[英]Which event to check if element is outside viewport on scroll?
I am trying to check if an element is inside my viewport when scrolling. 我正在尝试在滚动时检查元素是否在我的视口内。 If it is outside my viewport, I add a class that fixes the element to the top.
如果它在我的视口之外,我会添加一个将元素修复到顶部的类。
The function I use to determine if the element is outside the viewport is: 我用来确定元素是否在视口之外的函数是:
isElementInViewport : function(el) {
//special bonus for those using jQuery
if (typeof jQuery === "function" && el instanceof jQuery) {
el = el[0];
}
var rect = el.getBoundingClientRect();
return (
rect.top >= 0 &&
rect.left >= 0 &&
rect.bottom <= (window.innerHeight || document.documentElement.clientHeight) && /*or $(window).height() */
rect.right <= (window.innerWidth || document.documentElement.clientWidth) /*or $(window).width() */
);
}
I added a scroll event which fires this function : 我添加了一个触发此函数的scroll事件:
$(window).on('scroll', function(event){
testObject.isElementInViewport(event.target);
}
The problem here is that while scrolling, my Lenovo Yoga's CPU goes crazy. 这里的问题是,在滚动时,我的联想瑜伽的CPU变得疯狂。 I have tried:
我努力了:
polling
with an interval polling
timeout function
and a variable outside the event function scope
to toggle on a certain interval timeout function
和variable outside the event function scope
来切换特定间隔 Both methods work, but I need a way to minimize performance impacts, because the page I use this in already has LOADS of JS working. 这两种方法都有效,但我需要一种方法来最小化性能影响,因为我使用它的页面已经有JS的LOADS工作。 I also need to fix the bar to top as soon as it gets outside the viewport and not a few milliseconds later.
我还需要在它到达视口之外时将其固定到顶部,而不是几毫秒之后。
Are there any low-performance solutions for this? 有没有针对此的低性能解决方案? Can this be done in CSS only?
这可以只在CSS中完成吗?
I've noticed that I didn't asked my question right. 我注意到我没有问我的问题。 The current answers below are working, but give the same HUGE performance impact when I scroll up and down a bit:
下面的当前答案是有效的,但当我向上和向下滚动时,会产生相同的巨大性能影响:
I need to prevent the script from needing so much CPU power! 我需要防止脚本需要这么多的CPU能力!
copied from jQuery – test if element is in viewport (visible on screen) 从jQuery复制- 测试元素是否在视口中(在屏幕上可见)
HTML HTML
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box orange"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
<div class="box"></div>
JQuery JQuery的
$.fn.isOnScreen = 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));
};
$(window).scroll(function() {
if ($('.orange').isOnScreen() === true) {
console.log("Element is in viewport")
}
});
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