检查元素是否在屏幕上，而不是公认的检查div是否完全在屏幕上的方法(如果div比屏幕大，这将不起作用)。在纯Javascript中:

/**
 * Checks if element is on the screen (Y axis only), returning true
 * even if the element is only partially on screen.
 *
 * @param element
 * @returns {boolean}
 */
function isOnScreenY(element) {
    var screen_top_position = window.scrollY;
    var screen_bottom_position = screen_top_position + window.innerHeight;

    var element_top_position = element.offsetTop;
    var element_bottom_position = element_top_position + element.offsetHeight;

    return (inRange(element_top_position, screen_top_position, screen_bottom_position)
    || inRange(element_bottom_position, screen_top_position, screen_bottom_position));
}

/**
 * Checks if x is in range (in-between) the
 * value of a and b (in that order). Also returns true
 * if equal to either value.
 *
 * @param x
 * @param a
 * @param b
 * @returns {boolean}
 */
function inRange(x, a, b) {
    return (x >= a && x <= b);
}

我在我的应用程序中有这样一个方法，但它不使用jQuery:

/* Get the TOP position of a given element. */
function getPositionTop(element){
    var offset = 0;
    while(element) {
        offset += element["offsetTop"];
        element = element.offsetParent;
    }
    return offset;
}

/* Is a given element is visible or not? */
function isElementVisible(eltId) {
    var elt = document.getElementById(eltId);
    if (!elt) {
        // Element not found.
        return false;
    }
    // Get the top and bottom position of the given element.
    var posTop = getPositionTop(elt);
    var posBottom = posTop + elt.offsetHeight;
    // Get the top and bottom position of the *visible* part of the window.
    var visibleTop = document.body.scrollTop;
    var visibleBottom = visibleTop + document.documentElement.offsetHeight;
    return ((posBottom >= visibleTop) && (posTop <= visibleBottom));
}

编辑:此方法适用于I.E.(至少版本6)。请阅读评论以了解FF的兼容性。

制作了一个简单的插件，用于检测元素在可滚动容器中是否可见

    $.fn.isVisible = function(){

      var win;
      if(!arguments[0])
      {
        console.error('Specify a target;');
        return false;
      }
      else
      {
        win = $(arguments[0]);
      }
      var viewport = {};
      var bounds = this.offset();
      bounds.right = bounds.left + this.outerWidth();
      bounds.bottom = bounds.top + this.outerHeight();
      viewport.bottom = win.height() + win.offset().top;
      return (!( bounds.top > viewport.bottom) && (win.offset().top < bounds.bottom));
    };

像这样调用$('elem_to_check').isVisible('scrollable_container');

希望能有所帮助。

Javascript代码可以写成:

窗口。addEventListener('scroll'， function() { var element = document.querySelector('#main-container'); var position = element.getBoundingClientRect(); //检查是否完全可见 如果位置。顶部>= 0 &&位置。bottom <= window.innerHeight) { console.log('元素在屏幕上完全可见'); ｝ //检查部分可见性 如果位置。顶部<窗口。innerHeight && position。底部>= 0){ console.log('元素在屏幕上部分可见'); ｝ })；

在react js中写为:

componentDidMount () { 窗口。addEventListener(“滚动”,this.isScrolledIntoView); ｝ componentWillUnmount () { 窗口。removeEventListener(“滚动”,this.isScrolledIntoView); ｝ isScrolledIntoView () { var element = document.querySelector('.element'); var position = element.getBoundingClientRect(); //检查是否完全可见 如果位置。顶部>= 0 &&位置。bottom <= window.innerHeight) { console.log('元素在屏幕上完全可见'); ｝ //检查部分可见性 如果位置。顶部<窗口。innerHeight && position。底部>= 0){ console.log('元素在屏幕上部分可见'); ｝ ｝

我找到的最简单的解决方案是交集观察者API:

var observer = new IntersectionObserver(function(entries) {
    if(entries[0].isIntersecting === true)
        console.log('Element has just become visible in screen');
}, { threshold: [0] });

observer.observe(document.querySelector("#main-container"));

function isScrolledIntoView(elem) {
    var docViewTop = $(window).scrollTop(),
        docViewBottom = docViewTop + $(window).height(),
        elemTop = $(elem).offset().top,
     elemBottom = elemTop + $(elem).height();
   //Is more than half of the element visible
   return ((elemTop + ((elemBottom - elemTop)/2)) >= docViewTop && ((elemTop + ((elemBottom - elemTop)/2)) <= docViewBottom));
}