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Put clock forward by one hour [duplicate]

It amazes me that JavaScript's Date object does not implement an add function of any kind.
I simply want a function that can do this:
var now = Date.now();
var fourHoursLater = now.addHours(4);
function Date.prototype.addHours(h) {
// How do I implement this?
}
I would simply like some pointers in a direction.
Do I need to do string parsing?
Can I use setTime?
How about milliseconds?
Like this:
new Date(milliseconds + 4*3600*1000 /* 4 hours in ms */)?
This seems really hackish though - and does it even work?

JavaScript itself has terrible Date/Time API's. Nonetheless, you can do this in pure JavaScript:
Date.prototype.addHours = function(h) {
this.setTime(this.getTime() + (h*60*60*1000));
return this;
}

Date.prototype.addHours= function(h){
this.setHours(this.getHours()+h);
return this;
}
Test:
alert(new Date().addHours(4));

The below code will add 4 hours to a date (example, today's date):
var today = new Date();
today.setHours(today.getHours() + 4);
It will not cause an error if you try to add 4 to 23 (see the documentation):
If a parameter you specify is outside of the expected range, setHours() attempts to update the date information in the Date object accordingly

It is probably better to make the addHours method immutable by returning a copy of the Date object rather than mutating its parameter.
Date.prototype.addHours= function(h){
var copiedDate = new Date(this.getTime());
copiedDate.setHours(copiedDate.getHours()+h);
return copiedDate;
}
This way you can chain a bunch of method calls without worrying about state.

The version suggested by kennebec will fail when changing to or from DST, since it is the hour number that is set.
this.setUTCHours(this.getUTCHours()+h);
will add h hours to this independent of time system peculiarities.
Jason Harwig's method works as well.

Get a date exactly two hours from now, in one line.
You need to pass milliseconds to new Date.
let expiryDate = new Date(new Date().setHours(new Date().getHours() + 2));
or
let expiryDate2 = new Date(Date.now() + 2 * (60 * 60 * 1000) );
let nowDate = new Date();
let expiryDate = new Date(new Date().setHours(new Date().getHours() + 2));
let expiryDate2 = new Date(Date.now() + 2 * (60 * 60 * 1000) );
console.log('now', nowDate);
console.log('expiry', expiryDate);
console.log('expiry 2', expiryDate2);

You can use the Moment.js library.
var moment = require('moment');
foo = new moment(something).add(10, 'm').toDate();

I also think the original object should not be modified. So to save future manpower here's a combined solution based on Jason Harwig's and Tahir Hasan answers:
Date.prototype.addHours= function(h){
var copiedDate = new Date();
copiedDate.setTime(this.getTime() + (h*60*60*1000));
return copiedDate;
}

If you would like to do it in a more functional way (immutability) I would return a new date object instead of modifying the existing and I wouldn't alter the prototype but create a standalone function. Here is the example:
//JS
function addHoursToDate(date, hours) {
return new Date(new Date(date).setHours(date.getHours() + hours));
}
//TS
function addHoursToDate(date: Date, hours: number): Date {
return new Date(new Date(date).setHours(date.getHours() + hours));
}
let myDate = new Date();
console.log(myDate)
console.log(addHoursToDate(myDate,2))

There is an add in the Datejs library.
And here are the JavaScript date methods. kennebec wisely mentioned getHours() and setHours();

Check if it’s not already defined. Otherwise, define it in the Date prototype:
if (!Date.prototype.addHours) {
Date.prototype.addHours = function(h) {
this.setHours(this.getHours() + h);
return this;
};
}

This is an easy way to get an incremented or decremented data value.
const date = new Date()
const inc = 1000 * 60 * 60 // an hour
const dec = (1000 * 60 * 60) * -1 // an hour
const _date = new Date(date)
return new Date(_date.getTime() + inc)
return new Date(_date.getTime() + dec)

Another way to handle this is to convert the date to unixtime (epoch), then add the equivalent in (milli)seconds, then convert it back. This way you can handle day and month transitions, like adding 4 hours to 21, which should result in the next day, 01:00.

SPRBRN is correct. In order to account for the beginning/end of the month and year, you need to convert to Epoch and back.
Here's how you do that:
var milliseconds = 0; //amount of time from current date/time
var sec = 0; //(+): future
var min = 0; //(-): past
var hours = 2;
var days = 0;
var startDate = new Date(); //start date in local time (we'll use current time as an example)
var time = startDate.getTime(); //convert to milliseconds since epoch
//add time difference
var newTime = time + milliseconds + (1000*sec) + (1000*60*min) + (1000*60*60*hrs) + (1000*60*60*24*days);
var newDate = new Date(newTime); //convert back to date; in this example: 2 hours from right now
Or do it in one line (where variable names are the same as above:
var newDate =
new Date(startDate.getTime() + millisecond +
1000 * (sec + 60 * (min + 60 * (hours + 24 * days))));

For a simple add/subtract hour/minute function in JavaScript, try this:
function getTime (addHour, addMin){
addHour = (addHour ? addHour : 0);
addMin = (addMin ? addMin : 0);
var time = new Date(new Date().getTime());
var AM = true;
var ndble = 0;
var hours, newHour, overHour, newMin, overMin;
// Change form 24 to 12 hour clock
if(time.getHours() >= 13){
hours = time.getHours() - 12;
AM = (hours>=12 ? true : false);
}else{
hours = time.getHours();
AM = (hours>=12 ? false : true);
}
// Get the current minutes
var minutes = time.getMinutes();
// Set minute
if((minutes + addMin) >= 60 || (minutes + addMin) < 0){
overMin = (minutes + addMin) % 60;
overHour = Math.floor((minutes + addMin - Math.abs(overMin))/60);
if(overMin < 0){
overMin = overMin + 60;
overHour = overHour-Math.floor(overMin/60);
}
newMin = String((overMin<10 ? '0' : '') + overMin);
addHour = addHour + overHour;
}else{
newMin = minutes + addMin;
newMin = String((newMin<10 ? '0' : '') + newMin);
}
// Set hour
if((hours + addHour >= 13) || (hours + addHour <= 0)){
overHour = (hours + addHour) % 12;
ndble = Math.floor(Math.abs((hours + addHour)/12));
if(overHour <= 0){
newHour = overHour + 12;
if(overHour == 0){
ndble++;
}
}else{
if(overHour == 0){
newHour = 12;
ndble++;
}else{
ndble++;
newHour = overHour;
}
}
newHour = (newHour<10 ? '0' : '') + String(newHour);
AM = ((ndble + 1) % 2 === 0) ? AM : !AM;
}else{
AM = (hours + addHour == 12 ? !AM : AM);
newHour = String((Number(hours) + addHour < 10 ? '0': '') + (hours + addHour));
}
var am = (AM) ? 'AM' : 'PM';
return new Array(newHour, newMin, am);
};
This can be used without parameters to get the current time:
getTime();
Or with parameters to get the time with the added minutes/hours:
getTime(1, 30); // Adds 1.5 hours to current time
getTime(2); // Adds 2 hours to current time
getTime(0, 120); // Same as above
Even negative time works:
getTime(-1, -30); // Subtracts 1.5 hours from current time
This function returns an array of:
array([Hour], [Minute], [Meridian])

If you need it as a string, for example:
var defaultTime: new Date().getHours() + 1 + ":" + new Date().getMinutes();

I think this should do the trick
var nextHour = Date.now() + 1000 * 60 * 60;
console.log(nextHour)

You can even format the date in desired format using the moment function after adding 2 hours.
var time = moment(new Date(new Date().setHours(new Date().getHours() + 2))).format("YYYY-MM-DD");
console.log(time);

A little messy, but it works!
Given a date format like this: 2019-04-03T15:58
//Get the start date.
var start = $("#start_date").val();
//Split the date and time.
var startarray = start.split("T");
var date = startarray[0];
var time = startarray[1];
//Split the hours and minutes.
var timearray = time.split(":");
var hour = timearray[0];
var minute = timearray[1];
//Add an hour to the hour.
hour++;
//$("#end_date").val = start;
$("#end_date").val(""+date+"T"+hour+":"+minute+"");
Your output would be: 2019-04-03T16:58

The easiest way to do it is:
var d = new Date();
d = new Date(d.setHours(d.getHours() + 2));
It will add 2 hours to the current time.
The value of d = Sat Jan 30 2021 23:41:43 GMT+0500 (Pakistan Standard Time).
The value of d after adding 2 hours = Sun Jan 31 2021 01:41:43 GMT+0500 (Pakistan Standard Time).

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();

Subtract the date from current date javascript [duplicate]

I'm creating an application which lets you define events with a time frame. I want to automatically fill in the end date when the user selects or changes the start date. I can't quite figure out, however, how to get the difference between the two times, and then how to create a new end Date using that difference.

In JavaScript, dates can be transformed to the number of milliseconds since the epoc by calling the getTime() method or just using the date in a numeric expression.
So to get the difference, just subtract the two dates.
To create a new date based on the difference, just pass the number of milliseconds in the constructor.
var oldBegin = ...
var oldEnd = ...
var newBegin = ...
var newEnd = new Date(newBegin + oldEnd - oldBegin);
This should just work
EDIT: Fixed bug pointed by #bdukes
EDIT:
For an explanation of the behavior, oldBegin, oldEnd, and newBegin are Date instances. Calling operators + and - will trigger Javascript auto casting and will automatically call the valueOf() prototype method of those objects. It happens that the valueOf() method is implemented in the Date object as a call to getTime().
So basically: date.getTime() === date.valueOf() === (0 + date) === (+date)

JavaScript perfectly supports date difference out of the box
https://jsfiddle.net/b9chris/v5twbe3h/
var msMinute = 60*1000,
msDay = 60*60*24*1000,
a = new Date(2012, 2, 12, 23, 59, 59),
b = new Date("2013 march 12");
console.log(Math.floor((b - a) / msDay) + ' full days between'); // 364
console.log(Math.floor(((b - a) % msDay) / msMinute) + ' full minutes between'); // 0
Now some pitfalls. Try this:
console.log(a - 10); // 1331614798990
console.log(a + 10); // mixed string
So if you have risk of adding a number and Date, convert Date to number directly.
console.log(a.getTime() - 10); // 1331614798990
console.log(a.getTime() + 10); // 1331614799010
My fist example demonstrates the power of Date object but it actually appears to be a time bomb

See JsFiddle DEMO
var date1 = new Date();
var date2 = new Date("2025/07/30 21:59:00");
//Customise date2 for your required future time
showDiff();
function showDiff(date1, date2){
var diff = (date2 - date1)/1000;
diff = Math.abs(Math.floor(diff));
var days = Math.floor(diff/(24*60*60));
var leftSec = diff - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
document.getElementById("showTime").innerHTML = "You have " + days + " days " + hrs + " hours " + min + " minutes and " + leftSec + " seconds before death.";
setTimeout(showDiff,1000);
}
for your HTML Code:
<div id="showTime"></div>

If you don't care about the time component, you can use .getDate() and .setDate() to just set the date part.
So to set your end date to 2 weeks after your start date, do something like this:
function GetEndDate(startDate)
{
var endDate = new Date(startDate.getTime());
endDate.setDate(endDate.getDate()+14);
return endDate;
}
To return the difference (in days) between two dates, do this:
function GetDateDiff(startDate, endDate)
{
return endDate.getDate() - startDate.getDate();
}
Finally, let's modify the first function so it can take the value returned by 2nd as a parameter:
function GetEndDate(startDate, days)
{
var endDate = new Date(startDate.getTime());
endDate.setDate(endDate.getDate() + days);
return endDate;
}

Thanks #Vincent Robert, I ended up using your basic example, though it's actually newBegin + oldEnd - oldBegin. Here's the simplified end solution:
// don't update end date if there's already an end date but not an old start date
if (!oldEnd || oldBegin) {
var selectedDateSpan = 1800000; // 30 minutes
if (oldEnd) {
selectedDateSpan = oldEnd - oldBegin;
}
newEnd = new Date(newBegin.getTime() + selectedDateSpan));
}

Depending on your needs, this function will calculate the difference between the 2 days, and return a result in days decimal.
// This one returns a signed decimal. The sign indicates past or future.
this.getDateDiff = function(date1, date2) {
return (date1.getTime() - date2.getTime()) / (1000 * 60 * 60 * 24);
}
// This one always returns a positive decimal. (Suggested by Koen below)
this.getDateDiff = function(date1, date2) {
return Math.abs((date1.getTime() - date2.getTime()) / (1000 * 60 * 60 * 24));
}

If using moment.js, there is a simpler solution, which will give you the difference in days in one single line of code.
moment(endDate).diff(moment(beginDate), 'days');
Additional details can be found in the moment.js page
Cheers,
Miguel

function compare()
{
var end_actual_time = $('#date3').val();
start_actual_time = new Date();
end_actual_time = new Date(end_actual_time);
var diff = end_actual_time-start_actual_time;
var diffSeconds = diff/1000;
var HH = Math.floor(diffSeconds/3600);
var MM = Math.floor(diffSeconds%3600)/60;
var formatted = ((HH < 10)?("0" + HH):HH) + ":" + ((MM < 10)?("0" + MM):MM)
getTime(diffSeconds);
}
function getTime(seconds) {
var days = Math.floor(leftover / 86400);
//how many seconds are left
leftover = leftover - (days * 86400);
//how many full hours fits in the amount of leftover seconds
var hours = Math.floor(leftover / 3600);
//how many seconds are left
leftover = leftover - (hours * 3600);
//how many minutes fits in the amount of leftover seconds
var minutes = leftover / 60;
//how many seconds are left
//leftover = leftover - (minutes * 60);
alert(days + ':' + hours + ':' + minutes);
}

alternative modificitaion extended code..
http://jsfiddle.net/vvGPQ/48/
showDiff();
function showDiff(){
var date1 = new Date("2013/01/18 06:59:00");
var date2 = new Date();
//Customise date2 for your required future time
var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));
var years = Math.floor(diff/(365*24*60*60));
var leftSec = diff - years * 365*24*60*60;
var month = Math.floor(leftSec/((365/12)*24*60*60));
var leftSec = leftSec - month * (365/12)*24*60*60;
var days = Math.floor(leftSec/(24*60*60));
var leftSec = leftSec - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
document.getElementById("showTime").innerHTML = "You have " + years + " years "+ month + " month " + days + " days " + hrs + " hours " + min + " minutes and " + leftSec + " seconds the life time has passed.";
setTimeout(showDiff,1000);
}

Below code will return the days left from today to futures date.
Dependencies: jQuery and MomentJs.
var getDaysLeft = function (date) {
var today = new Date();
var daysLeftInMilliSec = Math.abs(new Date(moment(today).format('YYYY-MM-DD')) - new Date(date));
var daysLeft = daysLeftInMilliSec / (1000 * 60 * 60 * 24);
return daysLeft;
};
getDaysLeft('YYYY-MM-DD');

<html>
<head>
<script>
function dayDiff()
{
var start = document.getElementById("datepicker").value;
var end= document.getElementById("date_picker").value;
var oneDay = 24*60*60*1000;
var firstDate = new Date(start);
var secondDate = new Date(end);
var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
document.getElementById("leave").value =diffDays ;
}
</script>
</head>
<body>
<input type="text" name="datepicker"value=""/>
<input type="text" name="date_picker" onclick="function dayDiff()" value=""/>
<input type="text" name="leave" value=""/>
</body>
</html>

this code fills the duration of study years when you input the start date and end date(qualify accured date) of study and check if the duration less than a year if yes the alert a message
take in mind there are three input elements the first txtFromQualifDate and second txtQualifDate and third txtStudyYears
it will show result of number of years with fraction
function getStudyYears()
{
if(document.getElementById('txtFromQualifDate').value != '' && document.getElementById('txtQualifDate').value != '')
{
var d1 = document.getElementById('txtFromQualifDate').value;
var d2 = document.getElementById('txtQualifDate').value;
var one_day=1000*60*60*24;
var x = d1.split("/");
var y = d2.split("/");
var date1=new Date(x[2],(x[1]-1),x[0]);
var date2=new Date(y[2],(y[1]-1),y[0])
var dDays = (date2.getTime()-date1.getTime())/one_day;
if(dDays < 365)
{
alert("the date between start study and graduate must not be less than a year !");
document.getElementById('txtQualifDate').value = "";
document.getElementById('txtStudyYears').value = "";
return ;
}
var dMonths = Math.ceil(dDays / 30);
var dYears = Math.floor(dMonths /12) + "." + dMonths % 12;
document.getElementById('txtStudyYears').value = dYears;
}
}

If you use Date objects and then use the getTime() function for both dates it will give you their respective times since Jan 1, 1970 in a number value. You can then get the difference between these numbers.
If that doesn't help you out, check out the complete documentation: http://www.w3schools.com/jsref/jsref_obj_date.asp

var getDaysLeft = function (date1, date2) {
var daysDiffInMilliSec = Math.abs(new Date(date1) - new Date(date2));
var daysLeft = daysDiffInMilliSec / (1000 * 60 * 60 * 24);
return daysLeft;
};
var date1='2018-05-18';
var date2='2018-05-25';
var dateDiff = getDaysLeft(date1, date2);
console.log(dateDiff);

To get the date difference in milliseconds between two dates:
var diff = Math.abs(date1 - date2);
I'm not sure what you mean by converting the difference back into a date though.

Many answers here are based on a direct subtraction of Date objects like new Date(…) - new Date(…). This is syntactically wrong. Browsers still accept it because of backward compatibility. But modern JS linters will throw at you.
The right way to calculate date differences in milliseconds is new Date(…).getTime() - new Date(…).getTime():
// Time difference between two dates
let diffInMillis = new Date(…).getTime() - new Date(…).getTime()
If you want to calculate the time difference to now, you can just remove the argument from the first Date:
// Time difference between now and some date
let diffInMillis = new Date().getTime() - new Date(…).getTime()

function checkdate() {
var indate = new Date()
indate.setDate(dat)
indate.setMonth(mon - 1)
indate.setFullYear(year)
var one_day = 1000 * 60 * 60 * 24
var diff = Math.ceil((indate.getTime() - now.getTime()) / (one_day))
var str = diff + " days are remaining.."
document.getElementById('print').innerHTML = str.fontcolor('blue')
}

THIS IS WHAT I DID ON MY SYSTEM.
var startTime=("08:00:00").split(":");
var endTime=("16:00:00").split(":");
var HoursInMinutes=((parseInt(endTime[0])*60)+parseInt(endTime[1]))-((parseInt(startTime[0])*60)+parseInt(startTime[1]));
console.log(HoursInMinutes/60);

How to find difference between two DateTime Strings in Javascript

So I have two strings in javascript:
old_date = "2010-11-10 07:30:40";
new_date = "2010-11-15 08:03:22";
I want to find the difference between these two dates, but I am totally at a loss :(
I tried to do the following:
old_date_obj = new Date(Date.parse(old_date));
new_date_obj = new Date(Date.parse(new_date));
But it is giving me error. However that is only start of my woes...I need to show the difference as:
Difference is X Days, Y hours and Z minutes
Can JavaScript/jQuery gurus help me please? Much appreciated...

<script>
function Calculate() {
old_date = "2010-11-10 07:30:40";
new_date = "2010-11-15 08:03:22";
old_date_obj = new Date(Date.parse(old_date, "dd/mm/yyyy HH:mm:ss"));
new_date_obj = new Date(Date.parse(new_date, "dd/mm/yyyy HH:mm:ss"));
var utc1 = Date.UTC(new_date_obj.getFullYear(), new_date_obj.getMonth(), new_date_obj.getDate());
var utc2 = Date.UTC(old_date_obj.getFullYear(), old_date_obj.getMonth(), old_date_obj.getDate());
alert(Math.floor((utc2 - utc1) / (1000 * 60 * 60 * 24)));
}
</script>

simply add in you date and it will work for you.
"2010-11-10T07:30:40+01:00"
for more detail check this answer
Answer in detail

<script type="text/javascript">
// The number of milliseconds in one day, hour, and minute
var ONE_DAY = 1000 * 60 * 60 * 24;
var ONE_HOUR = 1000 * 60 * 60;
var ONE_MINUTE = 1000 * 60;
var old_date = "2010-11-10T07:30:40";
var new_date = "2010-11-15T08:03:22";
// Convert both dates to milliseconds
var old_date_obj = new Date(old_date).getTime();
var new_date_obj = new Date(new_date).getTime();
// Calculate the difference in milliseconds
var difference_ms = Math.abs(new_date_obj - old_date_obj)
// Convert back to days, hours, and minutes
var days = Math.round(difference_ms / ONE_DAY);
var hours = Math.round(difference_ms / ONE_HOUR) - (days * 24) - 1;
var minutes = Math.round(difference_ms / ONE_MINUTE) - (days * 24 * 60) - (hours * 60);
alert('Difference is ' + days + ' days, ' + hours + ' hours and ' + minutes + ' minutes.' );
</script>

<script type="text/javascript">
function getDates(strDate1, strDate2) {
/*Now strDate1 and strDate2 string. So we should convert them to javascript datetime value.*/
var tempDate1 = strDate1.split(/\-|\s/)
var date1 = new Date(tempDate1.slice(0,3).reverse().join('/')+' '+tempDate1[3]);
var tempDate2 = strDate2.split(/\-|\s/)
var date2 = new Date(tempDate2.slice(0,3).reverse().join('/')+' '+tempDate2[3]);
var obj1 = $.datepicker.parseDate('dd.mm.yy', $("#date1").val());
var obj2 = $.datepicker.parseDate('dd.mm.yy', $("#date2").val());
console.log(findDifferentDate(obj1, obj2));
}
function findDifferentDate(obj1, obj2){
var date1 = getFormattedDate(obj1);
var date2 = getFormattedDate(obj2);
var year = date1.getFullYear() - date2.getFullYear();
var day = date1.getDate() - date2.getDate();
var month = date1.getMonth() - date2.getMonth();
var seconds = date1.getSeconds() - date2.getSeconds();
var minutes = date1.getMinutes() - date2.getMinutes();
var hour = date1.getHours() - date2.getHours();
return 'Difference is' + day + 'Days' + month + 'Months' + year + 'Years' + seconds + 'Seconds' + minutes + 'Minutes' + hour + 'Hours';
}
function getFormattedDate(date) {
var year = date.getFullYear();
var month = (1 + date.getMonth()).toString();
month = month.length > 1 ? month : '0' + month;
var day = date.getDate().toString();
day = day.length > 1 ? day : '0' + day;
return day + '.' + month + '.' + year;
}
</script>
If you call getDates method with your dates and then u can see difference time in the console.

var old_date = "2010-11-15 07:30:40";
var new_date = "2010-11-15 08:03:22";
var old_date_obj = new Date(Date.parse(old_date));
var new_date_obj = new Date(Date.parse(new_date));
var diffMs = Math.abs(new_date_obj - old_date_obj);
var diffDays = Math.round(diffMs / 86400000); // days
var diffHrs = Math.round((diffMs % 86400000) / 3600000); // hours
var diffMins = Math.round(((diffMs % 86400000) % 3600000) / 60000); // minutes
https://jsfiddle.net/yps2wb58/1/

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.