js中如何获取某个元素到浏览器最左和最右的距离？

JavaScript06

js中如何获取某个元素到浏览器最左和最右的距离？,第1张

js中获取某个元素到浏览器最左和最右的距离的程序代码是：

<!doctype html><html><head><meta charset="UTF-8"><style>

body{margin: 0padding: 0}

.mdiv{width: 100pxheight: 100pxbackground-color: red}

</style></head><body><div style="height: 1000px"></div><div></div><script src="jquery.js"></script>//自行下载<script>//原生//获取div距离顶部的距离

var mTop = document.getElementsByClassName('banner')[0].offsetTop

//减去滚动条的高度var sTop = document.body.scrollTopvar result = mTop - sTopconsole.log(result)//Jquery

mTop = $('.banner')[0].offsetTop

sTop = $(window).scrollTop()

result = mTop - sTop

console.log(result)

</script></body>

getDistance: function (lat1, lng1, lat2, lng2) {

    lat1 = lat1 || 0

    lng1 = lng1 || 0

    lat2 = lat2 || 0

    lng2 = lng2 || 0

    var rad1 = lat1 * Math.PI / 180.0

    var rad2 = lat2 * Math.PI / 180.0

    var a = rad1 - rad2

    var b = lng1 * Math.PI / 180.0 - lng2 * Math.PI / 180.0

    var r = 6378137

    return r * 2 * Math.asin(Math.sqrt(Math.pow(Math.sin(a / 2), 2) + Math.cos(rad1) * Math.cos(rad2) * Math.pow(Math.sin(b / 2), 2)))

  }

let PI = 3.14159265358979323//圆周率

let R = 6371229//地球半径

function getDistance(lon1,lat1,lon2,lat2){

let x,y,distance

let lonres = lon1 >lon2?lon1-lon2:lon2-lon1

let latres = lat1 >lat2?lat1-lat2:lat2-lat1

x = (lonres) * PI * R * Math.cos(((lat1 + lat2) / 2) * PI / 180) / 180

y = (lat2 - lat1) * PI * R / 180

return distance = Math.hypot(x, y)

}