30

我需要为以下问题找到一个好的解决方案。我看到很多人询问跟踪元素是否在页面或浏览器窗口的视口之内或之外。我需要能够复制这个动作，但是在一个滚动的 DIV 中，例如溢出：滚动。有谁知道这个具体行动的好例子吗？

提前致谢。

4

8 回答 8

37

这是已接受答案的纯 javascript 版本，不依赖于 jQuery，并对部分视图检测和顶部视图外的支持进行了一些修复。

function checkInView(container, element, partial) {

    //Get container properties
    let cTop = container.scrollTop;
    let cBottom = cTop + container.clientHeight;

    //Get element properties
    let eTop = element.offsetTop;
    let eBottom = eTop + element.clientHeight;

    //Check if in view    
    let isTotal = (eTop >= cTop && eBottom <= cBottom);
    let isPartial = partial && (
      (eTop < cTop && eBottom > cTop) ||
      (eBottom > cBottom && eTop < cBottom)
    );

    //Return outcome
    return  (isTotal  || isPartial);
}

作为奖励，如果元素不是（部分或完整），此功能可确保元素在视图中：

function ensureInView(container, element) {

    //Determine container top and bottom
    let cTop = container.scrollTop;
    let cBottom = cTop + container.clientHeight;

    //Determine element top and bottom
    let eTop = element.offsetTop;
    let eBottom = eTop + element.clientHeight;

    //Check if out of view
    if (eTop < cTop) {
      container.scrollTop -= (cTop - eTop);
    }
    else if (eBottom > cBottom) {
      container.scrollTop += (eBottom - cBottom);
    }
}
于 2016-05-17T20:07:17.417 回答
36

我之前遇到过同样的问题，我最终得到了以下函数。第一个参数是要检查的元素，第二个是检查元素是否部分在视图中。它仅用于垂直检查，您可以扩展它检查水平滚动。

function checkInView(elem,partial)
{
    var container = $(".scrollable");
    var contHeight = container.height();
    var contTop = container.scrollTop();
    var contBottom = contTop + contHeight ;

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

    var isTotal = (elemTop >= 0 && elemBottom <=contHeight);
    var isPart = ((elemTop < 0 && elemBottom > 0 ) || (elemTop > 0 && elemTop <= container.height())) && partial ;

    return  isTotal  || isPart ;
}

在jsFiddle上检查它。

于 2013-04-30T21:19:39.207 回答
7

基于最佳答案。而不是仅仅告诉你一个元素是否部分可见。我添加了一些额外的内容，因此您可以传入一个百分比（0-100），告诉您元素是否超过 x% 可见。

function (container, element, partial) {
    var cTop = container.scrollTop;
    var cBottom = cTop + container.clientHeight;
    var eTop = element.offsetTop;
    var eBottom = eTop + element.clientHeight;
    var isTotal = (eTop >= cTop && eBottom <= cBottom);
    var isPartial;

    if (partial === true) {
        isPartial = (eTop < cTop && eBottom > cTop) || (eBottom > cBottom && eTop < cBottom);
    } else if(typeof partial === "number"){
        if (eTop < cTop && eBottom > cTop) {
            isPartial = ((eBottom - cTop) * 100) / element.clientHeight > partial;
        } else if (eBottom > cBottom && eTop < cBottom){ 
            isPartial = ((cBottom - eTop) * 100) / element.clientHeight > partial;
        }
    }
    return (isTotal || isPartial);
}
于 2017-04-13T08:07:18.910 回答
4

我能够通过对发布的纯 javascript 版本进行一些小改动来完成这项工作

function checkInView(container, element, partial) {

    //Get container properties
    let cTop = container.scrollTop;
    let cBottom = cTop + container.clientHeight;

    //Get element properties
    let eTop = element.offsetTop - container.offsetTop; // change here
    let eBottom = eTop + element.clientHeight;

    //Check if in view    
    let isTotal = (eTop >= cTop && eBottom <= cBottom);
    let isPartial = partial && (
      (eTop < cTop && eBottom > cTop) ||
      (eBottom > cBottom && eTop < cBottom)
    );

    //Return outcome
    return  (isTotal  || isPartial);
  }
于 2018-03-13T10:09:34.973 回答
1

为了我的目的而玩弄它。这是我的解决方案（香草）

Menu 是容器，el 是活动元素。

const isVisible = (menu, el) => {
      const menuHeight = menu.offsetHeight;
      const menuScrollOffset = menu.scrollTop;
    
      const elemTop = el.offsetTop - menu.offsetTop;
      const elemBottom = elemTop + el.offsetHeight;
      return (elemTop >= menuScrollOffset &&
     elemBottom <= menuScrollOffset + menuHeight);
    }
于 2020-11-30T13:11:20.007 回答
1

这是一个纯javascript解决方案。

function elementIsVisible(element, container, partial) {
    var contHeight = container.offsetHeight,
    elemTop = offset(element).top - offset(container).top,
    elemBottom = elemTop + element.offsetHeight;
    return (elemTop >= 0 && elemBottom <= contHeight) || 
    (partial && ((elemTop < 0 && elemBottom > 0 ) || (elemTop > 0 && elemTop <= contHeight)))
}

// checks window
function isWindow( obj ) {
    return obj != null && obj === obj.window;
}

// returns corresponding window
function getWindow( elem ) {
    return isWindow( elem ) ? elem : elem.nodeType === 9 && elem.defaultView;
}

// taken from jquery
// @returns {{top: number, left: number}} 
function offset( elem ) {

    var docElem, win,
        box = { top: 0, left: 0 },
        doc = elem && elem.ownerDocument;

    docElem = doc.documentElement;

    if ( typeof elem.getBoundingClientRect !== typeof undefined ) {
        box = elem.getBoundingClientRect();
    }
    win = getWindow( doc );
    return {
        top: box.top + win.pageYOffset - docElem.clientTop,
        left: box.left + win.pageXOffset - docElem.clientLeft
    };
};
于 2016-10-31T11:42:45.513 回答
0

我用最后一个答案制作了一个 jquery 插件：

(function($) {
    $.fn.reallyVisible = function(opt) {

        var options = $.extend({
            cssChanges:[
                { name : 'visibility', states : ['hidden','visible'] }
            ],
            childrenClass:'mentioners2',
            partialview : true
        }, opt);

        var container = $(this);
        var contHeight;
        var contTop;
        var contBottom;
        var _this = this;
        var _children;

        this.checkInView = function(elem,partial){

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

            var isTotal = (elemTop >= 0 && elemBottom <=contHeight);
            var isPart = ((elemTop < 0 && elemBottom > 0 ) || (elemTop > 0 && elemTop <= container.height())) && partial ;

            return  isTotal  || isPart ;
        }

        this.bind('restoreProperties',function(){
            $.each(_children,function(i,elem){
                $.each(options.cssChanges,function(i,_property){
                    $(elem).css(_property.name,_property.states[1]);        
                });
            });
            _children = null;
        });

        return this.each(function(){
            contHeight = container.height();
            contTop = container.scrollTop();
            contBottom = contTop + contHeight ;

            _children = container.children("."+options.childrenClass);

            $.each(_children,function(i,elem){
                var res = _this.checkInView(elem,options.partialview);
                if(  !res ){
                    $.each(options.cssChanges,function(i,_property){
                        $(elem).css(_property.name,_property.states[0]);        
                    });
                }
            });

        }); 
    }

})(jQuery);
于 2013-09-18T18:40:58.187 回答
0

你可以试试这个

function isScrolledIntoView(elem) {
    var docViewTop = $(window).scrollTop();
    var docViewBottom = docViewTop + window.innerHeight;
    var el = $(elem);
    var elemTop = el.offset().top;
    var elemBottom = elemTop + el.height();
    var elemDisplayNotNone = el.css("display") !== "none";

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

例如：

isScrolledIntoView('#button')
于 2019-05-08T05:17:37.587 回答