How to add days to Date?

Question

How to add days to current Date using JavaScript? Does JavaScript have a built in function like .NET's AddDay()?

Answer 1

You can create one with:-

Date.prototype.addDays = function(days) {
    var date = new Date(this.valueOf());
    date.setDate(date.getDate() + days);
    return date;
}

var date = new Date();

console.log(date.addDays(5));

This takes care of automatically incrementing the month if necessary. For example:

8/31 + 1 day will become 9/1.

The problem with using setDate directly is that it's a mutator and that sort of thing is best avoided. ECMA saw fit to treat Date as a mutable class rather than an immutable structure.

Answer 2

Correct Answer:

function addDays(date, days) {
  var result = new Date(date);
  result.setDate(result.getDate() + days);
  return result;
}

Incorrect Answer:

This answer sometimes provides the correct result but very often returns the wrong year and month. The only time this answer works is when the date that you are adding days to happens to have the current year and month.

// Don't do it this way!
function addDaysWRONG(date, days) {
  var result = new Date(); // not instatiated with date!!! DANGER
  result.setDate(date.getDate() + days);
  return result;
}

Proof / Example

Check this JsFiddle

// Correct
function addDays(date, days) {
    var result = new Date(date);
    result.setDate(result.getDate() + days);
    return result;
}

// Bad Year/Month
function addDaysWRONG(date, days) {
    var result = new Date();
    result.setDate(date.getDate() + days);
    return result;
}

// Bad during DST
function addDaysDstFail(date, days) {
    var dayms = (days * 24 * 60 * 60 * 1000);
    return new Date(date.getTime() + dayms);    
}

// TEST
function formatDate(date) {
    return (date.getMonth() + 1) + '/' + date.getDate() + '/' + date.getFullYear();
}

$('tbody tr td:first-child').each(function () {
    var $in = $(this);
    var $out = $('<td/>').insertAfter($in).addClass("answer");
    var $outFail = $('<td/>').insertAfter($out);
    var $outDstFail = $('<td/>').insertAfter($outFail);
    var date = new Date($in.text());
    var correctDate = formatDate(addDays(date, 1));
    var failDate = formatDate(addDaysWRONG(date, 1));
    var failDstDate = formatDate(addDaysDstFail(date, 1));

    $out.text(correctDate);
    $outFail.text(failDate);
    $outDstFail.text(failDstDate);
    $outFail.addClass(correctDate == failDate ? "right" : "wrong");
    $outDstFail.addClass(correctDate == failDstDate ? "right" : "wrong");
});

body {
    font-size: 14px;
}

table {
    border-collapse:collapse;
}
table, td, th {
    border:1px solid black;
}
td {
    padding: 2px;
}

.wrong {
    color: red;
}
.right {
    color: green;
}
.answer {
    font-weight: bold;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<table>
    <tbody>
        <tr>
            <th colspan="4">DST Dates</th>
        </tr>
        <tr>
            <th>Input</th>
            <th>+1 Day</th>
            <th>+1 Day Fail</th>
            <th>+1 Day DST Fail</th>
        </tr>
        <tr><td>03/10/2013</td></tr>
        <tr><td>11/03/2013</td></tr>
        <tr><td>03/09/2014</td></tr>
        <tr><td>11/02/2014</td></tr>
        <tr><td>03/08/2015</td></tr>
        <tr><td>11/01/2015</td></tr>
        <tr>
            <th colspan="4">2013</th>
        </tr>
        <tr>
            <th>Input</th>
            <th>+1 Day</th>
            <th>+1 Day Fail</th>
            <th>+1 Day DST Fail</th>
        </tr>
        <tr><td>01/01/2013</td></tr>
        <tr><td>02/01/2013</td></tr>
        <tr><td>03/01/2013</td></tr>
        <tr><td>04/01/2013</td></tr>
        <tr><td>05/01/2013</td></tr>
        <tr><td>06/01/2013</td></tr>
        <tr><td>07/01/2013</td></tr>
        <tr><td>08/01/2013</td></tr>
        <tr><td>09/01/2013</td></tr>
        <tr><td>10/01/2013</td></tr>
        <tr><td>11/01/2013</td></tr>
        <tr><td>12/01/2013</td></tr>
        <tr>
            <th colspan="4">2014</th>
        </tr>
        <tr>
            <th>Input</th>
            <th>+1 Day</th>
            <th>+1 Day Fail</th>
            <th>+1 Day DST Fail</th>
        </tr>
        <tr><td>01/01/2014</td></tr>
        <tr><td>02/01/2014</td></tr>
        <tr><td>03/01/2014</td></tr>
        <tr><td>04/01/2014</td></tr>
        <tr><td>05/01/2014</td></tr>
        <tr><td>06/01/2014</td></tr>
        <tr><td>07/01/2014</td></tr>
        <tr><td>08/01/2014</td></tr>
        <tr><td>09/01/2014</td></tr>
        <tr><td>10/01/2014</td></tr>
        <tr><td>11/01/2014</td></tr>
        <tr><td>12/01/2014</td></tr>
        <tr>
            <th colspan="4">2015</th>
        </tr>
        <tr>
            <th>Input</th>
            <th>+1 Day</th>
            <th>+1 Day Fail</th>
            <th>+1 Day DST Fail</th>
        </tr>
        <tr><td>01/01/2015</td></tr>
        <tr><td>02/01/2015</td></tr>
        <tr><td>03/01/2015</td></tr>
        <tr><td>04/01/2015</td></tr>
        <tr><td>05/01/2015</td></tr>
        <tr><td>06/01/2015</td></tr>
        <tr><td>07/01/2015</td></tr>
        <tr><td>08/01/2015</td></tr>
        <tr><td>09/01/2015</td></tr>
        <tr><td>10/01/2015</td></tr>
        <tr><td>11/01/2015</td></tr>
        <tr><td>12/01/2015</td></tr>
    </tbody>
</table>

Answer 3

var today = new Date();
var tomorrow = new Date();
tomorrow.setDate(today.getDate()+1);

Be careful, because this can be tricky. When setting tomorrow, it only works because its current value matches the year and month for today. However, setting to a date number like "32" normally will still work just fine to move it to the next month.

Answer 4

These answers seem confusing to me, I prefer:

var ms = new Date().getTime() + 86400000;
var tomorrow = new Date(ms);

getTime() gives us milliseconds since 1970, and 86400000 is the number of milliseconds in a day. Hence, ms contains milliseconds for the desired date.

Using the millisecond constructor gives the desired date object.

Answer 5

My simple solution is:

nextday=new Date(oldDate.getFullYear(),oldDate.getMonth(),oldDate.getDate()+1);

this solution does not have problem with daylight saving time. Also, one can add/sub any offset for years, months, days etc.

day=new Date(oldDate.getFullYear()-2,oldDate.getMonth()+22,oldDate.getDate()+61);

is correct code.

Answer 6

Here is the way that use to add days, months, and years for a particular date in Javascript.

// To add Days
var d = new Date();
d.setDate(d.getDate() + 5);

// To add Months
var m = new Date();
m.setMonth(m.getMonth() + 5);

// To add Years
var y = new Date();
y.setFullYear(y.getFullYear() + 5);

Answer 7

Try

var someDate = new Date();
var duration = 2; //In Days
someDate.setTime(someDate.getTime() +  (duration * 24 * 60 * 60 * 1000));

Using setDate() to add a date wont solve your problem, try adding some days to a Feb month, if you try to add new days to it, it wont result in what you expected.

Answer 8

Just spent ages trying to work out what the deal was with the year not adding when following the lead examples below.

If you want to just simply add n days to the date you have you are best to just go:

myDate.setDate(myDate.getDate() + n);

or the longwinded version

var theDate = new Date(2013, 11, 15);
var myNewDate = new Date(theDate);
myNewDate.setDate(myNewDate.getDate() + 30);
console.log(myNewDate);

This today/tomorrow stuff is confusing. By setting the current date into your new date variable you will mess up the year value. if you work from the original date you won't.

Answer 9

The simplest approach that I have implemented is to use Date() itself. `

const days = 15; 
// Date.now() gives the epoch date value (in milliseconds) of current date 
nextDate = new Date( Date.now() + days * 24 * 60 * 60 * 1000)

`

Answer 10

int days = 1;
var newDate = new Date(Date.now() + days * 24*60*60*1000);

CodePen

var days = 2;
var newDate = new Date(Date.now() + days * 24*60*60*1000);

document.write('Today: <em>');
document.write(new Date());
document.write('</em><br/> New: <strong>');
document.write(newDate);

Answer 11

If you can, use moment.js. JavaScript doesn't have very good native date/time methods. The following is an example Moment's syntax:

var nextWeek = moment().add(7, 'days');
alert(nextWeek);

<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.17.1/moment-with-locales.min.js"></script>

Reference: http://momentjs.com/docs/#/manipulating/add/

Answer 12

the simplest answer is, assuming the need is to add 1 day to the current date:

var currentDate = new Date();
var numberOfDayToAdd = 1;
currentDate.setDate(currentDate.getDate() + numberOfDayToAdd );

To explain to you, line by line, what this code does:

1. Create the current date variable named currentDate. By default "new Date()" automatically assigns the current date to the variable.
2. Create a variable to save the number of day(s) to add to the date (you can skip this variable and use directly the value in the third line)
3. Change the value of Date (because Date is the number of the month's day saved in the object) by giving the same value + the number you want. The switch to the next month will be automatic

Answer 13

I created these extensions last night:
you can pass either positive or negative values;

example:

var someDate = new Date();
var expirationDate = someDate.addDays(10);
var previous = someDate.addDays(-5);


Date.prototype.addDays = function (num) {
    var value = this.valueOf();
    value += 86400000 * num;
    return new Date(value);
}

Date.prototype.addSeconds = function (num) {
    var value = this.valueOf();
    value += 1000 * num;
    return new Date(value);
}

Date.prototype.addMinutes = function (num) {
    var value = this.valueOf();
    value += 60000 * num;
    return new Date(value);
}

Date.prototype.addHours = function (num) {
    var value = this.valueOf();
    value += 3600000 * num;
    return new Date(value);
}

Date.prototype.addMonths = function (num) {
    var value = new Date(this.valueOf());

    var mo = this.getMonth();
    var yr = this.getYear();

    mo = (mo + num) % 12;
    if (0 > mo) {
        yr += (this.getMonth() + num - mo - 12) / 12;
        mo += 12;
    }
    else
        yr += ((this.getMonth() + num - mo) / 12);

    value.setMonth(mo);
    value.setYear(yr);
    return value;
}

Answer 14

A solution designed for the pipeline operator:

const addDays = days => date => {
  const result = new Date(date);

  result.setDate(result.getDate() + days);

  return result;
};

Usage:

// Without the pipeline operator...
addDays(7)(new Date());

// And with the pipeline operator...
new Date() |> addDays(7);

If you need more functionality, I suggest looking into the date-fns library.

Answer 15

Without using the second variable, you can replace 7 with your next x days:

let d=new Date(new Date().getTime() + (7 * 24 * 60 * 60 * 1000));

Answer 16

Short:

function addDays(date, number) {
  const newDate = new Date(date);
  return new Date(newDate.setDate(newDate.getDate() + number));
}

console.log({
  tomorrow: addDays(new Date(), 1)
});

Advance:

function addDays(date, number) {
  const newDate = new Date(date);
  return new Date(newDate.setDate(date.getDate() + number));
}

function addMonths(date, number) {
  const newDate = new Date(date);
  return new Date(newDate.setMonth(newDate.getMonth() + number));
}

function addYears(date, number) {
  const newDate = new Date(date);
  return new Date(newDate.setFullYear(newDate.getFullYear() + number));
}

function getNewDate(dateTime) {
  let date = new Date();
  let number = parseInt(dateTime.match(/\d+/)[0]);

  if (dateTime.indexOf('-') != -1)
    number = (-number);

  if (dateTime.indexOf('day') != -1)
    date = addDays(date, number);
  else if (dateTime.indexOf('month') != -1)
    date = addMonths(date, number);
  else if (dateTime.indexOf('year') != -1)
    date = addYears(date, number);

  return date;
}

console.log({
  tomorrow: getNewDate('+1day'),
  yesterday: getNewDate('-1day'),
  nextMonth: getNewDate('+1month'),
  nextYear: getNewDate('+1year'),
});

With fix provide by jperl

Answer 17

to substract 30 days use (24h=86400000ms)

new Date(+yourDate - 30 *86400000)

var yourDate=new Date();
var d = new Date(+yourDate - 30 *86400000) 

console.log(d)

Answer 18

The simplest solution.

 Date.prototype.addDays = function(days) {
   this.setDate(this.getDate() + parseInt(days));
   return this;
 };

 // and then call

 var newDate = new Date().addDays(2); //+2 days
 console.log(newDate);

 // or

 var newDate1 = new Date().addDays(-2); //-2 days
 console.log(newDate1);

Answer 19

You can try:

var days = 50;

const d = new Date();

d.setDate(d.getDate() + days)

This should work well.

Answer 20

You can use JavaScript, no jQuery required:

var someDate = new Date();
var numberOfDaysToAdd = 6;
someDate.setDate(someDate.getDate() + numberOfDaysToAdd); 
Formatting to dd/mm/yyyy :

var dd = someDate.getDate();
var mm = someDate.getMonth() + 1;
var y = someDate.getFullYear();

var someFormattedDate = dd + '/'+ mm + '/'+ y;

Answer 21

Thanks Jason for your answer that works as expected, here is a mix from your code and the handy format of AnthonyWJones :

Date.prototype.addDays = function(days){
    var ms = new Date().getTime() + (86400000 * days);
    var added = new Date(ms);
    return added;
}

Answer 22

Late to the party, but if you use jQuery then there's an excellent plugin called Moment:

http://momentjs.com/

var myDateOfNowPlusThreeDays = moment().add(3, "days").toDate();

http://momentjs.com/docs/#/manipulating/

And lots of other good stuff in there!

Edit: jQuery reference removed thanks to aikeru's comment

Answer 23

As simple as this:

new Date((new Date()).getTime() + (60*60*24*1000));

Answer 24

Old I know, but sometimes I like this:

function addDays(days) {
    return new Date(Date.now() + 864e5 * days);
}

Answer 25

I had issues with daylight savings time with the proposed solution.

By using getUTCDate / setUTCDate instead, I solved my issue.

// Curried, so that I can create helper functions like `add1Day`
const addDays = num => date => {
  // Make a working copy so we don't mutate the supplied date.
  const d = new Date(date);

  d.setUTCDate(d.getUTCDate() + num);

  return d;
}

Answer 26

Why so complicated?

Let's assume you store the number of days to add in a variable called days_to_add.

Then this short one should do it:

calc_date = new Date(Date.now() +(days_to_add * 86400000));

With Date.now() you get the actual unix timestamp as milliseconds and then you add as many milliseconds as you want to add days to. One day is 24h60min60s*1000ms = 86400000 ms or 864E5.

Answer 27

No, javascript has no a built in function, but you can use a simple line of code

timeObject.setDate(timeObject.getDate() + countOfDays);

Answer 28

The mozilla docs for setDate() don't indicate that it will handle end of month scenarios. See https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Date

setDate()

• Sets the day of the month (1-31) for a specified date according to local time.

That is why I use setTime() when I need to add days.

Answer 29

Generic prototype with no variables, it applies on an existing Date value:

Date.prototype.addDays = function (days) {
    return new Date(this.valueOf() + days * 864e5);
}

Answer 30

I guess I'll give an answer as well:
Personally, I like to attempt to avoid gratuitous variable declaration, method calls, and constructor calls, as they are all expensive on performance. (within reason, of course)
I was going to leave this as just comment under the Answer given by @AnthonyWJones but thought better of it.

// Prototype usage...
Date.prototype.addDays = Date.prototype.addDays || function( days ) {
    return this.setTime( 864E5 * days + this.valueOf() ) && this;
};

// Namespace usage...
namespace.addDaysToDate = function( date, days ) {
    return date.setTime( 864E5 * days + date.valueOf() ) && date;
};

// Basic Function declaration...
function addDaysToDate( date, days ) {
    return date.setTime( 864E5 * days + date.valueOf() ) && date;
};

The above will respect DST. Meaning if you add a number of days that cross DST, the displayed time (hour) will change to reflect that.
Example:
Nov 2, 2014 02:00 was the end of DST.

var dt = new Date( 2014, 10, 1, 10, 30, 0 );
console.log( dt );                  // Sat Nov 01 2014 10:30:00
console.log( dt.addDays( 10 ) );    // Tue Nov 11 2014 09:30:00

If you're looking to retain the time across DST (so 10:30 will still be 10:30)...

// Prototype usage...
Date.prototype.addDays = Date.prototype.addDays || function( days ) {
    return this.setDate( this.getDate() + days ) && this;
};

// Namespace usage...
namespace.addDaysToDate = function( date, days ) {
    return date.setDate( date.getDate() + days ) && date;
};

// Basic Function declaration...
function addDaysToDate( date, days ) {
    return date.setDate( date.getDate() + days ) && date;
};

So, now you have...

var dt = new Date( 2014, 10, 1, 10, 30, 0 );
console.log( dt );                  // Sat Nov 01 2014 10:30:00
console.log( dt.addDays( 10 ) );    // Tue Nov 11 2014 10:30:00

Answer 31

Extending prototype in javascript may not be a good idea, especially in professional codebases.

What you want to do is extend the native Date class:

class MyCustomDate extends Date {

  addDays(days) {
    const date = new MyCustomDate(this.valueOf());
    date.setDate(date.getDate() + days);
    return date;
  }
  
}

const today = new MyCustomDate();

const nextWeek = today.addDays(7)

console.log(nextWeek)

This way, if one day Javascript implements a native addDays method, you won't break anything.

Answer 32

I use something like:

new Date(dateObject.getTime() + amountOfDays * 24 * 60 * 60 * 1000)

Works with day saving time:

new Date(new Date(2014, 2, 29, 20, 0, 0).getTime() + 1 * 24 * 60 * 60 * 1000)

Works with new year:

new Date(new Date(2014, 11, 31, 20, 0, 0).getTime() + 1 * 24 * 60 * 60 * 1000)

It can be parametrized:

function DateAdd(source, amount, step) {
  var factor = 1;
  if (step == "day") factor = 24 * 60 * 60 * 1000;
  else if (step == "hour") factor = 60 * 60 * 1000;
  ...
  new Date(source.getTime() + amount * factor);
}

Answer 33

Edit: Instead of setTime() (or setHours()) you could do it this way:

Date.prototype.addDays= function(d){
  this.setDate(this.getDate() + d);
  return this;
};

var tomorrow = new Date().addDays(1);

Old:

Instead of using setTime() you can use setHours():

Date.prototype.addDays= function(d){
    this.setHours(this.getHours() + d * 24);
    return this;
};

var tomorrow = new Date().addDays(1);

See the JSFiddle...

Answer 34

Very simple code to add days in date in java script.

var d = new Date();
d.setDate(d.getDate() + prompt('how many days you want to add write here'));
alert(d);

Answer 35

There's a setDate and a getDate method, which allow you to do something like this :

var newDate = aDate.setDate(aDate.getDate() + numberOfDays);

If you want to both subtract a number of days and format your date in a human readable format, you should consider creating a custom DateHelper object that looks something like this :

var DateHelper = {
    addDays : function(aDate, numberOfDays) {
        aDate.setDate(aDate.getDate() + numberOfDays); // Add numberOfDays
        return aDate;                                  // Return the date
    },
    format : function format(date) {
        return [
           ("0" + date.getDate()).slice(-2),           // Get day and pad it with zeroes
           ("0" + (date.getMonth()+1)).slice(-2),      // Get month and pad it with zeroes
           date.getFullYear()                          // Get full year
        ].join('/');                                   // Glue the pieces together
    }
}

// With this helper, you can now just use one line of readable code to :
// ---------------------------------------------------------------------
// 1. Get the current date
// 2. Add 20 days
// 3. Format it
// 4. Output it
// ---------------------------------------------------------------------
document.body.innerHTML = DateHelper.format(DateHelper.addDays(new Date(), 20));

(see also this Fiddle)

Answer 36

I am using the following solution.

var msInDay = 86400000;
var daysToAdd = 5;
var now = new Date();
var milliseconds = now.getTime();
var newMillisecods = milliseconds + msInDay * daysToAdd;
var newDate = new Date(newMillisecods);
//or now.setTime(newMillisecods);

Date has a constructor that accepts an int. This argument represents total milliseconds before/after Jan 1, 1970. It also has a method setTime which does the same without creating a new Date object.

What we do here is convert days to milliseconds and add this value to the value provided by getTime. Finally, we give the result to Date(milliseconds) constructor or setTime(milliseconds) method.

Answer 37

Our team considers date-fns the best library in this space. It treats dates as immutable (Moment.js will probably never adopt immutability), it's faster, and can be loaded modularly.

const newDate = DateFns.addDays(oldDate, 2);

Answer 38

There is a problem with this kind of functions, I solve it with parseInt()

Date.prototype.addDays = function(dias) {

    var date = new Date(this.valueOf());
    date.setDate(parseInt(date.getDate()) + parseInt(dias));
    return date;
}

Date.prototype.addMonths = function(months) {
    var date = new Date(this.valueOf());
    date.setMonth(parseInt(date.getMonth()) + parseInt(months));
    return date;
}


Date.prototype.addYears = function(years) {
    var date = new Date(this.valueOf());
    date.setFullYear(parseInt(date.getFullYear()) + parseInt(years));
    return date;
}

Answer 39

    //the_day is 2013-12-31
    var the_day = Date.UTC(2013, 11, 31); 
    // Now, the_day will be "1388448000000" in UTC+8; 
    var the_next_day = new Date(the_day + 24 * 60 * 60 * 1000);
    // Now, the_next_day will be "Wed Jan 01 2014 08:00:00 GMT+0800"

Answer 40

For those using Angular:

Just do:

$scope.booking.totTijd.setMinutes($scope.booking.totTijd.getMinutes()+15);
$scope.booking.totTijd.setDate($scope.booking.totTijd.getDate() + 1);

Answer 41

You can create your custom helper function here

function plusToDate(currentDate, unit, howMuch) {

    var config = {
        second: 1000, // 1000 miliseconds
        minute: 60000,
        hour: 3600000,
        day: 86400000,
        week: 604800000,
        month: 2592000000, // Assuming 30 days in a month
        year: 31536000000 // Assuming 365 days in year
    };

    var now = new Date(currentDate);

    return new Date(now + config[unit] * howMuch);
}

var today = new Date();
var theDayAfterTommorow = plusToDate(today, 'day', 2);

By the way, this is generic solution for adding seconds or minutes or days whatever you want.

Answer 42

The easiest way to get this done is using date-fns library.

var addDays = require('date-fns/add_days')
addDays(date, amount)

The documentation is available in this link here. You can also get this done using moment.js. The reference link is here

Hope it helps!

Answer 43

I've used this approach to get the right date in one line to get the time plus one day following what people were saying above.

((new Date()).setDate((new Date()).getDate()+1))

I just figured I would build off a normal (new Date()):

(new Date()).getDate()
> 21

Using the code above I can now set all of that within Date() in (new Date()) and it behaves normally.

(new Date(((new Date()).setDate((new Date()).getDate()+1)))).getDate()
> 22

or to get the Date object:

(new Date(((new Date()).setDate((new Date()).getDate()+1))))

Answer 44

I can't believe there's no cut'n'paste solution in this thread after 5 years!
SO: To get the same time-of-day regardless of summertime interference:

Date.prototype.addDays = function(days)
    {
    var dat = new Date( this.valueOf() )

    var hour1 = dat.getHours()
    dat.setTime( dat.getTime() + days * 86400000) // 24*60*60*1000 = 24 hours
    var hour2 = dat.getHours()

    if (hour1 != hour2) // summertime occured +/- a WHOLE number of hours thank god!
        dat.setTime( dat.getTime() + (hour1 - hour2) * 3600000) // 60*60*1000 = 1 hour

    return dat
or
    this.setTime( dat.getTime() ) // to modify the object directly
    }

There. Done!

Answer 45

function addDays(n){
    var t = new Date();
    t.setDate(t.getDate() + n); 
    var month = "0"+(t.getMonth()+1);
    var date = "0"+t.getDate();
    month = month.slice(-2);
    date = date.slice(-2);
     var date = date +"/"+month +"/"+t.getFullYear();
    alert(date);
}

addDays(5);

Answer 46

Some implementations to extend Date https://gist.github.com/netstart/c92e09730f3675ba8fb33be48520a86d

/**
 * just import, like
 *
 * import './../shared/utils/date.prototype.extendions.ts';
 */
declare global {
  interface Date {
    addDays(days: number, useThis?: boolean): Date;

    addSeconds(seconds: number): Date;

    addMinutes(minutes: number): Date;

    addHours(hours: number): Date;

    addMonths(months: number): Date;

    isToday(): boolean;

    clone(): Date;

    isAnotherMonth(date: Date): boolean;

    isWeekend(): boolean;

    isSameDate(date: Date): boolean;

    getStringDate(): string;
  }
}

Date.prototype.addDays = function(days: number): Date {
  if (!days) {
    return this;
  }
  this.setDate(this.getDate() + days);
  return this;
};

Date.prototype.addSeconds = function(seconds: number) {
  let value = this.valueOf();
  value += 1000 * seconds;
  return new Date(value);
};

Date.prototype.addMinutes = function(minutes: number) {
  let value = this.valueOf();
  value += 60000 * minutes;
  return new Date(value);
};

Date.prototype.addHours = function(hours: number) {
  let value = this.valueOf();
  value += 3600000 * hours;
  return new Date(value);
};

Date.prototype.addMonths = function(months: number) {
  const value = new Date(this.valueOf());

  let mo = this.getMonth();
  let yr = this.getYear();

  mo = (mo + months) % 12;
  if (0 > mo) {
    yr += (this.getMonth() + months - mo - 12) / 12;
    mo += 12;
  } else {
    yr += ((this.getMonth() + months - mo) / 12);
  }

  value.setMonth(mo);
  value.setFullYear(yr);
  return value;
};

Date.prototype.isToday = function(): boolean {
  const today = new Date();
  return this.isSameDate(today);
};

Date.prototype.clone = function(): Date {
  return new Date(+this);
};

Date.prototype.isAnotherMonth = function(date: Date): boolean {
  return date && this.getMonth() !== date.getMonth();
};

Date.prototype.isWeekend = function(): boolean {
  return this.getDay() === 0 || this.getDay() === 6;
};

Date.prototype.isSameDate = function(date: Date): boolean {
  return date && this.getFullYear() === date.getFullYear() && this.getMonth() === date.getMonth() && this.getDate() === date.getDate();
};

Date.prototype.getStringDate = function(): string {
  // Month names in Brazilian Portuguese
  const monthNames = ['Janeiro', 'Fevereiro', 'Março', 'Abril', 'Maio', 'Junho', 'Julho', 'Agosto', 'Setembro', 'Outubro', 'Novembro', 'Dezembro'];
  // Month names in English
  // let monthNames = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December'];
  const today = new Date();
  if (this.getMonth() === today.getMonth() && this.getDay() === today.getDay()) {
    return 'Hoje';
    // return "Today";
  } else if (this.getMonth() === today.getMonth() && this.getDay() === today.getDay() + 1) {
    return 'Amanhã';
    // return "Tomorrow";
  } else if (this.getMonth() === today.getMonth() && this.getDay() === today.getDay() - 1) {
    return 'Ontem';
    // return "Yesterday";
  } else {
    return this.getDay() + ' de ' + this.monthNames[this.getMonth()] + ' de ' + this.getFullYear();
    // return this.monthNames[this.getMonth()] + ' ' + this.getDay() + ', ' +  this.getFullYear();
  }
};


export {};

Answer 47

date d = new Date() // current date

date tomorrow = d.setMonth(d.getMonth(),d.getDate()+1) // return a date incremented by 0 months and 1 day

Answer 48

new Date(Date.now() + 2000 * 86400)

This snippet adds two days to the current date using the "2000" argument. You can tweak the number of days by updating the "2000" value in the second argument.

You can use this single line format to add days to the current date using the native JavaScript date.

Answer 49

2.39KB minified. One file. https://github.com/rhroyston/clock-js

console.log(clock.what.weekday(clock.now + clock.unit.days)); //"wednesday"
console.log(clock.what.weekday(clock.now + (clock.unit.days * 2))); //"thursday"
console.log(clock.what.weekday(clock.now + (clock.unit.days * 3))); //"friday"

<script src="https://raw.githubusercontent.com/rhroyston/clock-js/master/clock.min.js"></script>

Answer 50

try this

function addDays(date,days) {        
      var one_day=1000*60*60*24; 
      return new Date(date.getTime()+(days*one_day)).toLocaleDateString(); 
    }

Answer 51

Use js-joda. It is an awesome immutable date and time library for javascript. Here is an excerpt from its cheat-sheet.

Add 17 days to today

LocalDate.now().plusDays(17); 

You can also build the desired date from now multiple operations at once.

LocalDate.now()
  .plusMonths(1)
  .withDayOfMonth(1)
  .minusDays(17); 

Or:

var d = LocalDate.parse('2019-02-23'); 
d.minus(Period.ofMonths(3).plusDays(3)); // '2018-11-20'

Answer 52

My test exemple can do adition an minus in the same instance of Date Object.

Date.prototype.reset = function()
{
    let newDate = new Date(this.timeStamp)
    this.setFullYear        (newDate.getFullYear())
    this.setMonth           (newDate.getMonth())
    this.setDate            (newDate.getDate())
    this.setHours           (newDate.getHours())
    this.setMinutes     (newDate.getMinutes())
    this.setSeconds     (newDate.getSeconds())
    this.setMilliseconds    (newDate.getMilliseconds())
}

Date.prototype.addDays = function(days)
{
      this.timeStamp = this[Symbol.toPrimitive]('number')
      let daysInMiliseconds = (days * (1000 * 60 * 60 * 24))
      this.timeStamp = this.timeStamp + daysInMiliseconds
      this.reset()
}

Date.prototype.minusDays = function(days)
{
      this.timeStamp = this[Symbol.toPrimitive]('number')
      let daysInMiliseconds = (days * (1000 * 60 * 60 * 24))
      if(daysInMiliseconds <= this.timeStamp)
      {
          this.timeStamp = this.timeStamp - daysInMiliseconds
          this.reset()
      }
       
}

var temp = new Date(Date.now())// from now time

console.log(temp.toDateString())
temp.addDays(31)
console.log(temp.toDateString())
temp.minusDays(5)
console.log(temp.toDateString())

Answer 53

I sum hours and days...

Date.prototype.addDays = function(days){
    days = parseInt(days, 10)
    this.setDate(this.getUTCDate() + days);
    return this;
}

Date.prototype.addHours = function(hrs){
    var hr = this.getUTCHours() + parseInt(hrs  , 10);
    while(hr > 24){
      hr = hr - 24;
      this.addDays(1);
    }

    this.setHours(hr);
    return this;
}

Answer 54

For everybody who don't know how to make it work : there is a full working code it's not perfect but you can copy past it and it's working.

In InDesign creat a .jsx in the startup scripts folder in "Program Files\Adobe\Adobe InDesign 2021\Scripts\startup scripts".

You can us the Extendscript Toolkit CC in the creative cloud to make it and paste this:

The restart indesign and jjmmyyyy+30 should be in the texte variable. this will show the date like this jj/m/yyyy idk how to make it show 24/07/2021 insted of 24/7/2021 but goodenough for me .

    #targetengine 'usernameVariable'
    function addVariables(openEvent) 
    {  
    var doc = openEvent.parent;  
    while ( doc.constructor.name != "Document" )  
      {  
    if ( doc.constructor.name == "Application" ){ return; }  
        doc = doc.parent;  
      }  
    // from http://stackoverflow.com/questions/563406/add-days-to-datetime


    var someDate = new Date();
    var numberOfDaysToAdd = 30;
    someDate.setDate(someDate.getDate() + numberOfDaysToAdd); 


    var dd = someDate.getDate();
    var mm = someDate.getMonth() + 1;
    var y = someDate.getFullYear();

    var someFormattedDate = dd + '/'+ mm + '/'+ y;  

      createTextVariable(doc, "jjmmyyyy+30", someFormattedDate);  
    }
    function createTextVariable(target, variableName, variableContents)  
    {  
    var usernameVariable = target.textVariables.itemByName(variableName);  
    if (!usernameVariable.isValid)  
      {  
        usernameVariable = target.textVariables.add();  
        usernameVariable.variableType = VariableTypes.CUSTOM_TEXT_TYPE;  
        usernameVariable.name = variableName;  
      }  
      usernameVariable.variableOptions.contents = variableContents;  
    }  
    app.addEventListener('afterOpen', addVariables);

Answer 55

the same answer: How to add number of days to today's date?

    function DaysOfMonth(nYear, nMonth) {
        switch (nMonth) {
            case 0:     // January
                return 31; break;
            case 1:     // February
                if ((nYear % 4) == 0) {
                    return 29;
                }
                else {
                    return 28;
                };
                break;
            case 2:     // March
                return 31; break;
            case 3:     // April
                return 30; break;
            case 4:     // May
                return 31; break;
            case 5:     // June
                return 30; break;
            case 6:     // July
                return 31; break;
            case 7:     // August
                return 31; break;
            case 8:     // September
                return 30; break;
            case 9:     // October
                return 31; break;
            case 10:     // November
                return 30; break;
            case 11:     // December
                return 31; break;
        }
    };

    function SkipDate(dDate, skipDays) {
        var nYear = dDate.getFullYear();
        var nMonth = dDate.getMonth();
        var nDate = dDate.getDate();
        var remainDays = skipDays;
        var dRunDate = dDate;

        while (remainDays > 0) {
            remainDays_month = DaysOfMonth(nYear, nMonth) - nDate;
            if (remainDays > remainDays_month) {
                remainDays = remainDays - remainDays_month - 1;
                nDate = 1;
                if (nMonth < 11) { nMonth = nMonth + 1; }
                else {
                    nMonth = 0;
                    nYear = nYear + 1;
                };
            }
            else {
                nDate = nDate + remainDays;
                remainDays = 0;
            };
            dRunDate = Date(nYear, nMonth, nDate);
        }
        return new Date(nYear, nMonth, nDate);
    };