How do I get the number of days between two dates in JavaScript?
How do I get the number of days between two dates in JavaScript? For example, given two dates in input boxes:
<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>
<script>
alert(datediff("day", first, second)); // what goes here?
</script>
Here is a quick and dirty implementation of datediff , as a proof of concept to solve the problem as presented in the question. It relies on the fact that you can get the elapsed milliseconds between two dates by subtracting them, which coerces them into their primitive number value (milliseconds since the start of 1970).
// new Date("dateString") is browser-dependent and discouraged, so we'll write
// a simple parse function for U.S. date format (which does no error checking)
function parseDate(str) {
var mdy = str.split('/');
return new Date(mdy[2], mdy[0]-1, mdy[1]);
}
function datediff(first, second) {
// Take the difference between the dates and divide by milliseconds per day.
// Round to nearest whole number to deal with DST.
return Math.round((second-first)/(1000*60*60*24));
}
alert(datediff(parseDate(first.value), parseDate(second.value)));<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>As of this writing, only one of the other answers correctly handles DST (daylight saving time) transitions. Here are the results on a system located in California:
1/1/2013- 3/10/2013- 11/3/2013-
User Formula 2/1/2013 3/11/2013 11/4/2013 Result
--------- --------------------------- -------- --------- --------- ---------
Miles (d2 - d1) / N 31 0.9583333 1.0416666 Incorrect
some Math.floor((d2 - d1) / N) 31 0 1 Incorrect
fuentesjr Math.round((d2 - d1) / N) 31 1 1 Correct
toloco Math.ceiling((d2 - d1) / N) 31 1 2 Incorrect
N = 86400000
Although Math.round returns the correct results, I think it's somewhat clunky. Instead, by explicitly accounting for changes to the UTC offset when DST begins or ends, we can use exact arithmetic:
function treatAsUTC(date) {
var result = new Date(date);
result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
return result;
}
function daysBetween(startDate, endDate) {
var millisecondsPerDay = 24 * 60 * 60 * 1000;
return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}
alert(daysBetween($('#first').val(), $('#second').val()));
Explanation
JavaScript date calculations are tricky because Date objects store times internally in UTC, not local time. For example, 3/10/2013 12:00 AM Pacific Standard Time (UTC-08:00) is stored as 3/10/2013 8:00 AM UTC, and 3/11/2013 12:00 AM Pacific Daylight Time (UTC-07:00) is stored as 3/11/2013 7:00 AM UTC. On this day, midnight to midnight local time is only 23 hours in UTC!
Although a day in local time can have more or less than 24 hours, a day in UTC is always exactly 24 hours.1 The daysBetween method shown above takes advantage of this fact by first calling treatAsUTC to adjust both local times to midnight UTC, before subtracting and dividing.
1. JavaScript ignores leap seconds.
The easiest way to get the difference between two dates:
var diff = Math.floor(( Date.parse(str2) - Date.parse(str1) ) / 86400000);
You get the difference days (or NaN if one or both could not be parsed). The parse date gived the result in milliseconds and to get it by day you have to divided it by 24 * 60 * 60 * 1000
If you want it divided by days, hours, minutes, seconds and milliseconds:
function dateDiff( str1, str2 ) {
var diff = Date.parse( str2 ) - Date.parse( str1 );
return isNaN( diff ) ? NaN : {
diff : diff,
ms : Math.floor( diff % 1000 ),
s : Math.floor( diff / 1000 % 60 ),
m : Math.floor( diff / 60000 % 60 ),
h : Math.floor( diff / 3600000 % 24 ),
d : Math.floor( diff / 86400000 )
};
}
Here is my refactored version of James version:
function mydiff(date1,date2,interval) {
var second=1000, minute=second*60, hour=minute*60, day=hour*24, week=day*7;
date1 = new Date(date1);
date2 = new Date(date2);
var timediff = date2 - date1;
if (isNaN(timediff)) return NaN;
switch (interval) {
case "years": return date2.getFullYear() - date1.getFullYear();
case "months": return (
( date2.getFullYear() * 12 + date2.getMonth() )
-
( date1.getFullYear() * 12 + date1.getMonth() )
);
case "weeks" : return Math.floor(timediff / week);
case "days" : return Math.floor(timediff / day);
case "hours" : return Math.floor(timediff / hour);
case "minutes": return Math.floor(timediff / minute);
case "seconds": return Math.floor(timediff / second);
default: return undefined;
}
}
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