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

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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();

Calendar to show today's current date with the next 60 days

I have created a calendar that successfully displays todays current date. It also shows the remaining months however I am trying to change the functionality of the calendar so that it only shows the next 60 days from today's current date.
The code so far:
var todayDate = new Date();
var finalDate = new Date(todayDate)
finalDate.setDate(todayDate.getDate() + 60)
var dayNumber = todayDate.getDate();
var month = todayDate.getMonth();
var year = todayDate.getFullYear();
var months = ["January","February","March","April","May","June","July","August","September","October","November","December"];
var calendarTable = document.getElementById("calendar-body");
document.getElementById("month").innerHTML = months[month];
document.getElementById("year").innerHTML = year;
function createCalendar(month, year) {
var firstDay = new Date(year, month).getDay();
var totalDays = daysInMonth(month, year);
blankDates(firstDay);
for (var day = 1; day <= totalDays; day++) {
var cell = document.createElement("li");
var cellText = document.createTextNode(day);
if (
dayNumber === day &&
month === todayDate.getMonth() &&
year === todayDate.getFullYear()
) {
cell.classList.add("todays-day");
}
cell.setAttribute("data-day", day);
cell.setAttribute("data-month", month);
cell.setAttribute("data-year", year);
cell.classList.add("singleDay");
cell.appendChild(cellText);
calendarTable.appendChild(cell);
}
}
function daysInMonth(month, year) {
return new Date(year, month + 1, 0).getDate();
}
function blankDates(count) {
for (var x = 0; x < count; x++) {
var cell = document.createElement("li");
var cellText = document.createTextNode("");
cell.appendChild(cellText);
calendarTable.appendChild(cell);
}
}
createCalendar(month, year);
Any ideas would be great. Thanks :)

First you need to find the number of dates between two dates,
In this case is finalDate and todayDate
var diff = Math.floor(( Date.parse(finalDate) - Date.parse(todayDate)) / 86400000);
You get the difference days (or NaN if one or both could not be parsed). The parse date gived the result in milliseconds and to get it by day you have to divided it by 24 * 60 * 60 * 1000
The output of the diff now is 60
Then you add the diff variable into the for loop.
Woking example

Just an idea.
But you could generate an array of the dates between now and 60 days.
Then loop through that array to build the calendar.
This way the Date functions can be used for each date in the array.
Here's a simple function to generate such array.
function getDateRange(pStartDate, pDays) {
let days = pDays || 1;
let dates = [];
let n = 0;
let startDate = new Date(pStartDate.toISOString().slice(0,10));
let endDate = new Date(startDate.getTime() + (days * 86400000));
do {
dates.push(new Date(startDate.getTime() + (n * 86400000)));
n++;
} while (n <= days);
let obj = {
startDate : startDate,
endDate : endDate,
days : dates.length,
dates : dates
}
return obj;
}
var dateRange = getDateRange(new Date(), 60);
console.log(dateRange.dates);

Get only number of working days between two days in javascript

I have to calculate variables in javascript after x days from current day.I have to add some number of days based on some input parameter.
var currentDate = new Date();
var dd = currentDate.getDate();
var mm = currentDate.getMonth()+1;
var yyyy = currentDate.getFullYear();
if(dd<10) {
dd='0'+dd
}
if(mm<10) {
mm='0'+mm
}
currentDate= mm+'/'+dd+'/'+yyyy;
Now I want to get some date after 28 days from currentDate variable but it should not include Saturday and Sunday.
So my question is how to exclude the weekends (2 days) from the 28 (for example).
Any help will be appreciated.

Here's a generic function to add n business days to a date
function addDays(dt, n) {
var rem = n % 5;
var add = 7 * (n - rem) / 5 + rem;
var ret = new Date(dt);
ret.setDate(ret.getDate() + add);
if (ret.getDay() == 6) ret.setDate(ret.getDate() + 2);
else if (ret.getDay() == 0) ret.setDate(ret.getDate() + 1);
return ret;
}

This is really simple. Script below goes through all days between start and end date and checks if it isn't Saturday (tmpDate.getDay() != 6) or Sunday (tmpDate.getDay() != 0)
var start = new Date();
var end = new Date(2016, 6, 1);
var allDays = Math.floor((end.getTime() - start.getTime())/ 86400000);
var workingDays = 0;
for(var i = 0; i < allDays; i++){
var tmpDate = new Date();
tmpDate.setTime(today.getTime() + i * 24*60*60*1000);
if(tmpDate.getDay() != 0 && tmpDate.getDay() != 6)
wokringDays++;
}
alert(workingDays);

This will give you the number of working days:
function getWorkingDays(currentDateObj, numberOfDays)
{
if(numberOfDays < 0) return false;
var futureDateObj = new Date();
futureDateObj.setDate(currentDateObj.getDate() + numberOfDays);
var daysCnt = 1 + Math.round((futureDateObj.getTime()-currentDateObj.getTime())/(24*3600*1000));
var weekCnt = Math.floor( (currentDateObj.getDay() + daysCnt) / 7 );
var weekends = 2 * weekCnt + (currentDateObj.getDay()==0) - (futureDateObj.getDay()==6);
return numberOfDays - weekends;
}
console.log(getWorkingDays(new Date(), 28));

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/

get day of month difference javascript

I want to get the difference between 2 dates in days months and years.
The problem is that because there is not a set amount of days to any Month, I can't figure out how I can compare 2 dates with Months that are between 28 and 31 days. I have done it for Years and Months, because I know there are 12 Months to a year, so I can calculate it. Maybe someone knows a method of getting round this. here's what I have so far. Thanks for reading.
var date1 = new Date(2013,10,1);
var date2 = new Date();
var y1 = date1.getFullYear();
var m1 = date1.getMonth();
var y2 = date2.getFullYear();
var m2 = date2.getMonth();
var totDiff = (y1 - y2) * 12 + (m1 - m2 -1);
var yDiff = Math.floor(Math.abs(totDiff) / 12);
var mDiff = totDiff - (Math.floor((totDiff) / 12) * 12);
console.log("years: " + yDiff + " months: " + mDiff);

I would iterate a date in a 'while' loop, increasing the lower date and counting iterations needed to achieve the greater date. That way you can count the difference in any time units.
The method may look like following. Quite a lot of code, but i hope it's clear (see 'dateDiff' method below):
function addTime(date, unit){
var date = new Date(date);
switch (unit){
case "day":
date.setDate(date.getDate() + 1);
break;
case "month":
date.setMonth(date.getMonth() + 1);
break;
case "year":
date.setYear(date.getFullYear() + 1);
break;
}
return date;
}
function dateDiff(from, to){
var from = new Date(from),
to = new Date(to),
res = {
days:0,
months:0,
years:0
};
while(addTime(from, 'year').valueOf() <= to.valueOf()){
res.years++;
from = addTime(from, 'year');
}
while(addTime(from, 'month').valueOf() <= to.valueOf()){
res.months++;
from = addTime(from, 'month');
}
while(addTime(from, 'day').valueOf() <= to.valueOf()){
res.days++;
from = addTime(from, 'day');
}
return res;
}
and usage:
dateDiff(new Date(..some date..), new Date(..some greater date..))
Example
console.log( dateDiff(new Date(2013, 11, 1), new Date()) );
//Object {days: 26, months: 0, years: 0}
console.log( dateDiff(new Date(2000, 0, 1), new Date()) );
//Object {days: 26, months: 11, years: 13}

var today=new Date('January 01, 2014 00:00:00');
var tt=new Date();
var ss=(today-tt)/1000;
days = parseInt(ss / 86400);
ss = ss % 86400;
hours = parseInt(ss / 3600);
ss = ss % 3600;
minutes = parseInt(ss / 60);
seconds = parseInt(ss % 60);
document.getElementById('d').innerHTML=days;
document.getElementById('h').innerHTML=hours;
document.getElementById('m').innerHTML=minutes;
document.getElementById('s').innerHTML=seconds;
You can find the difference betwwen two dates in year,days,hours.mins even in secs also , by using some calculation as above.