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
这里有另一个解决方案:
<script type="text/javascript">
$.fn.is_on_screen = function(){
var win = $(window);
var viewport = {
top : win.scrollTop(),
left : win.scrollLeft()
};
viewport.right = viewport.left + win.width();
viewport.bottom = viewport.top + win.height();
var bounds = this.offset();
bounds.right = bounds.left + this.outerWidth();
bounds.bottom = bounds.top + this.outerHeight();
return (!(viewport.right < bounds.left || viewport.left > bounds.right || viewport.bottom < bounds.top || viewport.top > bounds.bottom));
};
if( $('.target').length > 0 ) { // if target element exists in DOM
if( $('.target').is_on_screen() ) { // if target element is visible on screen after DOM loaded
$('.log').html('<div class="alert alert-success">target element is visible on screen</div>'); // log info
} else {
$('.log').html('<div class="alert">target element is not visible on screen</div>'); // log info
}
}
$(window).on('scroll', function(){ // bind window scroll event
if( $('.target').length > 0 ) { // if target element exists in DOM
if( $('.target').is_on_screen() ) { // if target element is visible on screen after DOM loaded
$('.log').html('<div class="alert alert-success">target element is visible on screen</div>'); // log info
} else {
$('.log').html('<div class="alert">target element is not visible on screen</div>'); // log info
}
}
});
</script>
在JSFiddle中可以看到
其他回答
修改了已接受的答案,以便元素必须将其display属性设置为“none”以外的值,以便质量为可见。
function isScrolledIntoView(elem) {
var docViewTop = $(window).scrollTop();
var docViewBottom = docViewTop + $(window).height();
var elemTop = $(elem).offset().top;
var elemBottom = elemTop + $(elem).height();
var elemDisplayNotNone = $(elem).css("display") !== "none";
return ((elemBottom <= docViewBottom) && (elemTop >= docViewTop) && elemDisplayNotNone);
}
这应该可以达到目的:
function isScrolledIntoView(elem)
{
var docViewTop = $(window).scrollTop();
var docViewBottom = docViewTop + $(window).height();
var elemTop = $(elem).offset().top;
var elemBottom = elemTop + $(elem).height();
return ((elemBottom <= docViewBottom) && (elemTop >= docViewTop));
}
简单实用函数 这将允许您调用一个实用函数,该函数接受您正在寻找的元素,以及您希望元素完全或部分地显示在视图中。
function Utils() {
}
Utils.prototype = {
constructor: Utils,
isElementInView: function (element, fullyInView) {
var pageTop = $(window).scrollTop();
var pageBottom = pageTop + $(window).height();
var elementTop = $(element).offset().top;
var elementBottom = elementTop + $(element).height();
if (fullyInView === true) {
return ((pageTop < elementTop) && (pageBottom > elementBottom));
} else {
return ((elementTop <= pageBottom) && (elementBottom >= pageTop));
}
}
};
var Utils = new Utils();
使用
var isElementInView = Utils.isElementInView($('#flyout-left-container'), false);
if (isElementInView) {
console.log('in view');
} else {
console.log('out of view');
}
jquery scrollspy插件将允许您轻松做到这一点。 https://github.com/thesmart/jquery-scrollspy
$('.tile').on('scrollSpy:enter', function() {
console.log('enter:', $(this).attr('id'));
});
$('.tile').on('scrollSpy:exit', function() {
console.log('exit:', $(this).attr('id'));
});
$('.tile').scrollSpy();
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。
推特斯科特道丁的酷功能为我的要求- 这是用来查找元素是否刚刚滚动到屏幕上,即它的上边缘。
function isScrolledIntoView(elem)
{
var docViewTop = $(window).scrollTop();
var docViewBottom = docViewTop + $(window).height();
var elemTop = $(elem).offset().top;
return ((elemTop <= docViewBottom) && (elemTop >= docViewTop));
}