情况下,这将不一定工作,但在我的情况下,我正在使用这个,它为我所需要的工作。所以,如果你正在寻找一个基本的解决方案(不包括所有的可能性),如果这个简单的解决方案适合你的特殊需求,它“可能”对你有帮助。

var element= document.getElementById('elementId');

if (element.style.display == "block"){

<!-- element is visible -->

} else {

<!-- element is hidden-->

}