我通过AJAX加载元素。其中一些只有当你向下滚动页面时才能看到。有什么方法可以知道元素现在是否在页面的可见部分?


当前回答

isScrolledIntoView是一个非常必要的函数,所以我尝试了它,它适用于不高于视口的元素,但如果元素比视口大,它就不起作用了。要解决这个问题,只需改变条件

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

:

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

请看这里的演示:http://jsfiddle.net/RRSmQ/

其他回答

jQuery有一个名为inview的插件,它添加了一个新的inview事件。


下面是jQuery插件不使用事件的代码:

$.extend($.expr[':'],{
    inView: function(a) {
        var st = (document.documentElement.scrollTop || document.body.scrollTop),
            ot = $(a).offset().top,
            wh = (window.innerHeight && window.innerHeight < $(window).height()) ? window.innerHeight : $(window).height();
        return ot > st && ($(a).height() + ot) < (st + wh);
    }
});

(function( $ ) {
    $.fn.inView = function() {
        var st = (document.documentElement.scrollTop || document.body.scrollTop),
        ot = $(this).offset().top,
        wh = (window.innerHeight && window.innerHeight < $(window).height()) ? window.innerHeight : $(window).height();

        return ot > st && ($(this).height() + ot) < (st + wh);
    };
})( jQuery );

我在一个叫James的家伙的评论中发现了这一点(http://remysharp.com/2009/01/26/element-in-view-event-plugin/)

用香草语回答:

function isScrolledIntoView(el) {
    var rect = el.getBoundingClientRect();
    var elemTop = rect.top;
    var elemBottom = rect.bottom;

    // Only completely visible elements return true:
    var isVisible = (elemTop >= 0) && (elemBottom <= window.innerHeight);
    // Partially visible elements return true:
    //isVisible = elemTop < window.innerHeight && elemBottom >= 0;
    return isVisible;
}

在打印稿

  private readonly isElementInViewPort = (el: HTMLElement): boolean => {
      const rect = el.getBoundingClientRect();
      const elementTop = rect.top;
      const elementBottom = rect.bottom;
      const scrollPosition = el?.scrollTop || document.body.scrollTop;
      return (
        elementBottom >= 0 &&
        elementTop <= document.documentElement.clientHeight &&
        elementTop + rect.height > elementTop &&
        elementTop <= elementBottom &&
        elementTop >= scrollPosition
      );

};

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。

我找到的最简单的解决方案是交集观察者API:

var observer = new IntersectionObserver(function(entries) {
    if(entries[0].isIntersecting === true)
        console.log('Element has just become visible in screen');
}, { threshold: [0] });

observer.observe(document.querySelector("#main-container"));