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

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>

其他回答

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-->

}

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。

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)。

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)

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