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"));