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How to calculate automatically number of days between two dates in javascript? [duplicate]

This question already has answers here:
How to calculate number of days between two dates?
(42 answers)
Closed 3 months ago.
I want to calculate date difference in days, hours, minutes, seconds, milliseconds, nanoseconds. How can I do it?

Assuming you have two Date objects, you can just subtract them to get the difference in milliseconds:
var difference = date2 - date1;
From there, you can use simple arithmetic to derive the other values.

var DateDiff = {
inDays: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return Math.floor((t2-t1)/(24*3600*1000));
},
inWeeks: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000*7));
},
inMonths: function(d1, d2) {
var d1Y = d1.getFullYear();
var d2Y = d2.getFullYear();
var d1M = d1.getMonth();
var d2M = d2.getMonth();
return (d2M+12*d2Y)-(d1M+12*d1Y);
},
inYears: function(d1, d2) {
return d2.getFullYear()-d1.getFullYear();
}
}
var dString = "May, 20, 1984";
var d1 = new Date(dString);
var d2 = new Date();
document.write("<br />Number of <b>days</b> since "+dString+": "+DateDiff.inDays(d1, d2));
document.write("<br />Number of <b>weeks</b> since "+dString+": "+DateDiff.inWeeks(d1, d2));
document.write("<br />Number of <b>months</b> since "+dString+": "+DateDiff.inMonths(d1, d2));
document.write("<br />Number of <b>years</b> since "+dString+": "+DateDiff.inYears(d1, d2));
Code sample taken from here.

Another solution is convert difference to a new Date object and get that date's year(diff from 1970), month, day etc.
var date1 = new Date(2010, 6, 17);
var date2 = new Date(2013, 12, 18);
var diff = new Date(date2.getTime() - date1.getTime());
// diff is: Thu Jul 05 1973 04:00:00 GMT+0300 (EEST)
console.log(diff.getUTCFullYear() - 1970); // Gives difference as year
// 3
console.log(diff.getUTCMonth()); // Gives month count of difference
// 6
console.log(diff.getUTCDate() - 1); // Gives day count of difference
// 4
So difference is like "3 years and 6 months and 4 days". If you want to take difference in a human readable style, that can help you.

Expressions like "difference in days" are never as simple as they seem. If you have the following dates:
d1: 2011-10-15 23:59:00
d1: 2011-10-16 00:01:00
the difference in time is 2 minutes, should the "difference in days" be 1 or 0? Similar issues arise for any expression of the difference in months, years or whatever since years, months and days are of different lengths and different times (e.g. the day that daylight saving starts is 1 hour shorter than usual and two hours shorter than the day that it ends).
Here is a function for a difference in days that ignores the time, i.e. for the above dates it returns 1.
/*
Get the number of days between two dates - not inclusive.
"between" does not include the start date, so days
between Thursday and Friday is one, Thursday to Saturday
is two, and so on. Between Friday and the following Friday is 7.
e.g. getDaysBetweenDates( 22-Jul-2011, 29-jul-2011) => 7.
If want inclusive dates (e.g. leave from 1/1/2011 to 30/1/2011),
use date prior to start date (i.e. 31/12/2010 to 30/1/2011).
Only calculates whole days.
Assumes d0 <= d1
*/
function getDaysBetweenDates(d0, d1) {
var msPerDay = 8.64e7;
// Copy dates so don't mess them up
var x0 = new Date(d0);
var x1 = new Date(d1);
// Set to noon - avoid DST errors
x0.setHours(12,0,0);
x1.setHours(12,0,0);
// Round to remove daylight saving errors
return Math.round( (x1 - x0) / msPerDay );
}
This can be more concise:
/* Return number of days between d0 and d1.
** Returns positive if d0 < d1, otherwise negative.
**
** e.g. between 2000-02-28 and 2001-02-28 there are 366 days
** between 2015-12-28 and 2015-12-29 there is 1 day
** between 2015-12-28 23:59:59 and 2015-12-29 00:00:01 there is 1 day
** between 2015-12-28 00:00:01 and 2015-12-28 23:59:59 there are 0 days
**
** #param {Date} d0 - start date
** #param {Date} d1 - end date
** #returns {number} - whole number of days between d0 and d1
**
*/
function daysDifference(d0, d1) {
var diff = new Date(+d1).setHours(12) - new Date(+d0).setHours(12);
return Math.round(diff/8.64e7);
}
// Simple formatter
function formatDate(date){
return [date.getFullYear(),('0'+(date.getMonth()+1)).slice(-2),('0'+date.getDate()).slice(-2)].join('-');
}
// Examples
[[new Date(2000,1,28), new Date(2001,1,28)], // Leap year
[new Date(2001,1,28), new Date(2002,1,28)], // Not leap year
[new Date(2017,0,1), new Date(2017,1,1)]
].forEach(function(dates) {
document.write('From ' + formatDate(dates[0]) + ' to ' + formatDate(dates[1]) +
' is ' + daysDifference(dates[0],dates[1]) + ' days<br>');
});

<html lang="en">
<head>
<script>
function getDateDiff(time1, time2) {
var str1= time1.split('/');
var str2= time2.split('/');
// yyyy , mm , dd
var t1 = new Date(str1[2], str1[0]-1, str1[1]);
var t2 = new Date(str2[2], str2[0]-1, str2[1]);
var diffMS = t1 - t2;
console.log(diffMS + ' ms');
var diffS = diffMS / 1000;
console.log(diffS + ' ');
var diffM = diffS / 60;
console.log(diffM + ' minutes');
var diffH = diffM / 60;
console.log(diffH + ' hours');
var diffD = diffH / 24;
console.log(diffD + ' days');
alert(diffD);
}
//alert(getDateDiff('10/18/2013','10/14/2013'));
</script>
</head>
<body>
<input type="button"
onclick="getDateDiff('10/18/2013','10/14/2013')"
value="clickHere()" />
</body>
</html>

use Moment.js for all your JavaScript related date-time calculation
Answer to your question is:
var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b) // 86400000
Complete details can be found here

adding to #paresh mayani 's answer, to work like Facebook - showing how much time has passed in sec/min/hours/weeks/months/years
var DateDiff = {
inSeconds: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/1000);
},
inMinutes: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/60000);
},
inHours: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/3600000);
},
inDays: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000));
},
inWeeks: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000*7));
},
inMonths: function(d1, d2) {
var d1Y = d1.getFullYear();
var d2Y = d2.getFullYear();
var d1M = d1.getMonth();
var d2M = d2.getMonth();
return (d2M+12*d2Y)-(d1M+12*d1Y);
},
inYears: function(d1, d2) {
return d2.getFullYear()-d1.getFullYear();
}
}
var dString = "May, 20, 1984"; //will also get (Y-m-d H:i:s)
var d1 = new Date(dString);
var d2 = new Date();
var timeLaps = DateDiff.inSeconds(d1, d2);
var dateOutput = "";
if (timeLaps<60)
{
dateOutput = timeLaps+" seconds";
}
else
{
timeLaps = DateDiff.inMinutes(d1, d2);
if (timeLaps<60)
{
dateOutput = timeLaps+" minutes";
}
else
{
timeLaps = DateDiff.inHours(d1, d2);
if (timeLaps<24)
{
dateOutput = timeLaps+" hours";
}
else
{
timeLaps = DateDiff.inDays(d1, d2);
if (timeLaps<7)
{
dateOutput = timeLaps+" days";
}
else
{
timeLaps = DateDiff.inWeeks(d1, d2);
if (timeLaps<4)
{
dateOutput = timeLaps+" weeks";
}
else
{
timeLaps = DateDiff.inMonths(d1, d2);
if (timeLaps<12)
{
dateOutput = timeLaps+" months";
}
else
{
timeLaps = DateDiff.inYears(d1, d2);
dateOutput = timeLaps+" years";
}
}
}
}
}
}
alert (dateOutput);

With momentjs it's simple:
moment("2016-04-08").fromNow();

function DateDiff(date1, date2) {
date1.setHours(0);
date1.setMinutes(0, 0, 0);
date2.setHours(0);
date2.setMinutes(0, 0, 0);
var datediff = Math.abs(date1.getTime() - date2.getTime()); // difference
return parseInt(datediff / (24 * 60 * 60 * 1000), 10); //Convert values days and return value
}

var d1=new Date(2011,0,1); // jan,1 2011
var d2=new Date(); // now
var diff=d2-d1,sign=diff<0?-1:1,milliseconds,seconds,minutes,hours,days;
diff/=sign; // or diff=Math.abs(diff);
diff=(diff-(milliseconds=diff%1000))/1000;
diff=(diff-(seconds=diff%60))/60;
diff=(diff-(minutes=diff%60))/60;
days=(diff-(hours=diff%24))/24;
console.info(sign===1?"Elapsed: ":"Remains: ",
days+" days, ",
hours+" hours, ",
minutes+" minutes, ",
seconds+" seconds, ",
milliseconds+" milliseconds.");

I think this should do it.
let today = new Date();
let form_date=new Date('2019-10-23')
let difference=form_date>today ? form_date-today : today-form_date
let diff_days=Math.floor(difference/(1000*3600*24))

based on javascript runtime prototype implementation you can use simple arithmetic to subtract dates as in bellow
var sep = new Date(2020, 07, 31, 23, 59, 59);
var today = new Date();
var diffD = Math.floor((sep - today) / (1000 * 60 * 60 * 24));
console.log('Day Diff: '+diffD);
the difference return answer as milliseconds, then you have to convert it by division:
by 1000 to convert to second
by 1000×60 convert to minute
by 1000×60×60 convert to hour
by 1000×60×60×24 convert to day

function DateDiff(b, e)
{
let
endYear = e.getFullYear(),
endMonth = e.getMonth(),
years = endYear - b.getFullYear(),
months = endMonth - b.getMonth(),
days = e.getDate() - b.getDate();
if (months < 0)
{
years--;
months += 12;
}
if (days < 0)
{
months--;
days += new Date(endYear, endMonth, 0).getDate();
}
return [years, months, days];
}
[years, months, days] = DateDiff(
new Date("October 21, 1980"),
new Date("July 11, 2017")); // 36 8 20

Sorry but flat millisecond calculation is not reliable
Thanks for all the responses, but few of the functions I tried are failing either on
1. A date near today's date
2. A date in 1970 or
3. A date in a leap year.
Approach that best worked for me and covers all scenario e.g. leap year, near date in 1970, feb 29 etc.
var someday = new Date("8/1/1985");
var today = new Date();
var years = today.getFullYear() - someday.getFullYear();
// Reset someday to the current year.
someday.setFullYear(today.getFullYear());
// Depending on when that day falls for this year, subtract 1.
if (today < someday)
{
years--;
}
document.write("Its been " + years + " full years.");

This code will return the difference between two dates in days:
const previous_date = new Date("2019-12-23");
const current_date = new Date();
const current_year = current_date.getFullYear();
const previous_date_year =
previous_date.getFullYear();
const difference_in_years = current_year -
previous_date_year;
let months = current_date.getMonth();
months = months + 1; // for making the indexing
// of months from 1
for(let i = 0; i < difference_in_years; i++){
months = months + 12;
}
let days = current_date.getDate();
days = days + (months * 30.417);
console.log(`The days between ${current_date} and
${previous_date} are : ${days} (approximately)`);

If you are using moment.js then it is pretty simple to find date difference.
var now = "04/09/2013 15:00:00";
var then = "04/09/2013 14:20:30";
moment.utc(moment(now,"DD/MM/YYYY HH:mm:ss").diff(moment(then,"DD/MM/YYYY HH:mm:ss"))).format("HH:mm:ss")

This is how you can implement difference between dates without a framework.
function getDateDiff(dateOne, dateTwo) {
if(dateOne.charAt(2)=='-' & dateTwo.charAt(2)=='-'){
dateOne = new Date(formatDate(dateOne));
dateTwo = new Date(formatDate(dateTwo));
}
else{
dateOne = new Date(dateOne);
dateTwo = new Date(dateTwo);
}
let timeDiff = Math.abs(dateOne.getTime() - dateTwo.getTime());
let diffDays = Math.ceil(timeDiff / (1000 * 3600 * 24));
let diffMonths = Math.ceil(diffDays/31);
let diffYears = Math.ceil(diffMonths/12);
let message = "Difference in Days: " + diffDays + " " +
"Difference in Months: " + diffMonths+ " " +
"Difference in Years: " + diffYears;
return message;
}
function formatDate(date) {
return date.split('-').reverse().join('-');
}
console.log(getDateDiff("23-04-2017", "23-04-2018"));

function daysInMonth (month, year) {
return new Date(year, month, 0).getDate();
}
function getduration(){
let A= document.getElementById("date1_id").value
let B= document.getElementById("date2_id").value
let C=Number(A.substring(3,5))
let D=Number(B.substring(3,5))
let dif=D-C
let arr=[];
let sum=0;
for (let i=0;i<dif+1;i++){
sum+=Number(daysInMonth(i+C,2019))
}
let sum_alter=0;
for (let i=0;i<dif;i++){
sum_alter+=Number(daysInMonth(i+C,2019))
}
let no_of_month=(Number(B.substring(3,5)) - Number(A.substring(3,5)))
let days=[];
if ((Number(B.substring(3,5)) - Number(A.substring(3,5)))>0||Number(B.substring(0,2)) - Number(A.substring(0,2))<0){
days=Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter
}
if ((Number(B.substring(3,5)) == Number(A.substring(3,5)))){
console.log(Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter)
}
time_1=[]; time_2=[]; let hour=[];
time_1=document.getElementById("time1_id").value
time_2=document.getElementById("time2_id").value
if (time_1.substring(0,2)=="12"){
time_1="00:00:00 PM"
}
if (time_1.substring(9,11)==time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))
}
if (time_1.substring(9,11)!=time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))+12
}
let min=Math.abs(Number(time_1.substring(3,5))-Number(time_2.substring(3,5)))
document.getElementById("duration_id").value=days +" days "+ hour+" hour " + min+" min "
}
<input type="text" id="date1_id" placeholder="28/05/2019">
<input type="text" id="date2_id" placeholder="29/06/2019">
<br><br>
<input type="text" id="time1_id" placeholder="08:01:00 AM">
<input type="text" id="time2_id" placeholder="00:00:00 PM">
<br><br>
<button class="text" onClick="getduration()">Submit </button>
<br><br>
<input type="text" id="duration_id" placeholder="days hour min">

var date1 = new Date("06/30/2019");
var date2 = new Date("07/30/2019");
// To calculate the time difference of two dates
var Difference_In_Time = date2.getTime() - date1.getTime();
// To calculate the no. of days between two dates
var Difference_In_Days = Difference_In_Time / (1000 * 3600 * 24);
//To display the final no. of days (result)
document.write("Total number of days between dates <br>"
+ date1 + "<br> and <br>"
+ date2 + " is: <br> "
+ Difference_In_Days);

this should work just fine if you just need to show what time left, since JavaScript uses frames for its time you'll have get your End Time - The Time RN after that we can divide it by 1000 since apparently 1000 frames = 1 seconds, after that you can use the basic math of time, but there's still a problem to this code, since the calculation is static, it can't compensate for the different day total in a year (360/365/366), the bunch of IF after the calculation is to make it null if the time is lower than 0, hope this helps even though it's not exactly what you're asking :)
var now = new Date();
var end = new Date("End Time");
var total = (end - now) ;
var totalD = Math.abs(Math.floor(total/1000));
var years = Math.floor(totalD / (365*60*60*24));
var months = Math.floor((totalD - years*365*60*60*24) / (30*60*60*24));
var days = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24)/ (60*60*24));
var hours = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24)/ (60*60));
var minutes = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60)/ (60));
var seconds = Math.floor(totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60 - minutes*60);
var Y = years < 1 ? "" : years + " Years ";
var M = months < 1 ? "" : months + " Months ";
var D = days < 1 ? "" : days + " Days ";
var H = hours < 1 ? "" : hours + " Hours ";
var I = minutes < 1 ? "" : minutes + " Minutes ";
var S = seconds < 1 ? "" : seconds + " Seconds ";
var A = years == 0 && months == 0 && days == 0 && hours == 0 && minutes == 0 && seconds == 0 ? "Sending" : " Remaining";
document.getElementById('txt').innerHTML = Y + M + D + H + I + S + A;

Ok, there are a bunch of ways you can do that.
Yes, you can use plain old JS. Just try:
let dt1 = new Date()
let dt2 = new Date()
Let's emulate passage using Date.prototype.setMinutes and make sure we are in range.
dt1.setMinutes(7)
dt2.setMinutes(42)
console.log('Elapsed seconds:',(dt2-dt1)/1000)
Alternatively you could use some library like js-joda, where you can easily do things like this (directly from docs):
var dt1 = LocalDateTime.parse("2016-02-26T23:55:42.123");
var dt2 = dt1
.plusYears(6)
.plusMonths(12)
.plusHours(2)
.plusMinutes(42)
.plusSeconds(12);
// obtain the duration between the two dates
dt1.until(dt2, ChronoUnit.YEARS); // 7
dt1.until(dt2, ChronoUnit.MONTHS); // 84
dt1.until(dt2, ChronoUnit.WEEKS); // 356
dt1.until(dt2, ChronoUnit.DAYS); // 2557
dt1.until(dt2, ChronoUnit.HOURS); // 61370
dt1.until(dt2, ChronoUnit.MINUTES); // 3682242
dt1.until(dt2, ChronoUnit.SECONDS); // 220934532
There are plenty more libraries ofc, but js-joda has an added bonus of being available also in Java, where it has been extensively tested. All those tests have been migrated to js-joda, it's also immutable.

I made a below function to get the difference between now and "2021-02-26T21:50:42.123".
The difference return answer as milliseconds, so I convert it by using this formula:
(1000 * 3600 * 24).
function getDiff(dateAcquired) {
let calDiff = Math.floor(
(new Date() - new Date(dateAcquired)) / (1000 * 3600 * 24)
);
return calDiff;
}
console.log(getDiff("2021-02-26T21:50:42.123"));

Can be useful :
const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/24 * 60 * 60 * 1000)
or
const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/86400000)
where 24 * 60 * 60 * 1000 is (day * minutes * seconds * milliseconds) = 86400000 milliseconds in one day
Thank you

// the idea is to get time left for new year.
// Not considering milliseconds as of now, but that
// can be done
var newYear = '1 Jan 2023';
const secondsInAMin = 60;
const secondsInAnHour = 60 * secondsInAMin;
const secondsInADay = 24 * secondsInAnHour;
function DateDiffJs() {
var newYearDate = new Date(newYear);
var currDate = new Date();
var remainingSecondsInDateDiff = (newYearDate - currDate) / 1000;
var days = Math.floor(remainingSecondsInDateDiff / secondsInADay);
var remainingSecondsAfterDays = remainingSecondsInDateDiff - (days * secondsInADay);
var hours = Math.floor(remainingSecondsAfterDays / secondsInAnHour);
var remainingSecondsAfterhours = remainingSecondsAfterDays - (hours * secondsInAnHour);
var mins = Math.floor(remainingSecondsAfterhours / secondsInAMin);
var seconds = Math.floor(remainingSecondsAfterhours - (mins * secondsInAMin));
console.log(`days :: ${days}`)
console.log(`hours :: ${hours}`)
console.log(`mins :: ${mins}`)
console.log(`seconds :: ${seconds}`)
}
DateDiffJs();

Turning an input (amount of seconds) into an HH:MM:SS format for from API [duplicate]

How can I convert seconds to an HH-MM-SS string using JavaScript?

You can manage to do this without any external JavaScript library with the help of JavaScript Date method like following:
const date = new Date(null);
date.setSeconds(SECONDS); // specify value for SECONDS here
const result = date.toISOString().slice(11, 19);
Or, as per #Frank's comment; a one liner:
new Date(SECONDS * 1000).toISOString().slice(11, 19);

Updated (2020):
Please use #Frank's one line solution:
new Date(SECONDS * 1000).toISOString().substring(11, 16)
If SECONDS<3600 and if you want to show only MM:SS then use below code:
new Date(SECONDS * 1000).toISOString().substring(14, 19)
It is by far the best solution.
Old answer:
Use the Moment.js library.

I don't think any built-in feature of the standard Date object will do this for you in a way that's more convenient than just doing the math yourself.
hours = Math.floor(totalSeconds / 3600);
totalSeconds %= 3600;
minutes = Math.floor(totalSeconds / 60);
seconds = totalSeconds % 60;
Example:
let totalSeconds = 28565;
let hours = Math.floor(totalSeconds / 3600);
totalSeconds %= 3600;
let minutes = Math.floor(totalSeconds / 60);
let seconds = totalSeconds % 60;
console.log("hours: " + hours);
console.log("minutes: " + minutes);
console.log("seconds: " + seconds);
// If you want strings with leading zeroes:
minutes = String(minutes).padStart(2, "0");
hours = String(hours).padStart(2, "0");
seconds = String(seconds).padStart(2, "0");
console.log(hours + ":" + minutes + ":" + seconds);

I know this is kinda old, but...
ES2015:
var toHHMMSS = (secs) => {
var sec_num = parseInt(secs, 10)
var hours = Math.floor(sec_num / 3600)
var minutes = Math.floor(sec_num / 60) % 60
var seconds = sec_num % 60
return [hours,minutes,seconds]
.map(v => v < 10 ? "0" + v : v)
.filter((v,i) => v !== "00" || i > 0)
.join(":")
}
It will output:
toHHMMSS(129600) // 36:00:00
toHHMMSS(13545) // 03:45:45
toHHMMSS(180) // 03:00
toHHMMSS(18) // 00:18

As Cleiton pointed out in his answer, moment.js can be used for this:
moment().startOf('day')
.seconds(15457)
.format('H:mm:ss');

Here's a simple function for converting times that might help
function formatSeconds(seconds) {
var date = new Date(1970,0,1);
date.setSeconds(seconds);
return date.toTimeString().replace(/.*(\d{2}:\d{2}:\d{2}).*/, "$1");
}

This does the trick:
function secondstotime(secs)
{
var t = new Date(1970,0,1);
t.setSeconds(secs);
var s = t.toTimeString().substr(0,8);
if(secs > 86399)
s = Math.floor((t - Date.parse("1/1/70")) / 3600000) + s.substr(2);
return s;
}
(Sourced from here)

var timeInSec = "661"; //even it can be string
String.prototype.toHHMMSS = function () {
/* extend the String by using prototypical inheritance */
var seconds = parseInt(this, 10); // don't forget the second param
var hours = Math.floor(seconds / 3600);
var minutes = Math.floor((seconds - (hours * 3600)) / 60);
seconds = seconds - (hours * 3600) - (minutes * 60);
if (hours < 10) {hours = "0"+hours;}
if (minutes < 10) {minutes = "0"+minutes;}
if (seconds < 10) {seconds = "0"+seconds;}
var time = hours+':'+minutes+':'+seconds;
return time;
}
alert("5678".toHHMMSS()); // "01:34:38"
console.log(timeInSec.toHHMMSS()); //"00:11:01"
we can make this function lot shorter and crisp but that decreases the readability, so we will write it as simple as possible and as stable as possible.
or you can check this working here:

Try this:
function toTimeString(seconds) {
return (new Date(seconds * 1000)).toUTCString().match(/(\d\d:\d\d:\d\d)/)[0];
}

I think the most general (and cryptic) solution could be this
function hms(seconds) {
return [3600, 60]
.reduceRight(
(pipeline, breakpoint) => remainder =>
[Math.floor(remainder / breakpoint)].concat(pipeline(remainder % breakpoint)),
r => [r]
)(seconds)
.map(amount => amount.toString().padStart(2, '0'))
.join('-');
}
Or to copy & paste the shortest version
function hms(seconds) {
return [3600, 60]
.reduceRight(
(p, b) => r => [Math.floor(r / b)].concat(p(r % b)),
r => [r]
)(seconds)
.map(a => a.toString().padStart(2, '0'))
.join('-');
}
Some example outputs:
> hms(0)
< "00-00-00"
> hms(5)
< "00-00-05"
> hms(60)
< "00-01-00"
> hms(3785)
< "01-03-05"
> hms(37850)
< "10-30-50"
> hms(378500)
< "105-08-20"
How it works
Algorithm
To get hours you divide total seconds by 3600 and floor it.
To get minutes you divide remainder by 60 and floor it.
To get seconds you just use the remainder.
It would also be nice to keep individual amounts in an array for easier formatting.
For example given the input of 3785s the output should be [1, 3, 5], that is 1 hour, 3 minutes and 5 seconds.
Creating pipeline
Naming the 3600 and 60 constants "breakpoints" you can write this algorithm into function as this
function divideAndAppend(remainder, breakpoint, callback) {
return [Math.floor(remainder / breakpoint)].concat(callback(remainder % breakpoint));
}
It returns an array where first item is the amount for given breakpoint and the rest of the array is given by the callback.
Reusing the divideAndAppend in callback function will give you a pipeline of composed divideAndAppend functions. Each one of these
computes amount per given breakpoint and append it to the array making your desired output.
Then you also need the "final" callback that ends this pipeline. In another words you used all breakpoints and now you have only the remainder.
Since you have already the answer at 3) you should use some sort of identity function, in this case remainder => [remainder].
You can now write the pipeline like this
let pipeline = r3 => divideAndAppend(
r3,
3600,
r2 => divideAndAppend(
r2,
60,
r1 => [r1]));
> pipeline(3785)
< [1, 3, 5]
Cool right?
Generalizing using for-loop
Now you can generalize with a variable amount of breakpoints and create a for-loop that will compose individial divideAndAppend functions into
the pipeline.
You start with the identity function r1 => [r1], then use the 60 breakpoint and finally use the 3600 breakpoint.
let breakpoints = [60, 3600];
let pipeline = r => [r];
for (const b of breakpoints) {
const previousPipeline = pipeline;
pipeline = r => divideAndAppend(r, b, previousPipeline);
}
> pipeline(3785)
< [1, 3, 5]
Using Array.prototype.reduce()
Now you can rewrite this for-loop into reducer for shorter and more functional code. In other words rewrite function composition into the reducer.
let pipeline = [60, 3600].reduce(
(ppln, b) => r => divideAndAppend(r, b, ppln),
r => [r]
);
> pipeline(3785)
< [1, 3, 5]
The accumulator ppln is the pipeline and you are composing it using the previous version of it. The initial pipeline is r => [r].
You can now inline the function divideAndAppend and use Array.prototype.reduceRight which is the same as [].reverse().reduce(...) to make the breakpoints
definitions more natural.
let pipeline = [3600, 60]
.reduceRight(
(ppln, b) => r => [Math.floor(r / b)].concat(ppln(r % b)),
r => [r]
);
Which is the final form. Then you just appy mapping to string with padded 0's on left and join the strings with : separator;
More generalizations
Wrapping the reducer into function
function decompose(total, breakpoints) {
return breakpoints.reduceRight(
(p, b) => r => [Math.floor(r / b)].concat(p(r % b)),
r => [r]
)(total);
}
> decompose(3785, [3600, 60])
< [1, 3, 5]
you now have very general algorithm you can work with. For example:
Convert easily (the weird) us length standards
Given the standards
Unit
Divisions
1 foot
12 inches
1 yard
3 feet
1 mile
1760 yards
> decompose(123_456, [1760 * 3 * 12, 3 * 12, 12])
< [1, 1669, 1, 0]
123456 in = 1 mi, 1669 yd, 1 feet and 0 in
Or you can somewhat convert to decimal or binary representations
> decompose(123_456, [100_000, 10_000, 1000, 100, 10])
< [1, 2, 3, 4, 5, 6]
> decompose(127, [128, 64, 32, 16, 8, 4, 2])
< [0, 1, 1, 1, 1, 1, 1, 1]
Works also with floating point breakpoints
Since Javascript supports mod operator with floating point numbers, you can also do
> decompose(26.5, [20, 2.5])
< [1, 2, 1.5]
The edge case of no breakpoints is also naturally covered
> decompose(123, [])
< [123]

Here is an extension to Number class. toHHMMSS() converts seconds to an hh:mm:ss string.
Number.prototype.toHHMMSS = function() {
var hours = Math.floor(this / 3600) < 10 ? ("00" + Math.floor(this / 3600)).slice(-2) : Math.floor(this / 3600);
var minutes = ("00" + Math.floor((this % 3600) / 60)).slice(-2);
var seconds = ("00" + (this % 3600) % 60).slice(-2);
return hours + ":" + minutes + ":" + seconds;
}
// Usage: [number variable].toHHMMSS();
// Here is a simple test
var totalseconds = 1234;
document.getElementById("timespan").innerHTML = totalseconds.toHHMMSS();
// HTML of the test
<div id="timespan"></div>

Easy to follow version for noobies:
var totalNumberOfSeconds = YOURNUMBEROFSECONDS;
var hours = parseInt( totalNumberOfSeconds / 3600 );
var minutes = parseInt( (totalNumberOfSeconds - (hours * 3600)) / 60 );
var seconds = Math.floor((totalNumberOfSeconds - ((hours * 3600) + (minutes * 60))));
var result = (hours < 10 ? "0" + hours : hours) + ":" + (minutes < 10 ? "0" + minutes : minutes) + ":" + (seconds < 10 ? "0" + seconds : seconds);
console.log(result);

This function should do it :
var convertTime = function (input, separator) {
var pad = function(input) {return input < 10 ? "0" + input : input;};
return [
pad(Math.floor(input / 3600)),
pad(Math.floor(input % 3600 / 60)),
pad(Math.floor(input % 60)),
].join(typeof separator !== 'undefined' ? separator : ':' );
}
Without passing a separator, it uses : as the (default) separator :
time = convertTime(13551.9941351); // --> OUTPUT = 03:45:51
If you want to use - as a separator, just pass it as the second parameter:
time = convertTime(1126.5135155, '-'); // --> OUTPUT = 00-18-46
See also this Fiddle.

Chiming in on this old thread -- the OP stated HH:MM:SS, and many of the solutions work, until you realize you need more than 24 hours listed. And maybe you don't want more than a single line of code. Here you go:
d=(s)=>{f=Math.floor;g=(n)=>('00'+n).slice(-2);return f(s/3600)+':'+g(f(s/60)%60)+':'+g(s%60)}
It returns H+:MM:SS. To use it, simply use:
d(91260); // returns "25:21:00"
d(960); // returns "0:16:00"
...I tried to get it to use the least amount of code possible, for a nice one-liner approach.

For the special case of HH:MM:SS.MS (eq: "00:04:33.637") as used by FFMPEG to specify milliseconds.
[-][HH:]MM:SS[.m...]
HH expresses the number of hours, MM the number of minutes for a
maximum of 2 digits, and SS the number of seconds for a maximum of 2
digits. The m at the end expresses decimal value for SS.
/* HH:MM:SS.MS to (FLOAT)seconds ---------------*/
function timerToSec(timer){
let vtimer = timer.split(":")
let vhours = +vtimer[0]
let vminutes = +vtimer[1]
let vseconds = parseFloat(vtimer[2])
return vhours * 3600 + vminutes * 60 + vseconds
}
/* Seconds to (STRING)HH:MM:SS.MS --------------*/
function secToTimer(sec){
let o = new Date(0)
let p = new Date(sec*1000)
return new Date(p.getTime()-o.getTime())
.toISOString()
.split("T")[1]
.split("Z")[0]
}
/* Example: 7hours, 4 minutes, 33 seconds and 637 milliseconds */
const t = "07:04:33.637"
console.log(
t + " => " +
timerToSec(t) +
"s"
)
/* Test: 25473 seconds and 637 milliseconds */
const s = 25473.637 // "25473.637"
console.log(
s + "s => " +
secToTimer(s)
)
Example usage, a milliseconds transport timer:
/* Seconds to (STRING)HH:MM:SS.MS --------------*/
function secToTimer(sec){
let o = new Date(0)
let p = new Date(sec*1000)
return new Date(p.getTime()-o.getTime())
.toISOString()
.split("T")[1]
.split("Z")[0]
}
let job, origin = new Date().getTime()
const timer = () => {
job = requestAnimationFrame(timer)
OUT.textContent = secToTimer((new Date().getTime() - origin) / 1000)
}
requestAnimationFrame(timer)
span {font-size:4rem}
<span id="OUT"></span>
<br>
<button onclick="origin = new Date().getTime()">RESET</button>
<button onclick="requestAnimationFrame(timer)">RESTART</button>
<button onclick="cancelAnimationFrame(job)">STOP</button>
Example usage, binded to a media element
/* Seconds to (STRING)HH:MM:SS.MS --------------*/
function secToTimer(sec){
let o = new Date(0)
let p = new Date(sec*1000)
return new Date(p.getTime()-o.getTime())
.toISOString()
.split("T")[1]
.split("Z")[0]
}
VIDEO.addEventListener("timeupdate", function(e){
OUT.textContent = secToTimer(e.target.currentTime)
}, false)
span {font-size:4rem}
<span id="OUT"></span><br>
<video id="VIDEO" width="400" controls autoplay>
<source src="https://www.w3schools.com/html/mov_bbb.mp4" type="video/mp4">
</video>
Outside the question, those functions written in php:
<?php
/* HH:MM:SS to (FLOAT)seconds ------------------*/
function timerToSec($timer){
$vtimer = explode(":",$timer);
$vhours = (int)$vtimer[0];
$vminutes = (int)$vtimer[1];
$vseconds = (float)$vtimer[2];
return $vhours * 3600 + $vminutes * 60 + $vseconds;
}
/* Seconds to (STRING)HH:MM:SS -----------------*/
function secToTimer($sec){
return explode(" ", date("H:i:s", $sec))[0];
}

After looking at all the answers and not being happy with most of them, this is what I came up with. I know I am very late to the conversation, but here it is anyway.
function secsToTime(secs){
var time = new Date();
// create Date object and set to today's date and time
time.setHours(parseInt(secs/3600) % 24);
time.setMinutes(parseInt(secs/60) % 60);
time.setSeconds(parseInt(secs%60));
time = time.toTimeString().split(" ")[0];
// time.toString() = "HH:mm:ss GMT-0800 (PST)"
// time.toString().split(" ") = ["HH:mm:ss", "GMT-0800", "(PST)"]
// time.toTimeString().split(" ")[0]; = "HH:mm:ss"
return time;
}
I create a new Date object, change the time to my parameters, convert the Date Object to a time string, and removed the additional stuff by splitting the string and returning only the part that need.
I thought I would share this approach, since it removes the need for regex, logic and math acrobatics to get the results in "HH:mm:ss" format, and instead it relies on built in methods.
You may want to take a look at the documentation here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date

below is the given code which will convert seconds into hh-mm-ss format:
var measuredTime = new Date(null);
measuredTime.setSeconds(4995); // specify value of SECONDS
var MHSTime = measuredTime.toISOString().substr(11, 8);
Get alternative method from Convert seconds to HH-MM-SS format in JavaScript

var time1 = date1.getTime();
var time2 = date2.getTime();
var totalMilisec = time2 - time1;
alert(DateFormat('hh:mm:ss',new Date(totalMilisec)))
/* ----------------------------------------------------------
* Field | Full Form | Short Form
* -------------|--------------------|-----------------------
* Year | yyyy (4 digits) | yy (2 digits)
* Month | MMM (abbr.) | MM (2 digits)
| NNN (name) |
* Day of Month | dd (2 digits) |
* Day of Week | EE (name) | E (abbr)
* Hour (1-12) | hh (2 digits) |
* Minute | mm (2 digits) |
* Second | ss (2 digits) |
* ----------------------------------------------------------
*/
function DateFormat(formatString,date){
if (typeof date=='undefined'){
var DateToFormat=new Date();
}
else{
var DateToFormat=date;
}
var DAY = DateToFormat.getDate();
var DAYidx = DateToFormat.getDay();
var MONTH = DateToFormat.getMonth()+1;
var MONTHidx = DateToFormat.getMonth();
var YEAR = DateToFormat.getYear();
var FULL_YEAR = DateToFormat.getFullYear();
var HOUR = DateToFormat.getHours();
var MINUTES = DateToFormat.getMinutes();
var SECONDS = DateToFormat.getSeconds();
var arrMonths = new Array("January","February","March","April","May","June","July","August","September","October","November","December");
var arrDay=new Array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
var strMONTH;
var strDAY;
var strHOUR;
var strMINUTES;
var strSECONDS;
var Separator;
if(parseInt(MONTH)< 10 && MONTH.toString().length < 2)
strMONTH = "0" + MONTH;
else
strMONTH=MONTH;
if(parseInt(DAY)< 10 && DAY.toString().length < 2)
strDAY = "0" + DAY;
else
strDAY=DAY;
if(parseInt(HOUR)< 10 && HOUR.toString().length < 2)
strHOUR = "0" + HOUR;
else
strHOUR=HOUR;
if(parseInt(MINUTES)< 10 && MINUTES.toString().length < 2)
strMINUTES = "0" + MINUTES;
else
strMINUTES=MINUTES;
if(parseInt(SECONDS)< 10 && SECONDS.toString().length < 2)
strSECONDS = "0" + SECONDS;
else
strSECONDS=SECONDS;
switch (formatString){
case "hh:mm:ss":
return strHOUR + ':' + strMINUTES + ':' + strSECONDS;
break;
//More cases to meet your requirements.
}
}

I just wanted to give a little explanation to the nice answer above:
var totalSec = new Date().getTime() / 1000;
var hours = parseInt( totalSec / 3600 ) % 24;
var minutes = parseInt( totalSec / 60 ) % 60;
var seconds = totalSec % 60;
var result = (hours < 10 ? "0" + hours : hours) + "-" + (minutes < 10 ? "0" + minutes : minutes) + "-" + (seconds < 10 ? "0" + seconds : seconds);
On the second line, since there are 3600 seconds in 1 hour, we divide the total number of seconds by 3600 to get the total number of hours. We use parseInt to strip off any decimal. If totalSec was 12600 (3 and half hours), then parseInt( totalSec / 3600 ) would return 3, since we will have 3 full hours. Why do we need the % 24 in this case? If we exceed 24 hours, let's say we have 25 hours (90000 seconds), then the modulo here will take us back to 1 again, rather than returning 25. It is confining the result within a 24 hour limit, since there are 24 hours in one day.
When you see something like this:
25 % 24
Think of it like this:
25 mod 24 or what is the remainder when we divide 25 by 24

None of the answers here satisfies my requirements as I want to be able to handle
Large numbers of seconds (days), and
Negative numbers
Although those are not required by the OP, it's good practice to cover edge cases, especially when it takes little effort.
It's pretty obvious is that the OP means a NUMBER of seconds when he says seconds. Why would peg your function on String?
function secondsToTimeSpan(seconds) {
const value = Math.abs(seconds);
const days = Math.floor(value / 1440);
const hours = Math.floor((value - (days * 1440)) / 3600);
const min = Math.floor((value - (days * 1440) - (hours * 3600)) / 60);
const sec = value - (days * 1440) - (hours * 3600) - (min * 60);
return `${seconds < 0 ? '-':''}${days > 0 ? days + '.':''}${hours < 10 ? '0' + hours:hours}:${min < 10 ? '0' + min:min}:${sec < 10 ? '0' + sec:sec}`
}
secondsToTimeSpan(0); // => 00:00:00
secondsToTimeSpan(1); // => 00:00:01
secondsToTimeSpan(1440); // => 1.00:00:00
secondsToTimeSpan(-1440); // => -1.00:00:00
secondsToTimeSpan(-1); // => -00:00:01

Simple function to convert seconds into in hh:mm:ss format :
function getHHMMSSFromSeconds(totalSeconds) {
if (!totalSeconds) {
return '00:00:00';
}
const hours = Math.floor(totalSeconds / 3600);
const minutes = Math.floor(totalSeconds % 3600 / 60);
const seconds = totalSeconds % 60;
const hhmmss = padTo2(hours) + ':' + padTo2(minutes) + ':' + padTo2(seconds);
return hhmmss;
}
// function to convert single digit to double digit
function padTo2(value) {
if (!value) {
return '00';
}
return value < 10 ? String(value).padStart(2, '0') : value;
}

Here is a function to convert seconds to hh-mm-ss format based on powtac's answer here
jsfiddle
/**
* Convert seconds to hh-mm-ss format.
* #param {number} totalSeconds - the total seconds to convert to hh- mm-ss
**/
var SecondsTohhmmss = function(totalSeconds) {
var hours = Math.floor(totalSeconds / 3600);
var minutes = Math.floor((totalSeconds - (hours * 3600)) / 60);
var seconds = totalSeconds - (hours * 3600) - (minutes * 60);
// round seconds
seconds = Math.round(seconds * 100) / 100
var result = (hours < 10 ? "0" + hours : hours);
result += "-" + (minutes < 10 ? "0" + minutes : minutes);
result += "-" + (seconds < 10 ? "0" + seconds : seconds);
return result;
}
Example use
var seconds = SecondsTohhmmss(70);
console.log(seconds);
// logs 00-01-10

There are lots of options of solve this problem, and obvious there are good option suggested about, But I wants to add one more optimized code here
function formatSeconds(sec) {
return [(sec / 3600), ((sec % 3600) / 60), ((sec % 3600) % 60)]
.map(v => v < 10 ? "0" + parseInt(v) : parseInt(v))
.filter((i, j) => i !== "00" || j > 0)
.join(":");
}
if you don't wants formatted zero with less then 10 number, you can use
function formatSeconds(sec) {
return parseInt(sec / 3600) + ':' + parseInt((sec % 3600) / 60) + ':' + parseInt((sec % 3600) % 60);
}
Sample Code http://fiddly.org/1c476/1

In one line, using T.J. Crowder's solution :
secToHHMMSS = seconds => `${Math.floor(seconds / 3600)}:${Math.floor((seconds % 3600) / 60)}:${Math.floor((seconds % 3600) % 60)}`
In one line, another solution that also count days :
secToDHHMMSS = seconds => `${parseInt(seconds / 86400)}d ${new Date(seconds * 1000).toISOString().substr(11, 8)}`
Source : https://gist.github.com/martinbean/2bf88c446be8048814cf02b2641ba276

var sec_to_hms = function(sec){
var min, hours;
sec = sec - (min = Math.floor(sec/60))*60;
min = min - (hours = Math.floor(min/60))*60;
return (hours?hours+':':'') + ((min+'').padStart(2, '0')) + ':'+ ((sec+'').padStart(2, '0'));
}
alert(sec_to_hms(2442542));

Have you tried adding seconds to a Date object?
Date.prototype.addSeconds = function(seconds) {
this.setSeconds(this.getSeconds() + seconds);
};
var dt = new Date();
dt.addSeconds(1234);
A sample:
https://jsfiddle.net/j5g2p0dc/5/
Updated:
Sample link was missing so I created a new one.

You can also use below code:
int ss = nDur%60;
nDur = nDur/60;
int mm = nDur%60;
int hh = nDur/60;

For anyone using AngularJS, a simple solution is to filter the value with the date API, which converts milliseconds to a string based on the requested format. Example:
<div>Offer ends in {{ timeRemaining | date: 'HH:mm:ss' }}</div>
Note that this expects milliseconds, so you may want to multiply timeRemaining by 1000 if you are converting from seconds (as the original question was formulated).

I ran into the case some have mentioned where the number of seconds is more than a day. Here's an adapted version of #Harish Anchu's top-rated answer that accounts for longer periods of time:
function secondsToTime(seconds) {
const arr = new Date(seconds * 1000).toISOString().substr(11, 8).split(':');
const days = Math.floor(seconds / 86400);
arr[0] = parseInt(arr[0], 10) + days * 24;
return arr.join(':');
}
Example:
secondsToTime(101596) // outputs '28:13:16' as opposed to '04:13:16'

String.prototype.toHHMMSS = function () {
var sec_num = parseInt(this, 10); // don't forget the second param
var hours = Math.floor(sec_num / 3600);
var minutes = Math.floor((sec_num - (hours * 3600)) / 60);
var seconds = sec_num - (hours * 3600) - (minutes * 60);
if (hours < 10) {hours = "0"+hours;}
if (minutes < 10) {minutes = "0"+minutes;}
if (seconds < 10) {seconds = "0"+seconds;}
return hours+':'+minutes+':'+seconds;
}
Usage Example
alert("186".toHHMMSS());

how to get the days and exact minutes,exact hours from two dates in javascript?

I have start date time and end time,i need to split how many days , hours ,minutes in the two dates
for example ,
startdatetime = "09-06-2017 10:30"
enddatetime = "10-06-2017 11:45"
i need this result : 1 day 1 hour and 15 minutes
I try this one
var t = end - start;
var z = parseInt(t / 1000 / 60);
var time = display(z);
function display(a)
{
console.log(a);
var hours = Math.trunc(a/60);
var minutes = a % 60;
var one_day=1000*60*60*24
var days = Math.ceil(a/one_day)
var time = [hours,minutes,days];
return time;
}
i get the following 1day 24 hours and 15 minutes , can anyone help me , if its new logic means i will change into it,thanks in advance

Using momentjs, you can :
Parse your input string using moment(String, String)
Parse your input string using moment.utc
Get difference using diff() function
Create a duration from the difference value
Use duration days(), hours(), minutes() to get your result
Here a live sample:
var startdatetime = "2017-06-09T07:00:01.000Z";
var enddatetime = "2017-06-10T09:00:00.000Z";
// Parse input
var mStart = moment.utc(startdatetime);
var mEnd = moment.utc(enddatetime);
// Calculate difference and create duration
var dur = moment.duration( mEnd.diff(mStart) );
// Show the result
console.log(dur.days() + ' days ' + dur.hours() + ' hour ' + dur.minutes() + ' minutes');
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.18.1/moment.min.js"></script>
If you want you can use moment-duration-format plug-in to get the same result using format() method on duration. Here a working sample:
var startdatetime = "2017-06-09T07:00:01.000Z";
var enddatetime = "2017-06-10T09:00:00.000Z";
// Parse input
var mStart = moment.utc(startdatetime);
var mEnd = moment.utc(enddatetime);
// Calculate difference and create duration
var dur = moment.duration( mEnd.diff(mStart) );
// Show the result
console.log(dur.format('d [day] h [hour] m [minutes]'));
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.18.1/moment.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment-duration-format/1.3.0/moment-duration-format.min.js"></script>

Well, if you look at documentation for javascript Date objects, there is a getTime() method . You can also use the valueOf() method. They both return the number of milliseconds representing your Date object.
You can simply call that on both Date objects and then find the difference. Once you have the difference you can find the amount of secs, mins , hrs, days, etc. Here is an example:
var start = new Date(*some date*);
var end = new Date(*some date*);
var dif = end.valueOf() - start.valueOf();
if (dif >= 0) {
var secs = Math.floor(dif / 1000 % 60);
var mins = Math.floor(dif / 1000 / 60 % 60);
var hrs = Math.floor(dif / 1000 / 60 / 60 % 24);
var days =
Math.floor(dif / 1000 / 60 / 60 / 24 % 365);
var yrs =
Math.floor(dif / 1000 / 60 / 60 / 24 / 365);

Try the following:
var t = end - start;
var z = parseInt(t / 1000 / 60);
var time = display(z);
function display(minutes)
{
var hours = (minutes / 60 | 0) % 24;
var minutes = (minutes | 0) % 60;
var days = minutes / 60 / 24 | 0;
return [hours, minutes, days];
}
Note that in javascript, doing x | 0 is the same as Math.floor(x).

It looks to me like your calculation for hours still has the days in it. Once you have established the days, just subtract those out when you calculate the hours.
var start = new Date("June 09, 2017 10:30:00");
var end = new Date("June 10, 2017 11:45:00");
var t = end - start;
var z = parseInt(t / 1000 / 60);
var time = display(z);
console.log(time);
function display(a)
{
var minutes = a % 60;
var one_day=1000*60*60*24
var days = Math.ceil(a/one_day)
var hours = Math.trunc((a-(days*1440))/60);
var time = [hours,minutes,days];
return time;
}
Having said that, I highly recommend moment.js to handle this type of thing, if you can.

var startDateTime = 1497029400000;
var endDateTime = 1497120300000;
var timeDifference = endDateTime - startDateTime
// with the given dates, days equals 1.0520833333333333
// we want to extract the trailing decimal values using modulus to get the other times
function getTimeDifference(timeDifference) {
var days = timeDifference/1000/60/60/24
days >= 1
? var dayCount = Math.trunc(days); // store the day count
: var dayCount = 0; // it is less than one day
// get the remaining hours
var hours = (days % 1) * 24;
var hoursCount = Math.trunc((days % 1) * 24);
// get the remaining minutes
var minutesCount = Math.ceil((hours % 1) * 60);
}

calculate time difference between two date in HH:MM:SS javascript

I have created one timer application in javascript.
Firstly it takes the current UTC date to init timer with some reference. here's the code
on_timer: function(e) {
var self = this;
if ($(e.target).hasClass("pt_timer_start")) {
var current_date = this.get_current_UTCDate();
this.project_timesheet_db.set_current_timer_activity({date: current_date});
this.start_interval();
this.initialize_timer();
this.$el.find(".pt_timer_start,.pt_timer_stop").toggleClass("o_hidden");
Now, Once timer is started and after some time span timer has some elapsed time with reference to above on_timer: function(e) function.
This function is
start_interval: function() {
var timer_activity = this.project_timesheet_db.get_current_timer_activity();
var self = this;
this.intervalTimer = setInterval(function(){
self.$el.find(".pt_duration").each(function() {
var el_hour = $(this).find("span.hours");
var el_minute = $(this).find("span.minutes");
var minute = parseInt(el_minute.text());
if(minute >= 60) {
el_hour.text(_.str.sprintf("%02d", parseInt(el_hour.text()) + 1));
minute = 0;
}
el_minute.text(_.str.sprintf("%02d", minute));
var el_second = $(this).find("span.seconds");
var seconds = parseInt(el_second.text()) + 1;
if(seconds > 60) {
el_minute.text(_.str.sprintf("%02d", parseInt(el_minute.text()) + 1));
seconds = 0;
}
el_second.text(_.str.sprintf("%02d", seconds));
});
}, 1000);
},
Now, considering el_hour, el_minute, el_seconds How to can i count time difference between init time and current timer value in HH:MM:SS manner.
thanks in advance for help

To convert H:M:S to seconds, you can use a simple function like:
// Convert H:M:S to seconds
// Seconds are optional (i.e. n:n is treated as h:s)
function hmsToSeconds(s) {
var b = s.split(':');
return b[0]*3600 + b[1]*60 + (+b[2] || 0);
}
Then to convert seconds back to HMS:
// Convert seconds to hh:mm:ss
// Allow for -ve time values
function secondsToHMS(secs) {
function z(n){return (n<10?'0':'') + n;}
var sign = secs < 0? '-':'';
secs = Math.abs(secs);
return sign + z(secs/3600 |0) + ':' + z((secs%3600) / 60 |0) + ':' + z(secs%60);
}
var a = '01:43:28';
var b = '12:22:46';
console.log(secondsToHMS(hmsToSeconds(a) - hmsToSeconds(b))); // -10:39:18
console.log(secondsToHMS(hmsToSeconds(b) - hmsToSeconds(a))); // 10:39:18
You may want to abbreviate the function names to say:
toHMS(toSec(a) - toSec(b)); // -10:39:18
Note that this doesn't cover where the time may cross a daylight saving boundary. For that you need fully qualified dates that include the year, month and day. Use the values to create date objects, find the difference, convert to seconds and use the secondsToHMS function.
Edit
The question title mentions dates, however the content only seems to mention strings of hours, minutes and seconds.
If you have Date objects, you can get the difference between them in milliseconds using:
var diffMilliseconds = date0 - date1;
and convert to seconds:
var diffSeconds = diffMilliseconds / 1000;
and present as HH:MM:SS using the secondsToHMS function above:
secondsToHMS((date0 - date1) / 1000);
e.g.
var d0 = new Date(2014,10,10,1,43,28);
var d1 = new Date(2014,10,10,12,22,46);
console.log( secondsToHMS((d0 - d1) / 1000)); // -10:39:18

I think there is a simpler solution.
function dateDiffToString(a, b){
// make checks to make sure a and b are not null
// and that they are date | integers types
diff = Math.abs(a - b);
ms = diff % 1000;
diff = (diff - ms) / 1000
ss = diff % 60;
diff = (diff - ss) / 60
mm = diff % 60;
diff = (diff - mm) / 60
hh = diff % 24;
days = (diff - hh) / 24
return days + ":" + hh+":"+mm+":"+ss+"."+ms;
}
var today = new Date()
var yest = new Date()
yest = yest.setDate(today.getDate()-1)
console.log(dateDiffToString(yest, today))

const dateDiffToString = (a, b) => {
let diff = Math.abs(a - b);
let ms = diff % 1000;
diff = (diff - ms) / 1000;
let s = diff % 60;
diff = (diff - s) / 60;
let m = diff % 60;
diff = (diff - m) / 60;
let h = diff;
let ss = s <= 9 && s >= 0 ? `0${s}` : s;
let mm = m <= 9 && m >= 0 ? `0${m}` : m;
let hh = h <= 9 && h >= 0 ? `0${h}` : h;
return hh + ':' + mm + ':' + ss;
};

This may be the simple answer
var d1 = new Date(2014,10,11,1,43,28);
var d2 = new Date(2014,10,11,2,53,58);
var date = new Date(d2-d1);
var hour = date.getUTCHours();
var min = date.getUTCMinutes();
var sec = date.getUTCSeconds();
var day = date.getUTCDate() - 1;
console.log(day + ":" + hour + ":" + min + ":" + sec)

More intuitive and easier to read.
function hmsToSeconds(t) {
const [hours, minutes, seconds] = t.split(':')
return Number(hours) * 60 * 60 + Number(minutes) * 60 + Number(seconds)
}
function secondsToHMS(secs) {
return new Date(secs * 1000).toISOString().substr(11, 8)
}
var startTime = '01:43:28';
var endTime = '12:22:46';
console.log(secondsToHMS(hmsToSeconds(endTime) - hmsToSeconds(startTime))); //10:39:18

compare date in javascript format 06/01/2014 13:24

I am trying to compare two dates in javascript, which are in two input type="text", and that have the format "06/11/2013 13:24".
Any idea of how to compare them?
Thanks!

You need to parse string to Date - object:
var firstDate = new Date("06/11/2013 13:24");
var secondDate = new Date(youSecondDateString);

Age from Date of Birth using JQuery
Possibly a duplicate question to above but the answers is
var startDt=document.getElementById("startDateId").value;
var endDt=document.getElementById("endDateId").value;
if( (new Date(startDt).getTime() > new Date(endDt).getTime()))
{
//perform desired operation here
}

The Date object will do what you want - construct one for each date, then just compare them using the usual operators.
For example, subtracting date1 from date2 will give you the number of milliseconds between two dates.
You can get the number of seconds by dividing the milliseconds by 1000, and rounding the number:
var seconds = Math.round((date2-date1)/1000);
You could then divide by 60 to get the minutes, again by 60 to get the hours, then by 24 to get the days (and so on).
Here's how you might get a figure in the format dd:hh:mm
window.minutesPerDay = 60 * 24;
function pad(number) {
var result = "" + number;
if (result.length < 2) {
result = "0" + result;
}
return result;
}
function millisToDaysHoursMinutes(millis) {
var seconds = millis / 1000;
var totalMinutes = seconds / 60;
var days = totalMinutes / minutesPerDay;
totalMinutes -= minutesPerDay * days;
var hours = totalMinutes / 60;
totalMinutes -= hours * 60;
return days + ":" + pad(hours) + ":" + pad(totalMinutes);
}
var date1 = new Date("06/11/2013 13:24"),
date2 = new Date("07/11/2013 13:24"),
milliseconds = date2 - date1;
alert(millisToDaysHoursMinutes(milliseconds));
Fiddle
I took the millisToDaysHoursMinutes function from here.