是否有一种有效的方法来判断DOM元素(在HTML文档中)当前是否可见(出现在视口中)?

(这个问题指的是Firefox。)


当前回答

我使用这个函数(它只检查y是否在屏幕上,因为大多数时候x是不需要的)

function elementInViewport(el) {
    var elinfo = {
        "top":el.offsetTop,
        "height":el.offsetHeight,
    };

    if (elinfo.top + elinfo.height < window.pageYOffset || elinfo.top > window.pageYOffset + window.innerHeight) {
        return false;
    } else {
        return true;
    }

}

其他回答

新的交集观察者API非常直接地解决了这个问题。

这个解决方案将需要一个polyfill,因为Safari, Opera和Internet Explorer还不支持这个(polyfill包含在解决方案中)。

在这个解决方案中,在视图外有一个框,即目标(观察到的)。当它进入视图时,头部顶部的按钮是隐藏的。一旦框离开视图,就会显示它。

const buttonToHide = document.querySelector('button'); const hideWhenBoxInView = new IntersectionObserver((entries) => { if (entries[0].intersectionRatio <= 0) { // If not in view buttonToHide.style.display = "inherit"; } else { buttonToHide.style.display = "none"; } }); hideWhenBoxInView.observe(document.getElementById('box')); header { position: fixed; top: 0; width: 100vw; height: 30px; background-color: lightgreen; } .wrapper { position: relative; margin-top: 600px; } #box { position: relative; left: 175px; width: 150px; height: 135px; background-color: lightblue; border: 2px solid; } <script src="https://polyfill.io/v2/polyfill.min.js?features=IntersectionObserver"></script> <header> <button>NAVIGATION BUTTON TO HIDE</button> </header> <div class="wrapper"> <div id="box"> </div> </div>

我发现这里公认的答案对于大多数用例来说过于复杂。这段代码很好地完成了工作(使用jQuery),并区分了完全可见和部分可见的元素:

var element         = $("#element");
var topOfElement    = element.offset().top;
var bottomOfElement = element.offset().top + element.outerHeight(true);
var $window         = $(window);

$window.bind('scroll', function() {

    var scrollTopPosition   = $window.scrollTop()+$window.height();
    var windowScrollTop     = $window.scrollTop()

    if (windowScrollTop > topOfElement && windowScrollTop < bottomOfElement) {
        // Element is partially visible (above viewable area)
        console.log("Element is partially visible (above viewable area)");

    } else if (windowScrollTop > bottomOfElement && windowScrollTop > topOfElement) {
        // Element is hidden (above viewable area)
        console.log("Element is hidden (above viewable area)");

    } else if (scrollTopPosition < topOfElement && scrollTopPosition < bottomOfElement) {
        // Element is hidden (below viewable area)
        console.log("Element is hidden (below viewable area)");

    } else if (scrollTopPosition < bottomOfElement && scrollTopPosition > topOfElement) {
        // Element is partially visible (below viewable area)
        console.log("Element is partially visible (below viewable area)");

    } else {
        // Element is completely visible
        console.log("Element is completely visible");
    }
});

我尝试了Dan的答案,然而,用于确定边界的代数意味着元素必须既≤视口大小,又完全在视口内才能为真,很容易导致假否定。如果你想确定一个元素是否在视口中,ryanve的答案是接近的,但被测试的元素应该与视口重叠,所以试试这个:

function isElementInViewport(el) {
    var rect = el.getBoundingClientRect();

    return rect.bottom > 0 &&
        rect.right > 0 &&
        rect.left < (window.innerWidth || document.documentElement.clientWidth) /* or $(window).width() */ &&
        rect.top < (window.innerHeight || document.documentElement.clientHeight) /* or $(window).height() */;
}

对于类似的挑战,我非常喜欢这个要点,它为scrollIntoViewIfNeeded()暴露了一个填充。

所有必要的功夫都需要回答这个问题:

var parent = this.parentNode,
    parentComputedStyle = window.getComputedStyle(parent, null),
    parentBorderTopWidth = parseInt(parentComputedStyle.getPropertyValue('border-top-width')),
    parentBorderLeftWidth = parseInt(parentComputedStyle.getPropertyValue('border-left-width')),
    overTop = this.offsetTop - parent.offsetTop < parent.scrollTop,
    overBottom = (this.offsetTop - parent.offsetTop + this.clientHeight - parentBorderTopWidth) > (parent.scrollTop + parent.clientHeight),
    overLeft = this.offsetLeft - parent.offsetLeft < parent.scrollLeft,
    overRight = (this.offsetLeft - parent.offsetLeft + this.clientWidth - parentBorderLeftWidth) > (parent.scrollLeft + parent.clientWidth),
    alignWithTop = overTop && !overBottom;

这指的是你想知道的元素,例如,overTop或overBottom -你只需要得到漂移…

让我感到困扰的是,该功能没有以jquery为中心的版本可用。当我看到Dan的解决方案时,我发现有机会为那些喜欢用jQuery OO风格编程的人提供一些东西。它很漂亮,很时髦,对我来说很有魅力。

哒哒,哒哒,嘣

$.fn.inView = function(){
    if(!this.length) 
        return false;
    var rect = this.get(0).getBoundingClientRect();

    return (
        rect.top >= 0 &&
        rect.left >= 0 &&
        rect.bottom <= (window.innerHeight || document.documentElement.clientHeight) &&
        rect.right <= (window.innerWidth || document.documentElement.clientWidth)
    );

};

// Additional examples for other use cases
// Is true false whether an array of elements are all in view
$.fn.allInView = function(){
    var all = [];
    this.forEach(function(){
        all.push( $(this).inView() );
    });
    return all.indexOf(false) === -1;
};

// Only the class elements in view
$('.some-class').filter(function(){
    return $(this).inView();
});

// Only the class elements not in view
$('.some-class').filter(function(){
    return !$(this).inView();
});

使用

$(window).on('scroll',function(){

    if( $('footer').inView() ) {
        // Do cool stuff
    }
});