检查元素在DOM中是否可见

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

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.)

2013-10-29 22:04:49

其他回答

这就是我所做的:

HTML和CSS:默认情况下使元素隐藏

<html>
<body>

<button onclick="myFunction()">Click Me</button>

<p id="demo" style ="visibility: hidden;">Hello World</p> 

</body>
</html>

JavaScript:增加了一个代码来检查可见性是否被隐藏:

<script>
function myFunction() {
   if ( document.getElementById("demo").style.visibility === "hidden"){
   document.getElementById("demo").style.visibility = "visible";
   }
   else document.getElementById("demo").style.visibility = "hidden";
}
</script>

2018-05-27 09:18:36

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;
}

2017-12-05 06:29:38

下面是我编写的代码，用于在几个类似的元素中找到唯一可见的元素，并返回其“class”属性的值，而不使用jQuery:

  // Build a NodeList:
  var nl = document.querySelectorAll('.myCssSelector');

  // convert it to array:
  var myArray = [];for(var i = nl.length; i--; myArray.unshift(nl[i]));

  // now find the visible (= with offsetWidth more than 0) item:
  for (i =0; i < myArray.length; i++){
    var curEl = myArray[i];
    if (curEl.offsetWidth !== 0){
      return curEl.getAttribute("class");
    }
  }

2015-08-25 04:01:20

所以我找到了最可行的方法:

function visible(elm) {
  if(!elm.offsetHeight && !elm.offsetWidth) { return false; }
  if(getComputedStyle(elm).visibility === 'hidden') { return false; }
  return true;
}

这是基于以下事实:

显示:所有元素(即使是嵌套的元素)都没有宽度和高度。可见性即使对于嵌套的元素也是隐藏的。

因此不需要测试offsetParent或在DOM树中循环来测试哪个父对象具有可见性:hidden。这应该可以在ie9中工作。

你可能会说，如果透明度:0和折叠的元素(有宽度但没有高度-反之亦然)也不是真正可见的。但话说回来，它们并不是隐藏的。

2016-04-20 08:19:52

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-->

}

2022-01-10 22:47:05

检查元素在DOM中是否可见

推荐文章

最新文章

标签