<script> var n1 = new Date("2000/02/01").getTime(); var n2 = new Date("2000/03/02").getTime(); alert("相隔 "+ ((n2-n1)/(24*60*60*1000)) +" 天!"); </script>
刚才看错题意,没看清是javascrip的,现在改为如下:<script language=javascript> function datediff(date1,date2) { var d, s, t, t2; var MinMilli = 1000 * 60; var HrMilli = MinMilli * 60; var DyMilli = HrMilli * 24; d = new Date(); t = Date.parse(date1); t2= Date.parse(date2); s = Math.round((t2-t)/ DyMilli); return(s); }document.write(datediff("3-21-2005","3-25-2005")); //注意:这里的日期格式为“mm-dd-yyyy”或是:“mm/dd/yyyy” </script>
<script> var n1 = new Date("2000/02/01").getTime(); var n2 = new Date("2000/03/02").getTime(); alert("相隔 "+ ((n2-n1)/(24*60*60*1000)) +" 天!"); </script> 请教,如果("2000/02/01")换成一个录入的变量($date),又怎么实现?
var n1 = new Date("2000/02/01").getTime();
var n2 = new Date("2000/03/02").getTime();
alert("相隔 "+ ((n2-n1)/(24*60*60*1000)) +" 天!");
</script>
function datediff(date1,date2)
{ var d, s, t, t2;
var MinMilli = 1000 * 60;
var HrMilli = MinMilli * 60;
var DyMilli = HrMilli * 24;
d = new Date();
t = Date.parse(date1);
t2= Date.parse(date2);
s = Math.round((t2-t)/ DyMilli);
return(s);
}document.write(datediff("3-21-2005","3-25-2005"));
//注意:这里的日期格式为“mm-dd-yyyy”或是:“mm/dd/yyyy”
</script>
var n1 = new Date("2000/02/01").getTime();
var n2 = new Date("2000/03/02").getTime();
alert("相隔 "+ ((n2-n1)/(24*60*60*1000)) +" 天!");
</script>
请教,如果("2000/02/01")换成一个录入的变量($date),又怎么实现?