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
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。
其他回答
这个答案的一个更有效的版本:
/**
* Is element within visible region of a scrollable container
* @param {HTMLElement} el - element to test
* @returns {boolean} true if within visible region, otherwise false
*/
function isScrolledIntoView(el) {
var rect = el.getBoundingClientRect();
return (rect.top >= 0) && (rect.bottom <= window.innerHeight);
}
jQuery Waypoints插件在这里做得非常好。
$('.entry').waypoint(function() {
alert('You have scrolled to an entry.');
});
在插件的站点上有一些例子。
function isScrolledIntoView(elem) {
var docViewTop = $(window).scrollTop(),
docViewBottom = docViewTop + $(window).height(),
elemTop = $(elem).offset().top,
elemBottom = elemTop + $(elem).height();
//Is more than half of the element visible
return ((elemTop + ((elemBottom - elemTop)/2)) >= docViewTop && ((elemTop + ((elemBottom - elemTop)/2)) <= docViewBottom));
}
这是我的纯JavaScript解决方案,如果它隐藏在一个可滚动的容器。
演示在这里(尝试调整窗口的大小)
var visibleY = function(el){
var rect = el.getBoundingClientRect(), top = rect.top, height = rect.height,
el = el.parentNode
// Check if bottom of the element is off the page
if (rect.bottom < 0) return false
// Check its within the document viewport
if (top > document.documentElement.clientHeight) return false
do {
rect = el.getBoundingClientRect()
if (top <= rect.bottom === false) return false
// Check if the element is out of view due to a container scrolling
if ((top + height) <= rect.top) return false
el = el.parentNode
} while (el != document.body)
return true
};
编辑2016-03-26:我已经更新了解决方案,以考虑滚动过去的元素,所以它隐藏在可滚动容器的顶部。 编辑2018-10-08:更新到当滚动到屏幕上方的视图外时处理。
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。