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

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。

唯一适用于我的解决方案是(当$("#elementToCheck")可见时返回true):

$ (document) .scrollTop () + window.innerHeight + $ (" # elementToCheck ") .height () > $ (" # elementToCheck ") .offset直()上

如果你想在另一个div中滚动项目，

function isScrolledIntoView (elem, divID) 

{

    var docViewTop = $('#' + divID).scrollTop();


    var docViewBottom = docViewTop + $('#' + divID).height();

    var elemTop = $(elem).offset().top;
    var elemBottom = elemTop + $(elem).height();

    return ((elemBottom <= docViewBottom) && (elemTop >= docViewTop)); 
}

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。