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>