js example code for calculating the difference in days between two times_javascript skills

Copy code The code is as follows:

//Determine whether it is a leap year
function isLeapYear( year){
if(year % 4 == 0 && ((year % 100 != 0) || (year % 400 == 0)))
{
return true;
}
return false;
}
//Judge the two dates before and after
function validatePeriod(fyear,fmonth,fday,byear,bmonth,bday){
if(fyear < byear){
return true;
}else if(fyear == byear){
if(fmonth < bmonth){
return true;
} else if (fmonth == bmonth){
if(fday <= bday){
return true;
}else {
return false;
}
} else {
return false;
}
}else {
return false;
}
}
//Calculate the difference between two dates
function dateDiff(d1,d2){
var disNum=compareDate( d1,d2);
return disNum;
}
function compareDate(date1,date2)
{
var regexp=/^(d{1,4})[-|.] {1}(d{1,2})[-|.]{1}(d{1,2})$/;
var monthDays=[0,3,0,1,0,1,0 ,0,1,0,0,1];
regexp.test(date1);
var date1Year=RegExp.$1;
var date1Month=RegExp.$2;
var date1Day=RegExp. $3;

regexp.test(date2);
var date2Year=RegExp.$1;
var date2Month=RegExp.$2;
var date2Day=RegExp.$3;

if(validatePeriod(date1Year,date1Month,date1Day,date2Year,date2Month,date2Day)){
firstDate=new Date(date1Year,date1Month,date1Day);
secondDate=new Date(date2Year,date2Month,date2Day) ;

result=Math.floor((secondDate.getTime()-firstDate.getTime())/(1000*3600*24));
for(j=date1Year;j<=date2Year;j ){
if(isLeapYear(j)){
monthDays[1]=2;
}else{
monthDays[1]=3;
}
for(i=date1Month-1 ;i result=result-monthDays[i];
}
}
return result;
}else{
alert('Sorry for the first one The time must be less than the second time, thank you! ');
exit;
}
}

Call this function to pass two time values: 2013-01-19 2013-12-19

days = dateDiff(d1,d2);