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
检查元素是否在屏幕上,而不是公认的检查div是否完全在屏幕上的方法(如果div比屏幕大,这将不起作用)。在纯Javascript中:
/**
* Checks if element is on the screen (Y axis only), returning true
* even if the element is only partially on screen.
*
* @param element
* @returns {boolean}
*/
function isOnScreenY(element) {
var screen_top_position = window.scrollY;
var screen_bottom_position = screen_top_position + window.innerHeight;
var element_top_position = element.offsetTop;
var element_bottom_position = element_top_position + element.offsetHeight;
return (inRange(element_top_position, screen_top_position, screen_bottom_position)
|| inRange(element_bottom_position, screen_top_position, screen_bottom_position));
}
/**
* Checks if x is in range (in-between) the
* value of a and b (in that order). Also returns true
* if equal to either value.
*
* @param x
* @param a
* @param b
* @returns {boolean}
*/
function inRange(x, a, b) {
return (x >= a && x <= b);
}
其他回答
我正在寻找一种方法来查看元素是否即将进入视图,所以通过扩展上面的代码段,我设法做到了。我想我应该把这个留在这里,说不定能帮到谁
Elm =是视图中要检查的元素
scrollElement =你可以传递window或者带有滚动的父元素
Offset =如果你想让它在元素在屏幕前200px处触发,那么传递200
isscro冷景的功能(elem, scrole,抵消) { var $elem = $(elem); var $window = $); var docViewTop = $window.scrollTop(); var docViewBottom = docViewTop + $window.height(); var elemTop = $elem.抵消()top; var elemBottom = elemTop + $elem.height() 归来((elemBottom +) > = docViewBottom) &&偏移(elemTop-offset) < = docViewTop) | | ((elemBottom-offset) < = docViewBottom) && (elemTop +偏移)> = docViewTop); 的
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');
}
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"));
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;
}