js中将两个日期字符串相减

一直用后台Java代码处理时间，如果用Js怎么处理日期相减呢，在网上找到3个方法

知识啊，哎，还是不行啊。

1.

<script>
var sDate="2007-06-06";
var eDate="2007-07-07";
var sArr = sDate.split("-");
var eArr = eDate.split("-");
var sRDate = new Date(sArr[0], sArr[1], sArr[2]);
var eRDate = new Date(eArr[0], eArr[1], eArr[2]);
var result = (eRDate-sRDate)/(24*60*60*1000);
alert(result);
</script>

2.

<script>
//说明:这里用了Math.ceil()函数,向上取整,即零头算一天,^_^
var   strDate1   =   "2004-09-17   03:03:00.0";
var   strDate2   =   "2004-09-18   04:05:00.0";
strDate1=strDate1.substring(0,strDate1.lastIndexOf(".")).replace(/-/g, "/ ");
strDate2=strDate2.substring(0,strDate2.lastIndexOf(".")).replace(/-/g, "/ ");
//去掉毫秒 把-替换成/ 如果不替换转成时间戳类型火狐会出问题
var   date1   =  Date.parse(strDate1);
var   date2   =  Date.parse(strDate2);
alert( "strDate2与strDate1相差 "+(date2-date1)/(60*60*1000)+ "小时 ")
</script>

3.

function dateDiff(date1, date2) {
var type1 = typeof date1, type2 = typeof date2;
if (type1 == 'string')
date1 = stringToTime(date1);
else if (date1.getTime)
date1 = date1.getTime();
if (type2 == 'string')
date2 = stringToTime(date2);
else if (date2.getTime)
date2 = date2.getTime();
//alert((date1 - date2) / (1000*60*60));
return (date1 - date2) / (1000 * 60 * 60*24); //结果是小时
}
//字符串转成Time(dateDiff)所需方法
function stringToTime(string) {
var f = string.split(' ', 2);
var d = (f[0] ? f[0] : '').split('-', 3);
var t = (f[1] ? f[1] : '').split(':', 3);
return (new Date(
parseInt(d[0], 10) || null,
(parseInt(d[1], 10) || 1) - 1,
parseInt(d[2], 10) || null,
parseInt(t[0], 10) || null,
parseInt(t[1], 10) || null,
parseInt(t[2], 10) || null
)).getTime();

}
t=dateDiff("2007-02-11","2007-01-08");
alert(t);

原文地址：http://blog.csdn.net/feiliu010/archive/2006/12/18/1447843.aspx
http://topic.csdn.net/t/20040917/13/3382878.html http://jiangleikuku.blog.163.com/blog/static/16496971720112752039231/