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

改进了上面@Guy Messika的回答，如果中心点' X < 0是错误的，则中断并返回false，因为元素右侧可能会进入视图。这里有一个解决方案:

private isVisible(elem) {
    const style = getComputedStyle(elem);

    if (style.display === 'none') return false;
    if (style.visibility !== 'visible') return false;
    if ((style.opacity as any) === 0) return false;

    if (
        elem.offsetWidth +
        elem.offsetHeight +
        elem.getBoundingClientRect().height +
        elem.getBoundingClientRect().width === 0
    ) return false;

    const elementPoints = {
        center: {
            x: elem.getBoundingClientRect().left + elem.offsetWidth / 2,
            y: elem.getBoundingClientRect().top + elem.offsetHeight / 2,
        },
        topLeft: {
            x: elem.getBoundingClientRect().left,
            y: elem.getBoundingClientRect().top,
        },
        topRight: {
            x: elem.getBoundingClientRect().right,
            y: elem.getBoundingClientRect().top,
        },
        bottomLeft: {
            x: elem.getBoundingClientRect().left,
            y: elem.getBoundingClientRect().bottom,
        },
        bottomRight: {
            x: elem.getBoundingClientRect().right,
            y: elem.getBoundingClientRect().bottom,
        },
    };

    const docWidth = document.documentElement.clientWidth || window.innerWidth;
    const docHeight = document.documentElement.clientHeight || window.innerHeight;

    if (elementPoints.topLeft.x > docWidth) return false;
    if (elementPoints.topLeft.y > docHeight) return false;
    if (elementPoints.bottomRight.x < 0) return false;
    if (elementPoints.bottomRight.y < 0) return false;

    for (let index in elementPoints) {
        const point = elementPoints[index];
        let pointContainer = document.elementFromPoint(point.x, point.y);
        if (pointContainer !== null) {
            do {
                if (pointContainer === elem) return true;
            } while (pointContainer = pointContainer.parentNode);
        }
    }
    return false;
}

这是一种确定所有css属性(包括可见性)的方法:

html:

<div id="element">div content</div>

css:

#element
{
visibility:hidden;
}

javascript:

var element = document.getElementById('element');
 if(element.style.visibility == 'hidden'){
alert('hidden');
}
else
{
alert('visible');
}

它适用于任何css属性，非常通用和可靠。

2021的解决方案

根据MDN文档，交互观察器异步观察目标元素与祖先元素或顶级文档视口的交集中的变化。这意味着每当元素与视口相交时，交互观察器就会触发。

截至2021年，除IE外，目前所有浏览器都支持交集观测器。

实现

const el = document.getElementById("your-target-element");
const observer = new IntersectionObserver((entries) => {
    if(entries[0].isIntersecting){
         // el is visible
    } else {
         // el is not visible
    }
});

observer.observe(el); // Asynchronous call

The handler will fire when initially created. And then it will fire every time that it becomes slightly visible or becomes completely not visible. An element is deemed to be not-visible when it's not actually visible within the viewport. So if you scroll down and element goes off the screen, then the observer will trigger and the // el is not visible code will be triggered - even though the element is still "displayed" (i.e. doesn't have display:none or visibility:hidden). What matters is whether there are any pixels of the element that are actually visible within the viewport.

let element = document.getElementById('element');
let rect = element.getBoundingClientRect();

if(rect.top == 0 && 
  rect.bottom == 0 && 
  rect.left == 0 && 
  rect.right == 0 && 
  rect.width == 0 && 
  rect.height == 0 && 
  rect.x == 0 && 
  rect.y == 0)
{
  alert('hidden');
}
else
{
  alert('visible');
}

对我来说，所有其他的解决方案在某些情况下都失效了。

获胜的答案如下:

http://plnkr.co/edit/6CSCA2fe4Gqt4jCBP2wu?p=preview

最终，我认为最好的解决方案是$(elem).is(':visible')——然而，这不是纯javascript。它是jquery..

所以我偷看了他们的来源，找到了我想要的

jQuery.expr.filters.visible = function( elem ) {
    return !!( elem.offsetWidth || elem.offsetHeight || elem.getClientRects().length );
};

这是来源:https://github.com/jquery/jquery/blob/master/src/css/hiddenVisibleSelectors.js