日期处理问题

Question

<script>
function dur(d1,d2){
var o1=Date.parse(d1.replace("-","/"))
var o2=Date.parse(d2.replace("-","/"))
return Math.floor(Math.abs(o1-o2)/(1000*3600*24))
}
alert(dur("2002-1-1","2002-2-1"))
alert(dur("2002-3-1","2002-2-1"))
</script>

Answer 1

<script>
function dur(d1,d2){
return Math.floor(Math.abs(Date.parse(d1.replace("-","/"))-Date.parse(d2.replace("-","/")))/(1000*3600*24))
}
alert(dur("2002-1-1","2002-2-1"))
alert(dur("2002-3-1","2002-2-1"))
</script>

Answer 2

<script>
function Dayminus(asStartvalue,asEndvalue){
loinStart = asStartvalue.split("-");
loinEnd   = asEndvalue.split("-");
var loStart = new Date(loinStart[0],loinStart[1]-1,loinStart[2],0,0,0,0);
var loEnd   = new Date(loinEnd[0],loinEnd[1]-1,loinEnd[2],0,0,0,0);
var days=(loEnd.getTime()-loStart.getTime())/(1000*24*3600);
if(days<0) days=-days;
return days;

}
alert(Dayminus('1919-5-4','1989-6-4'));
</script>

Answer 3

sorry
<script>
function Dayminus(asStartvalue,asEndvalue){
loinStart = asStartvalue.split("-");
loinEnd   = asEndvalue.split("-");
var loStart = new Date(loinStart[0],loinStart[1]-1,loinStart[2],0,0,0,0);
var loEnd   = new Date(loinEnd[0],loinEnd[1]-1,loinEnd[2],0,0,0,0);
var days=(loEnd.getTime()-loStart.getTime())/(1000*24*3600);
if(days<0) days=-days;
return days;

}
if(Dayminus('1989-5-4','1989-6-4')>30)
alert("over 30 days!");
</script>