对于类似的挑战，我非常喜欢这个要点，它为scrollIntoViewIfNeeded()暴露了一个填充。

所有必要的功夫都需要回答这个问题:

var parent = this.parentNode,
    parentComputedStyle = window.getComputedStyle(parent, null),
    parentBorderTopWidth = parseInt(parentComputedStyle.getPropertyValue('border-top-width')),
    parentBorderLeftWidth = parseInt(parentComputedStyle.getPropertyValue('border-left-width')),
    overTop = this.offsetTop - parent.offsetTop < parent.scrollTop,
    overBottom = (this.offsetTop - parent.offsetTop + this.clientHeight - parentBorderTopWidth) > (parent.scrollTop + parent.clientHeight),
    overLeft = this.offsetLeft - parent.offsetLeft < parent.scrollLeft,
    overRight = (this.offsetLeft - parent.offsetLeft + this.clientWidth - parentBorderLeftWidth) > (parent.scrollLeft + parent.clientWidth),
    alignWithTop = overTop && !overBottom;

这指的是你想知道的元素，例如，overTop或overBottom -你只需要得到漂移…

我们现在有一个原生javascript交集观察者API 从中我们可以检测元素，无论它们是否在视口中。

这里有一个例子

const el = document.querySelector('#el') const observer = new window.IntersectionObserver(([entry]) => { if (entry. isintersection) { console.log(输入) 返回 ｝ console.log(离开) }, { 根:空, 阈值:0.1，//设置偏移量0.1表示如果元素在视口中至少占10%，则触发 }) observer.observe (el); 身体{ 身高:300 vh; ｝ # el { margin-top: 100 vh; ｝ <div id="el">这是元素</div>

在之前的回答中，大多数用法都没有做到这一点:

-当一个元素的任何像素都是可见的，但不是“一个角落”， -当一个元素大于viewport并且居中时， -大多数只检查文档或窗口内的单个元素。

好吧，对于所有这些问题，我都有一个解决方案，好的方面是:

-你可以返回可见时，只有一个像素从任何一方显示，而不是一个角落， -你仍然可以返回大于viewport的visible while元素， -你可以选择你的父元素，或者你可以自动让它选择。 -也适用于动态添加的元素

如果你检查下面的片段，你会发现在元素的容器中使用溢出滚动的区别不会造成任何麻烦，并且看到与其他答案不同的是，即使一个像素从任何一侧显示，或者当一个元素大于视口时，我们看到的是元素的内部像素，它仍然有效。

用法很简单:

// For checking element visibility from any sides
isVisible(element)

// For checking elements visibility in a parent you would like to check
var parent = document; // Assuming you check if 'element' inside 'document'
isVisible(element, parent)

// For checking elements visibility even if it's bigger than viewport
isVisible(element, null, true) // Without parent choice
isVisible(element, parent, true) // With parent choice

一个没有crossSearchAlgorithm的演示，对于大于viewport的元素，检查element3内部像素，可以看到:

function isVisible(element, parent, crossSearchAlgorithm) { var rect = element.getBoundingClientRect(), prect = (parent != undefined) ? parent.getBoundingClientRect() : element.parentNode.getBoundingClientRect(), csa = (crossSearchAlgorithm != undefined) ? crossSearchAlgorithm : false, efp = function (x, y) { return document.elementFromPoint(x, y) }; // Return false if it's not in the viewport if (rect.right < prect.left || rect.bottom < prect.top || rect.left > prect.right || rect.top > prect.bottom) { return false; } var flag = false; // Return true if left to right any border pixel reached for (var x = rect.left; x < rect.right; x++) { if (element.contains(efp(rect.top, x)) || element.contains(efp(rect.bottom, x))) { flag = true; break; } } // Return true if top to bottom any border pixel reached if (flag == false) { for (var y = rect.top; y < rect.bottom; y++) { if (element.contains(efp(rect.left, y)) || element.contains(efp(rect.right, y))) { flag = true; break; } } } if(csa) { // Another algorithm to check if element is centered and bigger than viewport if (flag == false) { var x = rect.left; var y = rect.top; // From top left to bottom right while(x < rect.right || y < rect.bottom) { if (element.contains(efp(x,y))) { flag = true; break; } if(x < rect.right) { x++; } if(y < rect.bottom) { y++; } } if (flag == false) { x = rect.right; y = rect.top; // From top right to bottom left while(x > rect.left || y < rect.bottom) { if (element.contains(efp(x,y))) { flag = true; break; } if(x > rect.left) { x--; } if(y < rect.bottom) { y++; } } } } } return flag; } // Check multiple elements visibility document.getElementById('container').addEventListener("scroll", function() { var elementList = document.getElementsByClassName("element"); var console = document.getElementById('console'); for (var i=0; i < elementList.length; i++) { // I did not define parent, so it will be element's parent if (isVisible(elementList[i])) { console.innerHTML = "Element with id[" + elementList[i].id + "] is visible!"; break; } else { console.innerHTML = "Element with id[" + elementList[i].id + "] is hidden!"; } } }); // Dynamically added elements for(var i=4; i <= 6; i++) { var newElement = document.createElement("div"); newElement.id = "element" + i; newElement.classList.add("element"); document.getElementById('container').appendChild(newElement); } #console { background-color: yellow; } #container { width: 300px; height: 100px; background-color: lightblue; overflow-y: auto; padding-top: 150px; margin: 45px; } .element { margin: 400px; width: 400px; height: 320px; background-color: green; } #element3 { position: relative; margin: 40px; width: 720px; height: 520px; background-color: green; } #element3::before { content: ""; position: absolute; top: -10px; left: -10px; margin: 0px; width: 740px; height: 540px; border: 5px dotted green; background: transparent; } <div id="console"></div> <div id="container"> <div id="element1" class="element"></div> <div id="element2" class="element"></div> <div id="element3" class="element"></div> </div>

你看，当你在element3内部时，它无法判断它是否可见，因为我们只检查元素是否从侧面或角落可见。

这个包含了crossSearchAlgorithm，它允许你在元素大于viewport时仍然返回visible:

function isVisible(element, parent, crossSearchAlgorithm) { var rect = element.getBoundingClientRect(), prect = (parent != undefined) ? parent.getBoundingClientRect() : element.parentNode.getBoundingClientRect(), csa = (crossSearchAlgorithm != undefined) ? crossSearchAlgorithm : false, efp = function (x, y) { return document.elementFromPoint(x, y) }; // Return false if it's not in the viewport if (rect.right < prect.left || rect.bottom < prect.top || rect.left > prect.right || rect.top > prect.bottom) { return false; } var flag = false; // Return true if left to right any border pixel reached for (var x = rect.left; x < rect.right; x++) { if (element.contains(efp(rect.top, x)) || element.contains(efp(rect.bottom, x))) { flag = true; break; } } // Return true if top to bottom any border pixel reached if (flag == false) { for (var y = rect.top; y < rect.bottom; y++) { if (element.contains(efp(rect.left, y)) || element.contains(efp(rect.right, y))) { flag = true; break; } } } if(csa) { // Another algorithm to check if element is centered and bigger than viewport if (flag == false) { var x = rect.left; var y = rect.top; // From top left to bottom right while(x < rect.right || y < rect.bottom) { if (element.contains(efp(x,y))) { flag = true; break; } if(x < rect.right) { x++; } if(y < rect.bottom) { y++; } } if (flag == false) { x = rect.right; y = rect.top; // From top right to bottom left while(x > rect.left || y < rect.bottom) { if (element.contains(efp(x,y))) { flag = true; break; } if(x > rect.left) { x--; } if(y < rect.bottom) { y++; } } } } } return flag; } // Check multiple elements visibility document.getElementById('container').addEventListener("scroll", function() { var elementList = document.getElementsByClassName("element"); var console = document.getElementById('console'); for (var i=0; i < elementList.length; i++) { // I did not define parent so it will be element's parent // and it will do crossSearchAlgorithm if (isVisible(elementList[i],null,true)) { console.innerHTML = "Element with id[" + elementList[i].id + "] is visible!"; break; } else { console.innerHTML = "Element with id[" + elementList[i].id + "] is hidden!"; } } }); // Dynamically added elements for(var i=4; i <= 6; i++) { var newElement = document.createElement("div"); newElement.id = "element" + i; newElement.classList.add("element"); document.getElementById('container').appendChild(newElement); } #console { background-color: yellow; } #container { width: 300px; height: 100px; background-color: lightblue; overflow-y: auto; padding-top: 150px; margin: 45px; } .element { margin: 400px; width: 400px; height: 320px; background-color: green; } #element3 { position: relative; margin: 40px; width: 720px; height: 520px; background-color: green; } #element3::before { content: ""; position: absolute; top: -10px; left: -10px; margin: 0px; width: 740px; height: 540px; border: 5px dotted green; background: transparent; } <div id="console"></div> <div id="container"> <div id="element1" class="element"></div> <div id="element2" class="element"></div> <div id="element3" class="element"></div> </div>

JSFiddle玩:http://jsfiddle.net/BerkerYuceer/grk5az2c/

这段代码是为了获得更精确的信息，无论元素的任何部分是否显示在视图中。对于性能选项或仅垂直幻灯片，不要使用此选项!这段代码在绘图情况下更有效。

作为Element.getBoundingClientRect()的支持，最简单的解决方案已经变得完美:

function isInView(el) {
  const box = el.getBoundingClientRect();
  return box.top < window.innerHeight && box.bottom >= 0;
}

这是我的解决方案。如果一个元素隐藏在一个可滚动的容器中，它将工作。

这里是一个演示(尝试调整窗口大小为)

var visibleY = function(el){
    var top = el.getBoundingClientRect().top, rect, el = el.parentNode;
    do {
        rect = el.getBoundingClientRect();
        if (top <= rect.bottom === false)
            return false;
        el = el.parentNode;
    } while (el != document.body);
    // Check it's within the document viewport
    return top <= document.documentElement.clientHeight;
};

我只需要检查它在Y轴上是否可见(用于滚动Ajax加载更多记录功能)。

更新:时间在流逝，我们的浏览器也是如此。这种方法不再被推荐，如果你不需要支持ie7之前的版本，你应该使用Dan的解决方案。

原来的解决方案(现已过时):

这将检查元素是否在当前视口中完全可见:

function elementInViewport(el) {
  var top = el.offsetTop;
  var left = el.offsetLeft;
  var width = el.offsetWidth;
  var height = el.offsetHeight;

  while(el.offsetParent) {
    el = el.offsetParent;
    top += el.offsetTop;
    left += el.offsetLeft;
  }

  return (
    top >= window.pageYOffset &&
    left >= window.pageXOffset &&
    (top + height) <= (window.pageYOffset + window.innerHeight) &&
    (left + width) <= (window.pageXOffset + window.innerWidth)
  );
}

你可以简单地修改它，以确定元素的任何部分在视口中是否可见:

function elementInViewport2(el) {
  var top = el.offsetTop;
  var left = el.offsetLeft;
  var width = el.offsetWidth;
  var height = el.offsetHeight;

  while(el.offsetParent) {
    el = el.offsetParent;
    top += el.offsetTop;
    left += el.offsetLeft;
  }

  return (
    top < (window.pageYOffset + window.innerHeight) &&
    left < (window.pageXOffset + window.innerWidth) &&
    (top + height) > window.pageYOffset &&
    (left + width) > window.pageXOffset
  );
}