这里有另一个解决方案:

<script type="text/javascript">
$.fn.is_on_screen = function(){
    var win = $(window);
    var viewport = {
        top : win.scrollTop(),
        left : win.scrollLeft()
    };
    viewport.right = viewport.left + win.width();
    viewport.bottom = viewport.top + win.height();

    var bounds = this.offset();
    bounds.right = bounds.left + this.outerWidth();
    bounds.bottom = bounds.top + this.outerHeight();

    return (!(viewport.right < bounds.left || viewport.left > bounds.right ||    viewport.bottom < bounds.top || viewport.top > bounds.bottom));
 };

if( $('.target').length > 0 ) { // if target element exists in DOM
    if( $('.target').is_on_screen() ) { // if target element is visible on screen after DOM loaded
        $('.log').html('<div class="alert alert-success">target element is visible on screen</div>'); // log info       
    } else {
        $('.log').html('<div class="alert">target element is not visible on screen</div>'); // log info
    }
}
$(window).on('scroll', function(){ // bind window scroll event
if( $('.target').length > 0 ) { // if target element exists in DOM
    if( $('.target').is_on_screen() ) { // if target element is visible on screen after DOM loaded
        $('.log').html('<div class="alert alert-success">target element is visible on screen</div>'); // log info
    } else {
        $('.log').html('<div class="alert">target element is not visible on screen</div>'); // log info
    }
}
});
</script>

在JSFiddle中可以看到

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。

这将考虑元素的任何填充、边框或边距，以及大于视口本身的元素。

function inViewport($ele) {
    var lBound = $(window).scrollTop(),
        uBound = lBound + $(window).height(),
        top = $ele.offset().top,
        bottom = top + $ele.outerHeight(true);

    return (top > lBound && top < uBound)
        || (bottom > lBound && bottom < uBound)
        || (lBound >= top && lBound <= bottom)
        || (uBound >= top && uBound <= bottom);
}

要调用它，可以使用这样的代码:

var $myElement = $('#my-element'),
    canUserSeeIt = inViewport($myElement);

console.log(canUserSeeIt); // true, if element is visible; false otherwise

这里的大多数答案都没有考虑到一个元素也可以被隐藏，因为它被滚动出div的视图，而不仅仅是整个页面。

为了排除这种可能性，基本上必须检查元素是否位于其每个父元素的边界内。

这个解决方案正是这样做的:

function(element, percentX, percentY){
    var tolerance = 0.01;   //needed because the rects returned by getBoundingClientRect provide the position up to 10 decimals
    if(percentX == null){
        percentX = 100;
    }
    if(percentY == null){
        percentY = 100;
    }

    var elementRect = element.getBoundingClientRect();
    var parentRects = [];

    while(element.parentElement != null){
        parentRects.push(element.parentElement.getBoundingClientRect());
        element = element.parentElement;
    }

    var visibleInAllParents = parentRects.every(function(parentRect){
        var visiblePixelX = Math.min(elementRect.right, parentRect.right) - Math.max(elementRect.left, parentRect.left);
        var visiblePixelY = Math.min(elementRect.bottom, parentRect.bottom) - Math.max(elementRect.top, parentRect.top);
        var visiblePercentageX = visiblePixelX / elementRect.width * 100;
        var visiblePercentageY = visiblePixelY / elementRect.height * 100;
        return visiblePercentageX + tolerance > percentX && visiblePercentageY + tolerance > percentY;
    });
    return visibleInAllParents;
};

它还允许您指定在每个方向上必须可见的百分比。 它不包括由于其他因素(如display: hidden)而隐藏的可能性。

这应该适用于所有主流浏览器，因为它只使用getBoundingClientRect。我个人在Chrome和Internet Explorer 11上测试了它。

仅限Javascript:)

function isInViewport(element) {
  var rect = element.getBoundingClientRect();
  var html = document.documentElement;
  return (
    rect.top >= 0 &&
    rect.left >= 0 &&
    rect.bottom <= (window.innerHeight || html.clientHeight) &&
    rect.right <= (window.innerWidth || html.clientWidth)
  );
}

我更喜欢使用jQuery expr

jQuery.extend(jQuery.expr[':'], {  
    inview: function (elem) {
        var t = $(elem);
        var offset = t.offset();
        var win = $(window); 
        var winST = win.scrollTop();
        var elHeight = t.outerHeight(true);

        if ( offset.top > winST - elHeight && offset.top < winST + elHeight + win.height()) {
            return true;    
        }    
        return false;  
    }
});

你可以这样用

$(".my-elem:inview"); //returns only element that is in view
$(".my-elem").is(":inview"); //check if element is in view
$(".my-elem:inview").length; //check how many elements are in view

你可以很容易地在滚动事件函数中添加这样的代码等，以检查它每次用户将滚动视图。