最新回答 / 写代码的阿木木
//获取相对于屏幕左边距离function getPosition(node){ var left = node.offsetLeft; var top = node.offsetTop; var parent = node.offsetParent; while(parent != null){ left = left + parent.offsetLeft; top = top + parent.offsetTop; parent = parent.offse...
2016-06-11