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。

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)); 
}

我们可以在使用ES6的现代浏览器中做这样的事情:

const isFullySeen = el => el &&
  typeof el.getBoundingClientRect === 'function' &&
  el.getBoundingClientRect()['bottom'] + window.scrollY <= 
    window.innerHeight + window.scrollY && 
  el.getBoundingClientRect()['top'] + window.scrollY <= 
    window.innerHeight + window.scrollY;

我正在寻找一种方法来查看元素是否即将进入视图，所以通过扩展上面的代码段，我设法做到了。我想我应该把这个留在这里，说不定能帮到谁

Elm =是视图中要检查的元素

scrollElement =你可以传递window或者带有滚动的父元素

Offset =如果你想让它在元素在屏幕前200px处触发，那么传递200

isscro冷景的功能(elem, scrole，抵消) { var $elem = $(elem); var $window = $); var docViewTop = $window.scrollTop(); var docViewBottom = docViewTop + $window.height(); var elemTop = $elem.抵消()top; var elemBottom = elemTop + $elem.height() 归来((elemBottom +) > = docViewBottom) &&偏移(elemTop-offset) < = docViewTop) | | ((elemBottom-offset) < = docViewBottom) && (elemTop +偏移)> = docViewTop); 的

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。