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How to round down getTime result (ignoring minutes and hours)

I need to get the number of milliseconds of a certain day (even today), but need the result rounded down.
For example the number of milliseconds until this moment by using getTime() method is 1432738826994.
I would like to round this down to get the number of milliseconds until the beginning of the day, I need to get rid of all the minutes and hours. Is there a clean and simple way of achieving this?

If you are happy with using an external library, the easiest way is with moment.js.
moment().startOf('day').valueOf();
Will give you the unix epoch value for the beginning of today's date.
If you want to use the Javascript built in date object, then you would either need to set the hours, minutes, seconds and milliseconds to 0:
var rightNow = new Date();
rightNow.setHours(0);
rightNow.setMinutes(0);
rightNow.setSeconds(0);
rightNow.setMilliseconds(0);
Or create a new object from just the year, month and day values:
var rightNow = new Date();
var earlierToday = new Date(rightNow.getFullYear(), rightNow.getMonth(), rightNow.getDate(), 0, 0, 0, 0);

Pure javascript solution
var d = new Date();
var h = d.getHours();
var m = d.getMinutes();
var s = d.getSeconds();
var mi = d.getMilliseconds();
var fromStart = mi + (s * 1000) + (m * 60 * 1000) + (h * 60 * 60 * 1000);
var roundedDown = Date.now() - fromStart;
To print the start of the day use new Date(Date.now() - fromStart)

Javascript, Hour comparisson

Having two strings (start and end time) in such form "16:30", "02:13" I want to compare them and check if the gap is greater than 5 mins.
How can this be achieved in Javascript in an easy way?

function parseTime(time) {
var timeArray = time.split(/:/);
// Using Jan 1st, 2010 as a "base date". Any other date should work.
return new Date(2010, 0, 1, +timeArray[0], +timeArray[1], 0);
}
var diff = Math.abs(parseTime("16:30").getTime() - parseTime("02:13").getTime());
if (diff > 5 * 60 * 1000) { // Difference is in milliseconds
alert("More that 5 mins.");
}
Do you need to wrap over midnight? Then this is more difficult. For example, 23:59 and 00:01 will produce a difference of 23 hours 58 minutes and not 2 minutes.
If that's the case you need to define your case more closely.

You can do as following:
if (((Date.parse("16:30") - Date.parse("02:13")) / 1000 / 60) > 5)
{
}

// time is a string having format "hh:mm"
function Time(time) {
var args = time.split(":");
var hours = args[0], minutes = args[1];
this.milliseconds = ((hours * 3600) + (minutes * 60)) * 1000;
}
Time.prototype.valueOf = function() {
return this.milliseconds;
}
// converts the given minutes to milliseconds
Number.prototype.minutes = function() {
return this * (1000 * 60);
}
Subtracting the times forces the object to evaluate it's value by calling the valueOf method that returns the given time in milliseconds. The minutes method is another convenience method to convert the given number of minutes to milliseconds, so we can use that as a base for comparison throughout.
new Time('16:30') - new Time('16:24') > (5).minutes() // true

This includes checking whether midnight is between the two times (as per your example).
var startTime = "16:30", endTime = "02:13";
var parsedStartTime = Date.parse("2010/1/1 " + startTime),
parsedEndTime = Date.parse("2010/1/1 " + endTime);
// if end date is parsed as smaller than start date, parse as the next day,
// to pick up on running over midnight
if ( parsedEndTime < parsedStartTime ) ed = Date.parse("2010/1/2 " + endTime);
var differenceInMinutes = ((parsedEndTime - parsedStartTime) / 60 / 1000);
if ( differenceInMinutes > 5 ) {
alert("More than 5 mins.");
}

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

This question already has answers here:
How to calculate number of days between two dates?
(42 answers)
Closed 8 years ago.
I am calculating the number of days between the 'from' and 'to' date. For example, if the from date is 13/04/2010 and the to date is 15/04/2010 the result should be
How do I get the result using JavaScript?

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));

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);
}

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;
}

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.

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;
}

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/

From my little date difference calculator:
var startDate = new Date(2000, 1-1, 1); // 2000-01-01
var endDate = new Date(); // Today
// Calculate the difference of two dates in total days
function diffDays(d1, d2)
{
var ndays;
var tv1 = d1.valueOf(); // msec since 1970
var tv2 = d2.valueOf();
ndays = (tv2 - tv1) / 1000 / 86400;
ndays = Math.round(ndays - 0.5);
return ndays;
}
So you would call:
var nDays = diffDays(startDate, endDate);
(Full source at http://david.tribble.com/src/javascript/jstimespan.html.)
Addendum
The code can be improved by changing these lines:
var tv1 = d1.getTime(); // msec since 1970
var tv2 = d2.getTime();

How to create a function which can return Date Difference in javascript?

i have created below function to get date difference between two dates.. pleas check it that its correct as well as how to find no of month (multiply by 30 or 31 ?) and year..
function days_between(date1, date2,datepart) {
// The number of milliseconds in one day
var ONE_DAY=0;
if ( datepart === undefined ) {
datepart = 'D';
}
if(datepart='Y')
{
ONE_DAY = 1000 * 60 * 60 * 24 * 30 *12
}
else if (datepart='M')
{
ONE_DAY = 1000 * 60 * 60 * 24 * 30
}
else
{
ONE_DAY = 1000 * 60 * 60 * 24 //for day
}
// 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)
}

If your application is going to require a lot of date manipulation methods, you may want to consider using something like the Datejs library.
If you include time.js from the Datejs library, you would be able to do the following:
var future = new Date().add({months: 5});
var now = new Date();
var span = new TimeSpan(future - now);
console.log("Days: ", span.getDays());
The time.js script is optional and is not included in the compiled /build/ versions. You can download it directly from the SVN repository.

You should transform in days after making the difference not before. Like:
var d1 = new Date().getTime(),
d2 = new Date(2010, 11, 31).getTime();
alert(Math.round((d2-d1)/24/60/60/1000) + ' days left for 2010');

How to add hours to a Date object?

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).