One: Calculate the number of days difference between two dates
| The code is as follows |
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var s = (new Date (ParamObj.end_date.replace (/-/g, "/"))-(new Date (ParamObj.start_date.replace (/-/g, "/")));
var day = s/1000/60/60/24
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Example 2
JS can actually directly date1-date2, but also converted to millisecond calculation time difference.
| code is as follows |
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//Two date Interval days
function DateDiff (sDate1, SDate2) {//sdate1 and sDate2 are 2002-12-18 format
var adate, ODate1, ODate2, idays
Adate = Sdate1.split ("-")
ODate1 = new Date (adate[1] + '-' + adate[2] + '- ' + adate[0])//Convert to 12-18-2002 format
Adate = Sdate2.split ("-")
ODate2 = new Date (adate[1] + ' -' + adate[2] + '-' + adate[0] '
Idays = parseint (Math.Abs (odate1-odate2)/1000/60/60/24)//Put the difference in milliseconds Convert number to days
return idays
} |
Second: Calculate the date after a certain number of days
In JavaScript, count the days after the day's date. Far from being in. NET, a function can solve the problem. On this issue, I was troubled for some time, finally through a Netizen's introduction to solve the problem. Put it together and share it.
| The code is as follows |
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<script language= "javascript" type= "Text/javascript" >
var startdate = new Date (); var intvalue = 0;
var enddate = null;
Intvalue = Startdate.gettime (); Intvalue + + 100 * (24 * 3600 * 1000);
EndDate = new Date (intvalue);
Alert (enddate.getfullyear () + "-" + (Enddate.getmonth () +1) + "-" + enddate.getdate ());
</script>
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The above 100 represents 100 days after the date that you can modify. JS in Date.gettime (), only after the date of 1970.01.01; there are 0-11 months, a little different, avoid oh. Of course, you can also calculate the date after a specific date.
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<script language= "javascript" type= "Text/javascript" >
var startdate = new Date (2007, (8-1), 1, 10, 10, 10);
var intvalue = 0;
var enddate = null;
Intvalue = Startdate.gettime (); Intvalue + + 100 * (24 * 3600 * 1000);
EndDate = new Date (intvalue);
Alert (enddate.getfullyear () + "-" + (Enddate.getmonth () +1) + "-" + enddate.getdate ());
</script>
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If none of the above is calculated for a leap year, the following example is finished
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Judge whether it is a leap year
function Isleapyear (year) {
if (year% 4 = 0 && (year% 100!= 0) | | (Year% 400 = 0))
{
return true;
}
return false;
}
Two dates before and after judgment
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 for 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<date2month;i++) {
Result=result-monthdays[i];
}
}
return result;
}else{
Alert (' Sorry the first 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); |
