How to calculate the number of days between two dates? [duplicate]

Answer 1

Here is a function that does this:

function days_between(date1, date2) {

    // The number of milliseconds in one day
    const ONE_DAY = 1000 * 60 * 60 * 24;

    // Calculate the difference in milliseconds
    const differenceMs = Math.abs(date1 - date2);

    // Convert back to days and return
    return Math.round(differenceMs / ONE_DAY);

}

Answer 2

Here is my implementation:

function daysBetween(one, another) {
  return Math.round(Math.abs((+one) - (+another))/8.64e7);
}

+<date> does the type coercion to the integer representation and has the same effect as <date>.getTime() and 8.64e7 is the number of milliseconds in a day.

Answer 3

const oneDay = 24 * 60 * 60 * 1000; // hours*minutes*seconds*milliseconds
const firstDate = new Date(2008, 1, 12);
const secondDate = new Date(2008, 1, 22);

const diffDays = Math.round(Math.abs((firstDate - secondDate) / oneDay));

Answer 4

I have written this solution for another post who asked, how to calculate the difference between two dates, so I share what I have prepared:

// Here are the two dates to compare
var date1 = '2011-12-24';
var date2 = '2012-01-01';

// First we split the values to arrays date1[0] is the year, [1] the month and [2] the day
date1 = date1.split('-');
date2 = date2.split('-');

// Now we convert the array to a Date object, which has several helpful methods
date1 = new Date(date1[0], date1[1], date1[2]);
date2 = new Date(date2[0], date2[1], date2[2]);

// We use the getTime() method and get the unixtime (in milliseconds, but we want seconds, therefore we divide it through 1000)
date1_unixtime = parseInt(date1.getTime() / 1000);
date2_unixtime = parseInt(date2.getTime() / 1000);

// This is the calculated difference in seconds
var timeDifference = date2_unixtime - date1_unixtime;

// in Hours
var timeDifferenceInHours = timeDifference / 60 / 60;

// and finaly, in days :)
var timeDifferenceInDays = timeDifferenceInHours  / 24;

alert(timeDifferenceInDays);

You can skip some steps in the code, I have written it so to make it easy to understand.

You'll find a running example here: http://jsfiddle.net/matKX/

Answer 5

Here's what I use. If you just subtract the dates, it won't work across the Daylight Savings Time Boundary (eg April 1 to April 30 or Oct 1 to Oct 31). This drops all the hours to make sure you get a day and eliminates any DST problem by using UTC.

var nDays = (    Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate()) -
                 Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate())) / 86400000;

as a function:

function DaysBetween(StartDate, EndDate) {
  // The number of milliseconds in all UTC days (no DST)
  const oneDay = 1000 * 60 * 60 * 24;

  // A day in UTC always lasts 24 hours (unlike in other time formats)
  const start = Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate());
  const end = Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate());

  // so it's safe to divide by 24 hours
  return (start - end) / oneDay;
}

Answer 6

Adjusted to allow for daylight saving differences. try this:

  function daysBetween(date1, date2) {

 // adjust diff for for daylight savings
 var hoursToAdjust = Math.abs(date1.getTimezoneOffset() /60) - Math.abs(date2.getTimezoneOffset() /60);
 // apply the tz offset
 date2.addHours(hoursToAdjust); 

    // The number of milliseconds in one day
    var ONE_DAY = 1000 * 60 * 60 * 24

    // Convert both dates to milliseconds
    var date1_ms = date1.getTime()
    var date2_ms = date2.getTime()

    // Calculate the difference in milliseconds
    var difference_ms = Math.abs(date1_ms - date2_ms)

    // Convert back to days and return
    return Math.round(difference_ms/ONE_DAY)

}

// you'll want this addHours function too 

Date.prototype.addHours= function(h){
    this.setHours(this.getHours()+h);
    return this;
}