Calculating hour difference in Javascript, but only certain hours - javascript

Unsure of the best way to describe this, but I need to calculate the difference in hours (rounded down) but only between 8pm and 6am (or rather 20:00 - 06:00 in this case!)
For example:
22:00 - 04:00 (6 hours)
02:40 - 10:20 (4 hours)
20:00 - 06:00 (10 hours)
Unfortunately I need to work on exact dates, because some will span over multiple days - and just to add to the confusion, I also need to exclude certain dates entirely for bank holidays (which I have a list of in an array) but have absolutely no idea how to implement this - any suggestions would be very welcome, thank you

Just going off what the inputs in your sample looks like:
// According to your example, your inputs are strings...
// t1 = "22:00"
// t2 = "04:00";
function hoursDiff(t1, t2){
// Parse out the times, using radix 10 to
// avoid octal edge cases ("08:00" & "09:00")
var time1 = parseInt( t1, 10 );
var time2 = parseInt( t2, 10 );
var hours = 0;
while ( time1 !== time2 ){
time1++;
hours++;
// If we've passed midnight, reset
// to 01:00AM
if ( time1 === 25 ){
time1 = 1;
}
}
return hours;
}

Ok, this was a nice puzzle. It really cost me too much time but it was fun.
Full working code below (jsfiddle here):
function isHoliday(/*Date*/ date) {
for(var i = 0; i < holidays.length; i++) {
if(holidays[i].getTime() == date.getTime()) {
return true;
}
}
return false;
}
function diffHours(/*Date*/ d1, /*Date*/ d2) {
var date1 = new Date(d1.getUTCFullYear()+"-"+(d1.getUTCMonth()+1)+"-"+d1.getUTCDate()+" UTC");
var date2 = new Date(d2.getUTCFullYear()+"-"+(d2.getUTCMonth()+1)+"-"+d2.getUTCDate()+" UTC");
var sum = 0;
var oneday = 24*3600*1000;
var hours, date;
// first day
if(!isHoliday(date1)) {
// decrease by a whole day first (will be added later)
sum -= 10;
// add real hours
hours = d1.getUTCHours() + d1.getUTCMinutes() / 60;
if(hours <= 6) {
sum += 10 - hours;
} else if(hours <= 20) {
sum += 4;
} else {
sum += 24 - hours;
}
}
// last day
if(!isHoliday(date2)) {
// decrease by a whole day first (will be added later)
sum -= 10;
// add real hours
hours = d2.getUTCHours() + d2.getUTCMinutes() / 60;
if(hours <= 6) {
sum += hours;
} else if(hours <= 20) {
sum += 6;
} else {
sum += hours - 14;
}
}
// whole days
while(date1 <= date2) {
if(!isHoliday(date1)) {
sum += 10;
}
// increase date by 1 day
date1.setTime(date1.getTime() + oneday);
}
return Math.floor(sum);
}
// ==============
// examples below
// --------------
// array of Dates (in UTC) to skip
var holidays = [
new Date("2012-01-04 UTC"),
];
for(var i = 0; i < holidays.length; i++) {
console.log('holiday: ', holidays[i].toUTCString());
}
a = new Date("2012-01-01 12:00 UTC");
b = new Date("2012-01-02 12:00 UTC");
c = new Date("2012-01-02 22:00 UTC");
d = new Date("2012-01-03 07:00 UTC");
e = new Date("2012-01-05 12:00 UTC");
console.log({d1: a.toUTCString(), d2: b.toUTCString(), hours: diffHours(a, b)});
console.log({d1: b.toUTCString(), d2: c.toUTCString(), hours: diffHours(b, c)});
console.log({d1: c.toUTCString(), d2: d.toUTCString(), hours: diffHours(c, d)});
console.log({d1: d.toUTCString(), d2: e.toUTCString(), hours: diffHours(d, e)});

Related

I want to input date calculated Year and month show two text box like txtYear and txtMonth [duplicate]

How to get the difference between two dates in years, months, and days in JavaScript, like: 10th of April 2010 was 3 years, x month and y days ago?
There are lots of solutions, but they only offer the difference in the format of either days OR months OR years, or they are not correct (meaning not taking care of actual number of days in a month or leap years, etc). Is it really that difficult to do that?
I've had a look at:
http://momentjs.com/ -> can only output the difference in either years, months, OR days
http://www.javascriptkit.com/javatutors/datedifference.shtml
http://www.javascriptkit.com/jsref/date.shtml
http://timeago.yarp.com/
www.stackoverflow.com -> Search function
In php it is easy, but unfortunately I can only use client-side script on that project. Any library or framework that can do it would be fine, too.
Here are a list of expected outputs for date differences:
//Expected output should be: "1 year, 5 months".
diffDate(new Date('2014-05-10'), new Date('2015-10-10'));
//Expected output should be: "1 year, 4 months, 29 days".
diffDate(new Date('2014-05-10'), new Date('2015-10-09'));
//Expected output should be: "1 year, 3 months, 30 days".
diffDate(new Date('2014-05-10'), new Date('2015-09-09'));
//Expected output should be: "9 months, 27 days".
diffDate(new Date('2014-05-10'), new Date('2015-03-09'));
//Expected output should be: "1 year, 9 months, 28 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-09'));
//Expected output should be: "1 year, 10 months, 1 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-11'));
How precise do you need to be? If you do need to take into account common years and leap years, and the exact difference in days between months then you'll have to write something more advanced but for a basic and rough calculation this should do the trick:
today = new Date()
past = new Date(2010,05,01) // remember this is equivalent to 06 01 2010
//dates in js are counted from 0, so 05 is june
function calcDate(date1,date2) {
var diff = Math.floor(date1.getTime() - date2.getTime());
var day = 1000 * 60 * 60 * 24;
var days = Math.floor(diff/day);
var months = Math.floor(days/31);
var years = Math.floor(months/12);
var message = date2.toDateString();
message += " was "
message += days + " days "
message += months + " months "
message += years + " years ago \n"
return message
}
a = calcDate(today,past)
console.log(a) // returns Tue Jun 01 2010 was 1143 days 36 months 3 years ago
Keep in mind that this is imprecise, in order to calculate the date with full precision one would have to have a calendar and know if a year is a leap year or not, also the way I'm calculating the number of months is only approximate.
But you can improve it easily.
Actually, there's a solution with a moment.js plugin and it's very easy.
You might use moment.js
Don't reinvent the wheel again.
Just plug Moment.js Date Range Plugin.
Example:
var starts = moment('2014-02-03 12:53:12');
var ends = moment();
var duration = moment.duration(ends.diff(starts));
// with ###moment precise date range plugin###
// it will tell you the difference in human terms
var diff = moment.preciseDiff(starts, ends, true);
// example: { "years": 2, "months": 7, "days": 0, "hours": 6, "minutes": 29, "seconds": 17, "firstDateWasLater": false }
// or as string:
var diffHuman = moment.preciseDiff(starts, ends);
// example: 2 years 7 months 6 hours 29 minutes 17 seconds
document.getElementById('output1').innerHTML = JSON.stringify(diff)
document.getElementById('output2').innerHTML = diffHuman
<html>
<head>
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.14.1/moment.min.js"></script>
<script src="https://raw.githubusercontent.com/codebox/moment-precise-range/master/moment-precise-range.js"></script>
</head>
<body>
<h2>Difference between "NOW and 2014-02-03 12:53:12"</h2>
<span id="output1"></span>
<br />
<span id="output2"></span>
</body>
</html>
Modified this to be a lot more accurate. It will convert dates to a 'YYYY-MM-DD' format, ignoring HH:MM:SS, and takes an optional endDate or uses the current date, and doesn't care about the order of the values.
function dateDiff(startingDate, endingDate) {
let startDate = new Date(new Date(startingDate).toISOString().substr(0, 10));
if (!endingDate) {
endingDate = new Date().toISOString().substr(0, 10); // need date in YYYY-MM-DD format
}
let endDate = new Date(endingDate);
if (startDate > endDate) {
const swap = startDate;
startDate = endDate;
endDate = swap;
}
const startYear = startDate.getFullYear();
const february = (startYear % 4 === 0 && startYear % 100 !== 0) || startYear % 400 === 0 ? 29 : 28;
const daysInMonth = [31, february, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
let yearDiff = endDate.getFullYear() - startYear;
let monthDiff = endDate.getMonth() - startDate.getMonth();
if (monthDiff < 0) {
yearDiff--;
monthDiff += 12;
}
let dayDiff = endDate.getDate() - startDate.getDate();
if (dayDiff < 0) {
if (monthDiff > 0) {
monthDiff--;
} else {
yearDiff--;
monthDiff = 11;
}
dayDiff += daysInMonth[startDate.getMonth()];
}
return yearDiff + 'Y ' + monthDiff + 'M ' + dayDiff + 'D';
}
// Examples
let dates = [
['2019-05-10','2019-05-10'], // 0Y 0M 0D
['2019-05-09','2019-05-10'], // 0Y 0M 1D
['2018-05-09','2019-05-10'], // 1Y 0M 1D
['2018-05-18','2019-05-10'], // 0Y 11M 23D
['2019-01-09','2019-05-10'], // 0Y 4M 1D
['2019-02-10','2019-05-10'], // 0Y 3M 0D
['2019-02-11','2019-05-10'], // 0Y 2M 27D
['2016-02-11','2019-05-10'], // 3Y 2M 28D - leap year
['1972-11-30','2019-05-10'], // 46Y 5M 10D
['2016-02-11','2017-02-11'], // 1Y 0M 0D
['2016-02-11','2016-03-10'], // 0Y 0M 28D - leap year
['2100-02-11','2100-03-10'], // 0Y 0M 27D - not a leap year
['2017-02-11','2016-02-11'], // 1Y 0M 0D - swapped dates to return correct result
[new Date() - 1000 * 60 * 60 * 24] // 0Y 0M 1D - uses current date
].forEach(([s, e]) => console.log(dateDiff(s, e)));
Older less accurate but much simpler version
#RajeevPNadig's answer was what I was looking for, but his code returns incorrect values as written. This code is not very accurate because it assumes that the sequence of dates from 1 January 1970 is the same as any other sequence of the same number of days. E.g. it calculates the difference from 1 July to 1 September (62 days) as 0Y 2M 3D and not 0Y 2M 0D because 1 Jan 1970 plus 62 days is 3 March.
// startDate must be a date string
function dateAgo(date) {
var startDate = new Date(date);
var diffDate = new Date(new Date() - startDate);
return ((diffDate.toISOString().slice(0, 4) - 1970) + "Y " +
diffDate.getMonth() + "M " + (diffDate.getDate()-1) + "D");
}
Then you can use it like this:
// based on a current date of 2018-03-09
dateAgo('1972-11-30'); // "45Y 3M 9D"
dateAgo('2017-03-09'); // "1Y 0M 0D"
dateAgo('2018-01-09'); // "0Y 2M 0D"
dateAgo('2018-02-09'); // "0Y 0M 28D" -- a little odd, but not wrong
dateAgo('2018-02-01'); // "0Y 1M 5D" -- definitely "feels" wrong
dateAgo('2018-03-09'); // "0Y 0M 0D"
If your use case is just date strings, then this works okay if you just want a quick and dirty 4 liner.
I used this simple code to get difference in Years, Months, days with current date.
var sdt = new Date('1972-11-30');
var difdt = new Date(new Date() - sdt);
alert((difdt.toISOString().slice(0, 4) - 1970) + "Y " + (difdt.getMonth()+1) + "M " + difdt.getDate() + "D");
I think you are looking for the same thing that I wanted. I tried to do this using the difference in milliseconds that javascript provides, but those results do not work in the real world of dates. If you want the difference between Feb 1, 2016 and January 31, 2017 the result I would want is 1 year, 0 months, and 0 days. Exactly one year (assuming you count the last day as a full day, like in a lease for an apartment). However, the millisecond approach would give you 1 year 0 months and 1 day, since the date range includes a leap year. So here is the code I used in javascript for my adobe form (you can name the fields): (edited, there was an error that I corrected)
var f1 = this.getField("LeaseExpiration");
var g1 = this.getField("LeaseStart");
var end = f1.value
var begin = g1.value
var e = new Date(end);
var b = new Date(begin);
var bMonth = b.getMonth();
var bYear = b.getFullYear();
var eYear = e.getFullYear();
var eMonth = e.getMonth();
var bDay = b.getDate();
var eDay = e.getDate() + 1;
if ((eMonth == 0)||(eMonth == 2)||(eMonth == 4)|| (eMonth == 6) || (eMonth == 7) ||(eMonth == 9)||(eMonth == 11))
{
var eDays = 31;
}
if ((eMonth == 3)||(eMonth == 5)||(eMonth == 8)|| (eMonth == 10))
{
var eDays = 30;
}
if (eMonth == 1&&((eYear % 4 == 0) && (eYear % 100 != 0)) || (eYear % 400 == 0))
{
var eDays = 29;
}
if (eMonth == 1&&((eYear % 4 != 0) || (eYear % 100 == 0)))
{
var eDays = 28;
}
if ((bMonth == 0)||(bMonth == 2)||(bMonth == 4)|| (bMonth == 6) || (bMonth == 7) ||(bMonth == 9)||(bMonth == 11))
{
var bDays = 31;
}
if ((bMonth == 3)||(bMonth == 5)||(bMonth == 8)|| (bMonth == 10))
{
var bDays = 30;
}
if (bMonth == 1&&((bYear % 4 == 0) && (bYear % 100 != 0)) || (bYear % 400 == 0))
{
var bDays = 29;
}
if (bMonth == 1&&((bYear % 4 != 0) || (bYear % 100 == 0)))
{
var bDays = 28;
}
var FirstMonthDiff = bDays - bDay + 1;
if (eDay - bDay < 0)
{
eMonth = eMonth - 1;
eDay = eDay + eDays;
}
var daysDiff = eDay - bDay;
if(eMonth - bMonth < 0)
{
eYear = eYear - 1;
eMonth = eMonth + 12;
}
var monthDiff = eMonth - bMonth;
var yearDiff = eYear - bYear;
if (daysDiff == eDays)
{
daysDiff = 0;
monthDiff = monthDiff + 1;
if (monthDiff == 12)
{
monthDiff = 0;
yearDiff = yearDiff + 1;
}
}
if ((FirstMonthDiff != bDays)&&(eDay - 1 == eDays))
{
daysDiff = FirstMonthDiff;
}
event.value = yearDiff + " Year(s)" + " " + monthDiff + " month(s) " + daysDiff + " days(s)"
I have created, yet another one, function for this purpose:
function dateDiff(date) {
date = date.split('-');
var today = new Date();
var year = today.getFullYear();
var month = today.getMonth() + 1;
var day = today.getDate();
var yy = parseInt(date[0]);
var mm = parseInt(date[1]);
var dd = parseInt(date[2]);
var years, months, days;
// months
months = month - mm;
if (day < dd) {
months = months - 1;
}
// years
years = year - yy;
if (month * 100 + day < mm * 100 + dd) {
years = years - 1;
months = months + 12;
}
// days
days = Math.floor((today.getTime() - (new Date(yy + years, mm + months - 1, dd)).getTime()) / (24 * 60 * 60 * 1000));
//
return {years: years, months: months, days: days};
}
Doesn't require any 3rd party libraries. Takes one argument -- date in YYYY-MM-DD format.
https://gist.github.com/lemmon/d27c2d4a783b1cf72d1d1cc243458d56
With dayjs we did it in that way:
export const getAgeDetails = (oldDate: dayjs.Dayjs, newDate: dayjs.Dayjs) => {
const years = newDate.diff(oldDate, 'year');
const months = newDate.diff(oldDate, 'month') - years * 12;
const days = newDate.diff(oldDate.add(years, 'year').add(months, 'month'), 'day');
return {
years,
months,
days,
allDays: newDate.diff(oldDate, 'day'),
};
};
It calculates it perfectly including leap years and different month amount of days.
For quick and easy use I wrote this function some time ago. It returns the diff between two dates in a nice format. Feel free to use it (tested on webkit).
/**
* Function to print date diffs.
*
* #param {Date} fromDate: The valid start date
* #param {Date} toDate: The end date. Can be null (if so the function uses "now").
* #param {Number} levels: The number of details you want to get out (1="in 2 Months",2="in 2 Months, 20 Days",...)
* #param {Boolean} prefix: adds "in" or "ago" to the return string
* #return {String} Diffrence between the two dates.
*/
function getNiceTime(fromDate, toDate, levels, prefix){
var lang = {
"date.past": "{0} ago",
"date.future": "in {0}",
"date.now": "now",
"date.year": "{0} year",
"date.years": "{0} years",
"date.years.prefixed": "{0} years",
"date.month": "{0} month",
"date.months": "{0} months",
"date.months.prefixed": "{0} months",
"date.day": "{0} day",
"date.days": "{0} days",
"date.days.prefixed": "{0} days",
"date.hour": "{0} hour",
"date.hours": "{0} hours",
"date.hours.prefixed": "{0} hours",
"date.minute": "{0} minute",
"date.minutes": "{0} minutes",
"date.minutes.prefixed": "{0} minutes",
"date.second": "{0} second",
"date.seconds": "{0} seconds",
"date.seconds.prefixed": "{0} seconds",
},
langFn = function(id,params){
var returnValue = lang[id] || "";
if(params){
for(var i=0;i<params.length;i++){
returnValue = returnValue.replace("{"+i+"}",params[i]);
}
}
return returnValue;
},
toDate = toDate ? toDate : new Date(),
diff = fromDate - toDate,
past = diff < 0 ? true : false,
diff = diff < 0 ? diff * -1 : diff,
date = new Date(new Date(1970,0,1,0).getTime()+diff),
returnString = '',
count = 0,
years = (date.getFullYear() - 1970);
if(years > 0){
var langSingle = "date.year" + (prefix ? "" : ""),
langMultiple = "date.years" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (years > 1 ? langFn(langMultiple,[years]) : langFn(langSingle,[years]));
count ++;
}
var months = date.getMonth();
if(count < levels && months > 0){
var langSingle = "date.month" + (prefix ? "" : ""),
langMultiple = "date.months" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (months > 1 ? langFn(langMultiple,[months]) : langFn(langSingle,[months]));
count ++;
} else {
if(count > 0)
count = 99;
}
var days = date.getDate() - 1;
if(count < levels && days > 0){
var langSingle = "date.day" + (prefix ? "" : ""),
langMultiple = "date.days" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (days > 1 ? langFn(langMultiple,[days]) : langFn(langSingle,[days]));
count ++;
} else {
if(count > 0)
count = 99;
}
var hours = date.getHours();
if(count < levels && hours > 0){
var langSingle = "date.hour" + (prefix ? "" : ""),
langMultiple = "date.hours" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (hours > 1 ? langFn(langMultiple,[hours]) : langFn(langSingle,[hours]));
count ++;
} else {
if(count > 0)
count = 99;
}
var minutes = date.getMinutes();
if(count < levels && minutes > 0){
var langSingle = "date.minute" + (prefix ? "" : ""),
langMultiple = "date.minutes" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (minutes > 1 ? langFn(langMultiple,[minutes]) : langFn(langSingle,[minutes]));
count ++;
} else {
if(count > 0)
count = 99;
}
var seconds = date.getSeconds();
if(count < levels && seconds > 0){
var langSingle = "date.second" + (prefix ? "" : ""),
langMultiple = "date.seconds" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (seconds > 1 ? langFn(langMultiple,[seconds]) : langFn(langSingle,[seconds]));
count ++;
} else {
if(count > 0)
count = 99;
}
if(prefix){
if(returnString == ""){
returnString = langFn("date.now");
} else if(past)
returnString = langFn("date.past",[returnString]);
else
returnString = langFn("date.future",[returnString]);
}
return returnString;
}
If you are using date-fns and if you dont want to install the Moment.js or the moment-precise-range-plugin. You can use the following date-fns function to get the same result as moment-precise-range-plugin
intervalToDuration({
start: new Date(),
end: new Date("24 Jun 2020")
})
This will give output in a JSON object like below
{
"years": 0,
"months": 0,
"days": 0,
"hours": 19,
"minutes": 35,
"seconds": 24
}
Live Example https://stackblitz.com/edit/react-wvxvql
Link to Documentation https://date-fns.org/v2.14.0/docs/intervalToDuration
Some math is in order.
You can subtract one Date object from another in Javascript, and you'll get the difference between them in milisseconds. From this result you can extract the other parts you want (days, months etc.)
For example:
var a = new Date(2010, 10, 1);
var b = new Date(2010, 9, 1);
var c = a - b; // c equals 2674800000,
// the amount of milisseconds between September 1, 2010
// and August 1, 2010.
Now you can get any part you want. For example, how many days have elapsed between the two dates:
var days = (a - b) / (60 * 60 * 24 * 1000);
// 60 * 60 * 24 * 1000 is the amount of milisseconds in a day.
// the variable days now equals 30.958333333333332.
That's almost 31 days. You can then round down for 30 days, and use whatever remained to get the amounts of hours, minutes etc.
Get the difference between two dates in a human way
This function is capable of returning natural-language-like text. Use it to get responses like:
"4 years, 1 month and 11 days"
"1 year and 2 months"
"11 months and 20 days"
"12 days"
IMPORTANT: date-fns is a dependency
Just copy the code below and plug in a past date into our getElapsedTime function! It will compare the entered date against the present time and return your human-like responses.
import * as dateFns from "https://cdn.skypack.dev/date-fns#2.22.1";
function getElapsedTime(pastDate) {
const duration = dateFns.intervalToDuration({
start: new Date(pastDate),
end: new Date(),
});
let [years, months, days] = ["", "", ""];
if (duration.years > 0) {
years = duration.years === 1 ? "1 year" : `${duration.years} years`;
}
if (duration.months > 0) {
months = duration.months === 1 ? "1 month" : `${duration.months} months`;
}
if (duration.days > 0) {
days = duration.days === 1 ? "1 day" : `${duration.days} days`;
}
let response = [years, months, days].filter(Boolean);
switch (response.length) {
case 3:
response[1] += " and";
response[0] += ",";
break;
case 2:
response[0] += " and";
break;
}
return response.join(" ");
}
Yet another solution, based on some PHP code.
The strtotime function, also based on PHP, can be found here: http://phpjs.org/functions/strtotime/.
Date.dateDiff = function(d1, d2) {
d1 /= 1000;
d2 /= 1000;
if (d1 > d2) d2 = [d1, d1 = d2][0];
var diffs = {
year: 0,
month: 0,
day: 0,
hour: 0,
minute: 0,
second: 0
}
$.each(diffs, function(interval) {
while (d2 >= (d3 = Date.strtotime('+1 '+interval, d1))) {
d1 = d3;
++diffs[interval];
}
});
return diffs;
};
Usage:
> d1 = new Date(2000, 0, 1)
Sat Jan 01 2000 00:00:00 GMT+0100 (CET)
> d2 = new Date(2013, 9, 6)
Sun Oct 06 2013 00:00:00 GMT+0200 (CEST)
> Date.dateDiff(d1, d2)
Object {
day: 5
hour: 0
minute: 0
month: 9
second: 0
year: 13
}
Very old thread, I know, but here's my contribution, as the thread is not solved yet.
It takes leap years into consideration and does not asume any fixed number of days per month or year.
It might be flawed in border cases as I haven't tested it thoroughly, but it works for all the dates provided in the original question, thus I'm confident.
function calculate() {
var fromDate = document.getElementById('fromDate').value;
var toDate = document.getElementById('toDate').value;
try {
document.getElementById('result').innerHTML = '';
var result = getDateDifference(new Date(fromDate), new Date(toDate));
if (result && !isNaN(result.years)) {
document.getElementById('result').innerHTML =
result.years + ' year' + (result.years == 1 ? ' ' : 's ') +
result.months + ' month' + (result.months == 1 ? ' ' : 's ') + 'and ' +
result.days + ' day' + (result.days == 1 ? '' : 's');
}
} catch (e) {
console.error(e);
}
}
function getDateDifference(startDate, endDate) {
if (startDate > endDate) {
console.error('Start date must be before end date');
return null;
}
var startYear = startDate.getFullYear();
var startMonth = startDate.getMonth();
var startDay = startDate.getDate();
var endYear = endDate.getFullYear();
var endMonth = endDate.getMonth();
var endDay = endDate.getDate();
// We calculate February based on end year as it might be a leep year which might influence the number of days.
var february = (endYear % 4 == 0 && endYear % 100 != 0) || endYear % 400 == 0 ? 29 : 28;
var daysOfMonth = [31, february, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
var startDateNotPassedInEndYear = (endMonth < startMonth) || endMonth == startMonth && endDay < startDay;
var years = endYear - startYear - (startDateNotPassedInEndYear ? 1 : 0);
var months = (12 + endMonth - startMonth - (endDay < startDay ? 1 : 0)) % 12;
// (12 + ...) % 12 makes sure index is always between 0 and 11
var days = startDay <= endDay ? endDay - startDay : daysOfMonth[(12 + endMonth - 1) % 12] - startDay + endDay;
return {
years: years,
months: months,
days: days
};
}
<p><input type="text" name="fromDate" id="fromDate" placeholder="yyyy-mm-dd" value="1999-02-28" /></p>
<p><input type="text" name="toDate" id="toDate" placeholder="yyyy-mm-dd" value="2000-03-01" /></p>
<p><input type="button" name="calculate" value="Calculate" onclick="javascript:calculate();" /></p>
<p />
<p id="result"></p>
let startDate = moment(new Date('2017-05-12')); // yyyy-MM-dd
let endDate = moment(new Date('2018-09-14')); // yyyy-MM-dd
let Years = newDate.diff(date, 'years');
let months = newDate.diff(date, 'months');
let days = newDate.diff(date, 'days');
console.log("Year: " + Years, ", Month: " months-(Years*12), ", Days: " days-(Years*365.25)-((365.25*(days- (Years*12)))/12));
Above snippet will print: Year: 1, Month: 4, Days: 2
Using Plane Javascript:
function dateDiffInDays(start, end) {
var MS_PER_DAY = 1000 * 60 * 60 * 24;
var a = new Date(start);
var b = new Date(end);
const diffTime = Math.abs(a - b);
const diffDays = Math.ceil(diffTime / MS_PER_DAY);
console.log("Days: ", diffDays);
// Discard the time and time-zone information.
const utc1 = Date.UTC(a.getFullYear(), a.getMonth(), a.getDate());
const utc2 = Date.UTC(b.getFullYear(), b.getMonth(), b.getDate());
return Math.floor((utc2 - utc1) / MS_PER_DAY);
}
function dateDiffInDays_Months_Years(start, end) {
var m1 = new Date(start);
var m2 = new Date(end);
var yDiff = m2.getFullYear() - m1.getFullYear();
var mDiff = m2.getMonth() - m1.getMonth();
var dDiff = m2.getDate() - m1.getDate();
if (dDiff < 0) {
var daysInLastFullMonth = getDaysInLastFullMonth(start);
if (daysInLastFullMonth < m1.getDate()) {
dDiff = daysInLastFullMonth + dDiff + (m1.getDate() -
daysInLastFullMonth);
} else {
dDiff = daysInLastFullMonth + dDiff;
}
mDiff--;
}
if (mDiff < 0) {
mDiff = 12 + mDiff;
yDiff--;
}
console.log('Y:', yDiff, ', M:', mDiff, ', D:', dDiff);
}
function getDaysInLastFullMonth(day) {
var d = new Date(day);
console.log(d.getDay() );
var lastDayOfMonth = new Date(d.getFullYear(), d.getMonth() + 1, 0);
console.log('last day of month:', lastDayOfMonth.getDate() ); //
return lastDayOfMonth.getDate();
}
Using moment.js:
function dateDiffUsingMoment(start, end) {
var a = moment(start,'M/D/YYYY');
var b = moment(end,'M/D/YYYY');
var diffDaysMoment = b.diff(a, 'days');
console.log('Moments.js : ', diffDaysMoment);
preciseDiffMoments(a,b);
}
function preciseDiffMoments( a, b) {
var m1= a, m2=b;
m1.add(m2.utcOffset() - m1.utcOffset(), 'minutes'); // shift timezone of m1 to m2
var yDiff = m2.year() - m1.year();
var mDiff = m2.month() - m1.month();
var dDiff = m2.date() - m1.date();
if (dDiff < 0) {
var daysInLastFullMonth = moment(m2.year() + '-' + (m2.month() + 1),
"YYYY-MM").subtract(1, 'M').daysInMonth();
if (daysInLastFullMonth < m1.date()) { // 31/01 -> 2/03
dDiff = daysInLastFullMonth + dDiff + (m1.date() -
daysInLastFullMonth);
} else {
dDiff = daysInLastFullMonth + dDiff;
}
mDiff--;
}
if (mDiff < 0) {
mDiff = 12 + mDiff;
yDiff--;
}
console.log('getMomentum() Y:', yDiff, ', M:', mDiff, ', D:', dDiff);
}
Tested the above functions using following samples:
var sample1 = all('2/13/2018', '3/15/2018'); // {'M/D/YYYY'} 30, Y: 0 , M: 1 , D: 2
console.log(sample1);
var sample2 = all('10/09/2019', '7/7/2020'); // 272, Y: 0 , M: 8 , D: 29
console.log(sample2);
function all(start, end) {
dateDiffInDays(start, end);
dateDiffInDays_Months_Years(start, end);
try {
dateDiffUsingMoment(start, end);
} catch (e) {
console.log(e);
}
}
by using Moment library and some custom logic, we can get the exact date difference
var out;
out = diffDate(new Date('2014-05-10'), new Date('2015-10-10'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2015-10-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2015-09-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2015-03-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2016-03-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2016-03-11'));
display(out);
function diffDate(startDate, endDate) {
var b = moment(startDate),
a = moment(endDate),
intervals = ['years', 'months', 'weeks', 'days'],
out = {};
for (var i = 0; i < intervals.length; i++) {
var diff = a.diff(b, intervals[i]);
b.add(diff, intervals[i]);
out[intervals[i]] = diff;
}
return out;
}
function display(obj) {
var str = '';
for (key in obj) {
str = str + obj[key] + ' ' + key + ' '
}
console.log(str);
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.24.0/moment.js"></script>
I did it using a bunch of functions. Pure JavaScript and precise.
This code includes functions that calculate time difference in days, months and years. One of them can be used to get precise time difference for example X years, Y months, Z days. At the end of code I provided some tests.
How it works:
getDaysDiff():
Transforms time difference from milliseconds to days.
getYearsDiff():
There is no worries for effect of months and days of both dates. The function calculates difference in years by moving dates back and forward.
getMonthsDiff() (This one has nothing to do with question, but the concept is used in calExactTimeDiff() and I thought someone may need such a function so I insert it):
This one is a little tricky. The hard work is to deal with month and day of both dates.
If the endDate's month is more than startDate's, this means another year (12 months) is passed. But this is being taken care of in monthsOfFullYears, so the only thing is needed is to add subtraction of month of endDate and startDate.
If the startDate's month is more than endDate's then there is no another year. So we should get the difference between them. Imagine we want to go from month 10 of the current year to 2 of the next year. We can go like this: 11, 12, 1, 2. So we passed 4 months. This is equal to 12 - (10 - 2). We get difference between the months and subtract it from months of a whole year.
Next step is to take care of days of months. If day of endDate is more than or equal to startDate this means another month is passed. So we add 1 to it. But if it's less, then there is nothing to worry about. But in my code I did not do this. Because when I added difference between months I assumed that the days of months are equal. So I already added 1. Thus if day of endDate is less than startDate, I have to decrease months by 1.
There is an exception: if months are equal and endDate's day is less than startDate's, month should be 11.
I used the same concept in calExactTimeDiff().
Hope to be useful :)
// time difference in Days
function getDaysDiff(startDate = new Date(), endDate = new Date()) {
if (startDate > endDate) [startDate, endDate] = [endDate, startDate];
let timeDiff = endDate - startDate;
let timeDiffInDays = Math.floor(timeDiff / (1000 * 3600 * 24));
return timeDiffInDays;
}
// time difference in Months
function getMonthsDiff(startDate = new Date(), endDate = new Date()) {
let monthsOfFullYears = getYearsDiff(startDate, endDate) * 12;
let months = monthsOfFullYears;
// the variable below is not necessary, but I kept it for understanding of code
// we can use "startDate" instead of it
let yearsAfterStart = new Date(
startDate.getFullYear() + getYearsDiff(startDate, endDate),
startDate.getMonth(),
startDate.getDate()
);
let isDayAhead = endDate.getDate() >= yearsAfterStart.getDate();
if (startDate.getMonth() == endDate.getMonth() && !isDayAhead) {
months = 11;
return months;
}
if (endDate.getMonth() >= yearsAfterStart.getMonth()) {
let diff = endDate.getMonth() - yearsAfterStart.getMonth();
months += (isDayAhead) ? diff : diff - 1;
}
else {
months += isDayAhead
? 12 - (startDate.getMonth() - endDate.getMonth())
: 12 - (startDate.getMonth() - endDate.getMonth()) - 1;
}
return months;
}
// time difference in Years
function getYearsDiff(startDate = new Date(), endDate = new Date()) {
if (startDate > endDate) [startDate, endDate] = [endDate, startDate];
let yearB4End = new Date(
endDate.getFullYear() - 1,
endDate.getMonth(),
endDate.getDate()
);
let year = 0;
year = yearB4End > startDate
? yearB4End.getFullYear() - startDate.getFullYear()
: 0;
let yearsAfterStart = new Date(
startDate.getFullYear() + year + 1,
startDate.getMonth(),
startDate.getDate()
);
if (endDate >= yearsAfterStart) year++;
return year;
}
// time difference in format: X years, Y months, Z days
function calExactTimeDiff(firstDate, secondDate) {
if (firstDate > secondDate)
[firstDate, secondDate] = [secondDate, firstDate];
let monthDiff = 0;
let isDayAhead = secondDate.getDate() >= firstDate.getDate();
if (secondDate.getMonth() >= firstDate.getMonth()) {
let diff = secondDate.getMonth() - firstDate.getMonth();
monthDiff += (isDayAhead) ? diff : diff - 1;
}
else {
monthDiff += isDayAhead
? 12 - (firstDate.getMonth() - secondDate.getMonth())
: 12 - (firstDate.getMonth() - secondDate.getMonth()) - 1;
}
let dayDiff = 0;
if (isDayAhead) {
dayDiff = secondDate.getDate() - firstDate.getDate();
}
else {
let b4EndDate = new Date(
secondDate.getFullYear(),
secondDate.getMonth() - 1,
firstDate.getDate()
)
dayDiff = getDaysDiff(b4EndDate, secondDate);
}
if (firstDate.getMonth() == secondDate.getMonth() && !isDayAhead)
monthDiff = 11;
let exactTimeDiffUnits = {
yrs: getYearsDiff(firstDate, secondDate),
mths: monthDiff,
dys: dayDiff,
};
return `${exactTimeDiffUnits.yrs} years, ${exactTimeDiffUnits.mths} months, ${exactTimeDiffUnits.dys} days`
}
let s = new Date(2012, 4, 12);
let e = new Date(2008, 5, 24);
console.log(calExactTimeDiff(s, e));
s = new Date(2001, 7, 4);
e = new Date(2016, 6, 9);
console.log(calExactTimeDiff(s, e));
s = new Date(2011, 11, 28);
e = new Date(2021, 3, 6);
console.log(calExactTimeDiff(s, e));
s = new Date(2020, 8, 7);
e = new Date(2021, 8, 6);
console.log(calExactTimeDiff(s, e));
There a a couple of npm packages that help in doing this. Below is a list gathered from various sources. I find the date-fns version to be the most simplest.
1. date-fns
You can use intervalToDuration, formatDuration from date-fns to humanize a duration in desired format like below:
import { intervalToDuration, formatDuration } from 'date-fns'
let totalDuration = intervalToDuration({
start: new Date(1929, 0, 15, 12, 0, 0),
end: new Date(1968, 3, 4, 19, 5, 0)
});
let textDuration = formatDuration(totalDuration, { format: ['years', 'months'], delimiter: ', ' })
// Output: "39 years, 2 months"
clone the above code from here for trying it yourself: https://runkit.com/embed/diu9o3qe53j4
2. luxon + humanize-duration
you can use luxon to extract the duration between dates and humanize that using humanize-duration like below:
const DateTime = luxon.DateTime;
const Interval = luxon.Interval;
const start = DateTime.fromSQL("2020-06-19 11:14:00");
const finish = DateTime.fromSQL("2020-06-21 13:11:00");
const formatted = Interval
.fromDateTimes(start, finish)
.toDuration()
.valueOf();
console.log(humanizeDuration(formatted))
// output: 2 days, 1 hour, 57 minutes
console.log(humanizeDuration(formatted, { language: 'es' }))
// output: 2 días, 1 hora, 57 minutos
console.log(humanizeDuration(formatted, { language: 'ru' }))
// output: 2 дня, 1 час, 57 минут
<script src="https://cdn.jsdelivr.net/npm/luxon#1.25.0/build/global/luxon.min.js"></script>
<script src="https://cdn.jsdelivr.net/npm/humanize-duration#3.25.1/humanize-duration.min.js"></script>
reference to above code: https://stackoverflow.com/a/65651515/6908282
I would personally use http://www.datejs.com/, really handy. Specifically, look at the time.js file: http://code.google.com/p/datejs/source/browse/trunk/src/time.js
Time span in full Days, Hours, Minutes, Seconds, Milliseconds:
// Extension for Date
Date.difference = function (dateFrom, dateTo) {
var diff = { TotalMs: dateTo - dateFrom };
diff.Days = Math.floor(diff.TotalMs / 86400000);
var remHrs = diff.TotalMs % 86400000;
var remMin = remHrs % 3600000;
var remS = remMin % 60000;
diff.Hours = Math.floor(remHrs / 3600000);
diff.Minutes = Math.floor(remMin / 60000);
diff.Seconds = Math.floor(remS / 1000);
diff.Milliseconds = Math.floor(remS % 1000);
return diff;
};
// Usage
var a = new Date(2014, 05, 12, 00, 5, 45, 30); //a: Thu Jun 12 2014 00:05:45 GMT+0400
var b = new Date(2014, 02, 12, 00, 0, 25, 0); //b: Wed Mar 12 2014 00:00:25 GMT+0400
var diff = Date.difference(b, a);
/* diff: {
Days: 92
Hours: 0
Minutes: 5
Seconds: 20
Milliseconds: 30
TotalMs: 7949120030
} */
Neither of the codes work for me, so I use this instead for months and days:
function monthDiff(d2, d1) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
months += d2.getMonth() + 1;
return months <= 0 ? 0 : months;
}
function daysInMonth(date) {
return new Date(date.getYear(), date.getMonth() + 1, 0).getDate();
}
function diffDate(date1, date2) {
if (date2 && date2.getTime() && !isNaN(date2.getTime())) {
var months = monthDiff(date1, date2);
var days = 0;
if (date1.getUTCDate() >= date2.getUTCDate()) {
days = date1.getUTCDate() - date2.getUTCDate();
}
else {
months--;
days = date1.getUTCDate() - date2.getUTCDate() + daysInMonth(date2);
}
// Use the variables months and days how you need them.
}
}
The following is an algorithm which gives correct but not totally precise since it does not take into account leap year. It also assumes 30 days in a month. A good usage for example is if someone lives in an address from 12/11/2010 to 11/10/2011, it can quickly tells that the person lives there for 10 months and 29 days. From 12/11/2010 to 11/12/2011 is 11 months and 1 day. For certain types of applications, that kind of precision is sufficient. This is for those types of applications because it aims for simplicity:
var datediff = function(start, end) {
var diff = { years: 0, months: 0, days: 0 };
var timeDiff = end - start;
if (timeDiff > 0) {
diff.years = end.getFullYear() - start.getFullYear();
diff.months = end.getMonth() - start.getMonth();
diff.days = end.getDate() - start.getDate();
if (diff.months < 0) {
diff.years--;
diff.months += 12;
}
if (diff.days < 0) {
diff.months = Math.max(0, diff.months - 1);
diff.days += 30;
}
}
return diff;
};
Unit tests
To calculate the difference between two dates in Years, Months, Days, Minutes, Seconds, Milliseconds using TypeScript/ JavaScript
dateDifference(actualDate) {
// Calculate time between two dates:
const date1 = actualDate; // the date you already commented/ posted
const date2: any = new Date(); // today
let r = {}; // object for clarity
let message: string;
const diffInSeconds = Math.abs(date2 - date1) / 1000;
const days = Math.floor(diffInSeconds / 60 / 60 / 24);
const hours = Math.floor(diffInSeconds / 60 / 60 % 24);
const minutes = Math.floor(diffInSeconds / 60 % 60);
const seconds = Math.floor(diffInSeconds % 60);
const milliseconds =
Math.round((diffInSeconds - Math.floor(diffInSeconds)) * 1000);
const months = Math.floor(days / 31);
const years = Math.floor(months / 12);
// the below object is just optional
// if you want to return an object instead of a message
r = {
years: years,
months: months,
days: days,
hours: hours,
minutes: minutes,
seconds: seconds,
milliseconds: milliseconds
};
// check if difference is in years or months
if (years === 0 && months === 0) {
// show in days if no years / months
if (days > 0) {
if (days === 1) {
message = days + ' day';
} else { message = days + ' days'; }
} else if (hours > 0) {
if (hours === 1) {
message = hours + ' hour';
} else {
message = hours + ' hours';
}
} else {
// show in minutes if no years / months / days
if (minutes === 1) {
message = minutes + ' minute';
} else {message = minutes + ' minutes';}
}
} else if (years === 0 && months > 0) {
// show in months if no years
if (months === 1) {
message = months + ' month';
} else {message = months + ' months';}
} else if (years > 0) {
// show in years if years exist
if (years === 1) {
message = years + ' year';
} else {message = years + ' years';}
}
return 'Posted ' + message + ' ago';
// this is the message a user see in the view
}
However, you can update the above logic for the message to show seconds and milliseconds too or else use the object 'r' to format the message whatever way you want.
If you want to directly copy the code, you can view my gist with the above code here
I know it is an old thread, but I'd like to put my 2 cents based on the answer by #Pawel Miech.
It is true that you need to convert the difference into milliseconds, then you need to make some math. But notice that, you need to do the math in backward manner, i.e. you need to calculate years, months, days, hours then minutes.
I used to do some thing like this:
var mins;
var hours;
var days;
var months;
var years;
var diff = new Date() - new Date(yourOldDate);
// yourOldDate may be is coming from DB, for example, but it should be in the correct format ("MM/dd/yyyy hh:mm:ss:fff tt")
years = Math.floor((diff) / (1000 * 60 * 60 * 24 * 365));
diff = Math.floor((diff) % (1000 * 60 * 60 * 24 * 365));
months = Math.floor((diff) / (1000 * 60 * 60 * 24 * 30));
diff = Math.floor((diff) % (1000 * 60 * 60 * 24 * 30));
days = Math.floor((diff) / (1000 * 60 * 60 * 24));
diff = Math.floor((diff) % (1000 * 60 * 60 * 24));
hours = Math.floor((diff) / (1000 * 60 * 60));
diff = Math.floor((diff) % (1000 * 60 * 60));
mins = Math.floor((diff) / (1000 * 60));
But, of course, this is not precise because it assumes that all years have 365 days and all months have 30 days, which is not true in all cases.
Its very simple please use the code below and it will give the difference in that format according to this //3 years 9 months 3 weeks 5 days 15 hours 50 minutes
Date.getFormattedDateDiff = function(date1, date2) {
var b = moment(date1),
a = moment(date2),
intervals = ['years','months','weeks','days'],
out = [];
for(var i=0; i<intervals.length; i++){
var diff = a.diff(b, intervals[i]);
b.add(diff, intervals[i]);
out.push(diff + ' ' + intervals[i]);
}
return out.join(', ');
};
var today = new Date(),
newYear = new Date(today.getFullYear(), 0, 1),
y2k = new Date(2000, 0, 1);
//(AS OF NOV 29, 2016)
//Time since New Year: 0 years, 10 months, 4 weeks, 0 days
console.log( 'Time since New Year: ' + Date.getFormattedDateDiff(newYear, today) );
//Time since Y2K: 16 years, 10 months, 4 weeks, 0 days
console.log( 'Time since Y2K: ' + Date.getFormattedDateDiff(y2k, today) );
This code should give you desired results
//************************** Enter your dates here **********************//
var startDate = "10/05/2014";
var endDate = "11/3/2016"
//******* and press "Run", you will see the result in a popup *********//
var noofdays = 0;
var sdArr = startDate.split("/");
var startDateDay = parseInt(sdArr[0]);
var startDateMonth = parseInt(sdArr[1]);
var startDateYear = parseInt(sdArr[2]);
sdArr = endDate.split("/")
var endDateDay = parseInt(sdArr[0]);
var endDateMonth = parseInt(sdArr[1]);
var endDateYear = parseInt(sdArr[2]);
console.log(startDateDay+' '+startDateMonth+' '+startDateYear);
var yeardays = 365;
var monthArr = [31,,31,30,31,30,31,31,30,31,30,31];
var noofyears = 0
var noofmonths = 0;
if((startDateYear%4)==0) monthArr[1]=29;
else monthArr[1]=28;
if(startDateYear == endDateYear){
noofyears = 0;
noofmonths = getMonthDiff(startDate,endDate);
if(noofmonths < 0) noofmonths = 0;
noofdays = getDayDiff(startDate,endDate);
}else{
if(endDateMonth < startDateMonth){
noofyears = (endDateYear - startDateYear)-1;
if(noofyears < 1) noofyears = 0;
}else{
noofyears = endDateYear - startDateYear;
}
noofmonths = getMonthDiff(startDate,endDate);
if(noofmonths < 0) noofmonths = 0;
noofdays = getDayDiff(startDate,endDate);
}
alert(noofyears+' year, '+ noofmonths+' months, '+ noofdays+' days');
function getDayDiff(startDate,endDate){
if(endDateDay >=startDateDay){
noofdays = 0;
if(endDateDay > startDateDay) {
noofdays = endDateDay - startDateDay;
}
}else{
if((endDateYear%4)==0) {
monthArr[1]=29;
}else{
monthArr[1] = 28;
}
if(endDateMonth != 1)
noofdays = (monthArr[endDateMonth-2]-startDateDay) + endDateDay;
else
noofdays = (monthArr[11]-startDateDay) + endDateDay;
}
return noofdays;
}
function getMonthDiff(startDate,endDate){
if(endDateMonth > startDateMonth){
noofmonths = endDateMonth - startDateMonth;
if(endDateDay < startDateDay){
noofmonths--;
}
}else{
noofmonths = (12-startDateMonth) + endDateMonth;
if(endDateDay < startDateDay){
noofmonths--;
}
}
return noofmonths;
}
https://jsfiddle.net/moremanishk/hk8c419f/
You should try using date-fns. Here's how I did it using intervalToDuration and formatDuration functions from date-fns.
let startDate = Date.parse("2010-10-01 00:00:00 UTC");
let endDate = Date.parse("2020-11-01 00:00:00 UTC");
let duration = intervalToDuration({start: startDate, end: endDate});
let durationInWords = formatDuration(duration, {format: ["years", "months", "days"]}); //output: 10 years 1 month
since I had to use moment-hijri (hijri calendar) and couldn't use moment.diff() method, I came up with this solution. can also be used with moment.js
var momenti = require('moment-hijri')
//calculate hijri
var strt = await momenti(somedateobject)
var until = await momenti()
var years = await 0
var months = await 0
var days = await 0
while(strt.valueOf() < until.valueOf()){
await strt.add(1, 'iYear');
await years++
}
await strt.subtract(1, 'iYear');
await years--
while(strt.valueOf() < until.valueOf()){
await strt.add(1, 'iMonth');
await months++
}
await strt.subtract(1, 'iMonth');
await months--
while(strt.valueOf() < until.valueOf()){
await strt.add(1, 'day');
await days++
}
await strt.subtract(1, 'day');
await days--
await console.log(years)
await console.log(months)
await console.log(days)
A solution with the ECMAScript "Temporal API" which is currently (as of 5th March 2022) in Stage 3 of Active Proposals, which will the method we will do this in the future (soon).
Here is a solution with the current temporal-polyfill
<script type='module'>
import * as TemporalModule from 'https://cdn.jsdelivr.net/npm/#js-temporal/polyfill#0.3.0/dist/index.umd.js'
const Temporal = temporal.Temporal;
//----------------------------------------
function dateDiff(start, end, maxUnit) {
return (Temporal.PlainDate.from(start).until(Temporal.PlainDate.from(end),{largestUnit:maxUnit}).toString()).match(/(\d*Y)|(\d*M)|(\d*D)/g).join(" ");
}
//----------------------------------------
console.log("Diff in (years, months, days): ",dateDiff("1963-02-03","2022-03-06","year"))
console.log("Diff in (months, days) : ",dateDiff("1963-02-03","2022-03-06","month"))
console.log("Diff in (days) : ",dateDiff("1963-02-03","2022-03-06","day"))
</script>
Your expected output is not correct. For example difference between '2014-05-10' and '2015-03-09' is not 9 months, 27 days
the correct answer is
(05-10 to 05-31) = 21 days
(2014-06 to 2015-03) = 9 months
(03-01 to 03-09) = 9 days
total is 9 months and 30 days
WARNING: An ideal function would be aware of leap years and days count in every month, but I found the results of this function accurate enough for my current task, so I shared it with you
function diffDate(date1, date2)
{
var daysDiff = Math.ceil((Math.abs(date1 - date2)) / (1000 * 60 * 60 * 24));
var years = Math.floor(daysDiff / 365.25);
var remainingDays = Math.floor(daysDiff - (years * 365.25));
var months = Math.floor((remainingDays / 365.25) * 12);
var days = Math.ceil(daysDiff - (years * 365.25 + (months / 12 * 365.25)));
return {
daysAll: daysDiff,
years: years,
months: months,
days:days
}
}
console.log(diffDate(new Date('2014-05-10'), new Date('2015-10-10')));
console.log(diffDate(new Date('2014-05-10'), new Date('2015-10-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2015-09-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2015-03-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2016-03-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2016-03-11')));

Calculate the minutes between to dates excluding evenings and weekends?

I'm trying to calculate the business time between two dates, down to the minute. Business hours being defined as between 8 AM and 6 PM EST (or EDT if its daylight savings).
I've found this answer, which performs this in hours, but am unsure how to convert to minutes, and ensure my timezones wont be messed up:
https://stackoverflow.com/a/11092865/104998
function isHoliday( /*Date*/ date) {
for (var i = 0; i < holidays.length; i++) {
if (holidays[i].getTime() == date.getTime()) {
return true;
}
}
return false;
}
function diffHours( /*Date*/ d1, /*Date*/ d2) {
var date1 = new Date(d1.getUTCFullYear() + "-" + (d1.getUTCMonth() + 1) + "-" + d1.getUTCDate() + " UTC");
var date2 = new Date(d2.getUTCFullYear() + "-" + (d2.getUTCMonth() + 1) + "-" + d2.getUTCDate() + " UTC");
var sum = 0;
var oneday = 24 * 3600 * 1000;
var hours, date;
// first day
if (!isHoliday(date1)) {
// decrease by a whole day first (will be added later)
sum -= 10;
// add real hours
hours = d1.getUTCHours() + d1.getUTCMinutes() / 60;
if (hours <= 6) {
sum += 10 - hours;
} else if (hours <= 20) {
sum += 4;
} else {
sum += 24 - hours;
}
}
// last day
if (!isHoliday(date2)) {
// decrease by a whole day first (will be added later)
sum -= 10;
// add real hours
hours = d2.getUTCHours() + d2.getUTCMinutes() / 60;
if (hours <= 6) {
sum += hours;
} else if (hours <= 20) {
sum += 6;
} else {
sum += hours - 14;
}
}
// whole days
while (date1 <= date2) {
if (!isHoliday(date1)) {
sum += 10;
}
// increase date by 1 day
date1.setTime(date1.getTime() + oneday);
}
return Math.floor(sum);
}
// ==============
// examples below
// --------------
// array of Dates (in UTC) to skip
var holidays = [
new Date("2012-01-04 UTC"),
];
for (var i = 0; i < holidays.length; i++) {
console.log('holiday: ', holidays[i].toUTCString());
}
a = new Date("2012-01-01 12:00 UTC");
b = new Date("2012-01-02 12:00 UTC");
c = new Date("2012-01-02 22:00 UTC");
d = new Date("2012-01-03 07:00 UTC");
e = new Date("2012-01-05 12:00 UTC");
console.log({
d1: a.toUTCString(),
d2: b.toUTCString(),
hours: diffHours(a, b)
});
console.log({
d1: b.toUTCString(),
d2: c.toUTCString(),
hours: diffHours(b, c)
});
console.log({
d1: c.toUTCString(),
d2: d.toUTCString(),
hours: diffHours(c, d)
});
console.log({
d1: d.toUTCString(),
d2: e.toUTCString(),
hours: diffHours(d, e)
});
Any help would be much appreciated.
Here is the class I came up with to do this, there must be a better way (this is my first time delving into nodejs)
var dateFuncs = DateFuncs.prototype;
function DateFuncs() {
}
DateFuncs.prototype.isWeekend = function(pDate) {
return (pDate.getDay() == 0 || pDate.getDay() == 6);
}
DateFuncs.prototype.isBusinessMinute = function(pDate) {
var hours = pDate.getHours() + pDate.getMinutes()/60;
return (hours >= 8 && hours < 18); //business hours are 8AM-6PM
}
DateFuncs.prototype.addMinutes = function(date, minutes) {
return new Date(date.getTime() + minutes*60000);
}
DateFuncs.prototype.diffBusinessMins = function(/*Date*/ startDate, /*Date*/ endDate) {
var minutesDiff = 0;
startDate.setSeconds(0,0);
endDate.setSeconds(0,0);
var curDate = new Date(startDate.getTime());
while(curDate.getTime() != endDate.getTime())
{
if(!this.isWeekend(curDate) && this.isBusinessMinute(curDate))
{
minutesDiff += 1;
}
curDate = this.addMinutes(curDate, 1);
}
return minutesDiff;
}
module.exports = DateFuncs;

Converting UTC to Decimal time

Background
I want to create a new date/time system based on an old French version with some modifications.
This involves converting UTC date/times to new quantities:
12 months => 10 months
52 weeks => 36.5 weeks
28/31 days per month => 36/37 days per month
24 hours => 20 hours
60 minutes => 100 minutes
60 seconds => 100 seconds
I've coded a clock in JavaScript as proof of concept, but unsure as to whether I have correctly calculated everything, additionally whether it's the best approach:
Code
1) getDecimalDate() calculates the day of the year, then works out which month it exists within a new calendar of 36 or 37 days per month. Then calculates the new date of the month.
function getDecimalDate(date) {
var oldDay = 1000 * 60 * 60 * 24,
startYear = new Date(Date.UTC(date.getUTCFullYear(), 0, 0)),
day = Math.floor((date - startYear) / oldDay),
num = 0,
month = 1;
if (day > 36) { num += 36; month = 2; }
if (day > 73) { num += 37; month = 3; }
if (day > 109) { num += 36; month = 4; }
if (day > 146) { num += 37; month = 5; }
if (day > 182) { num += 36; month = 6; }
if (day > 219) { num += 37; month = 7; }
if (day > 255) { num += 36; month = 8; }
if (day > 292) { num += 37; month = 9; }
if (day > 328) { num += 36; month = 10; }
return { day: day - num, month: month, year: date.getUTCFullYear(), num: num };
}
2) getDecimalTime() calculates the number of milliseconds since midnight, then changes it from old milliseconds per day to new totals, then calculates hours, mins etc
function getDecimalTime(date) {
var oldDay = 1000 * 60 * 60 * 24,
newDay = 1000 * 100 * 100 * 20,
startDay = new Date(Date.UTC(date.getUTCFullYear(), date.getUTCMonth(), date.getUTCDate())),
delta = ((date - startDay) / oldDay) * newDay;
var hours = Math.floor(delta / 10000000) % 20;
delta -= hours * 10000000;
var minutes = Math.floor(delta / 100000) % 100;
delta -= minutes * 100000;
var seconds = Math.floor(delta / 1000) % 100;
delta -= seconds * 1000;
var milliseconds = Math.floor(delta) % 1000;
return { milliseconds: milliseconds, seconds: seconds, minutes: minutes, hours: hours };
}
You can see a working version here:
https://jsfiddle.net/kmturley/7mrwc3x3/9/
Results
Bear in mind i've made up day/month names using Latin (Nov = 9, die = day, dec = 10, mense = month)
String - Saturday December 3 => Novdie Decmense 10
Date - 03-12-2016 => 10-10-2016
Time - 22:47:52 => 18:98:43
Questions
Is the math correct?
Are there any issues with timezones? i've
tried converting all Date objects to UTC but JavaScript can be
tricky
Can I improve the code? the month selection seems like it
could be improved but I couldn't figure out a better way to count 36
and 37 day months. if (num % 36.5 === 1) wouldn't work?
Thanks!
Update - 7th December 2016 - new versions based on solution:
https://jsfiddle.net/kmturley/7mrwc3x3/10/
https://github.com/kmturley/decimal-time
Is the math correct?
You didn't say which months have 35 days and which have 36 so we have to accept that the if statements are correct. You don't show how date is created so it may or may not be OK. And you don't say what happens for leap years, this system seems to only have 365 days per year.
The following:
24 hours => 20 hours
60 minutes => 100 minutes
60 seconds => 100 seconds
doesn't seem correct. Do you actually mean:
1 day = 20 decimal hours
1 decimal hour = 100 decimal minutes
1 decimal minute = 100 decimal seconds
1 decimal second = 1000 decimal milliseconds
Your strategy of getting the time in ms and scaling to decimal ms seems fine, I'll just make the following comments.
In getDecimalTime it is simpler to calculate startDay by first copying date then setting its UTC hours to zero:
startDay = new Date(+date);
startDate.setUTCHours(0,0,0,0);
Then scale:
var diffMilliseconds = date - startDate;
var decimalMilliseconds = diffMilliseconds / 8.64e7 * 2.0e8;
so 1 standard millisecond = 2.314814814814815 decimal milliseconds
In the date function, the expression:
new Date(date.getUTCFullYear(), 0, 0)
will create a date for 31 December the previous year (i.e. date of 0), if you're after 1 January then it should be:
new Date(date.getUTCFullYear(), 0, 1);
So likely you're one day out. Otherwise, the code seems to be correct. For me, the get decimal time function would be simpler as:
function getDecimalTime(date) {
// Pad numbers < 10
function z(n){return (n<10?'0':'')+n;}
// Copy date so don't modify original
var dayStart = new Date(+date);
var diffMs = date - dayStart.setUTCHours(0,0,0,0);
// Scale to decimal milliseconds
var decMs = Math.round(diffMs / 8.64e7 * 2.0e8);
// Get decimal hours, etc.
var decHr = decMs / 1.0e7 | 0;
var decMin = decMs % 1.0e7 / 1.0e5 | 0;
var decSec = decMs % 1.0e5 / 1.0e3 | 0;
decMs = decMs % 1.0e3;
return z(decHr) + ':' + z(decMin) + ':' + z(decSec) + '.' + ('0' + z(decMs)).slice(-3);
}
// Helper to format the time part of date
// as UTC hh:mm:ss.sss
function formatUTCTime(date) {
function z(n){return (n<10?'0':'')+n;}
return z(date.getUTCHours()) + ':' +
z(date.getUTCMinutes()) + ':' +
z(date.getUTCSeconds()) + '.' +
('00' + date.getUTCMilliseconds()).slice(-3);
}
// Test 00:00:00.003 => 00:00:00.007
// i.e. 3ms * 2.31decms => 6.93decms
var d = new Date(Date.UTC(2016,0,1,0,0,0,3));
console.log(getDecimalTime(d));
// Test 12:00:00.000 => 10:00:00.000
// i.e. noon to decimal noon
var d = new Date(Date.UTC(2016,0,1,12,0,0,0));
console.log(getDecimalTime(d));
// Test current time
d = new Date();
console.log(formatUTCTime(d));
console.log(getDecimalTime(d));

Increase number twice a week

How would you increase a a number on a webpage twice a week on specific days and times?
For example the webpage would read:
"2 Apples"
However every Tuesday & Thursday at 9:00pm the number should increase by two.
So by Friday the number should have increased to 6 "Apples"
What's a simple way to increment this in Javascript, php or Jquery?
As it was mentionned in the comments, I advise you to use a cron to handle this.
First, you have to store your value somewhere (file, databse, ...). Then, you should create a cron job, that runs a code that increase and update your value at the given days of the week.
Some help about crons : https://en.wikipedia.org/wiki/Cron#Examples
I would add this post about cron and php : Executing a PHP script with a CRON Job
Hope it helps
Here a Javascript version, you can pass it the date it should start counting, the date you want to know the ammount of apples of, the days of the week apples will be added, the hours of the day the updat will take place and the ammount of apples that will be added on each of these dates.
Date.prototype.addDays = function (days) {
var result = new Date(this);
result.setDate(result.getDate() + days);
return result;
}
Date.prototype.addHours = function (hours) {
var result = new Date(this);
result.setHours(result.getHours() + hours);
return result;
}
function getApples(startdate, date, updateDays, updateTime, applesPerUpdate) {
var startDay = startdate.getDay();
var firstUpdateDate;
for(day of updateDays) {
if (day >= startDay) {//assumes startdate has no time added
firstUpdateDate = startdate.addDays(day - startDay).addHours(updateTime);
break;
}
}
if (!firstUpdateDate)
firstUpdateDate = startdate.addDays(7 - (startDay - updateDays[0])).addHours(updateTime);
var updateDaysReverse = updateDays.slice(0).reverse();//clones the array
var dateDay = date.getDay();
var lastUpdateDate;
for(day of updateDaysReverse) {
if (day < dateDay || day == dateDay && date.getHours() > updateTime) {
lastUpdateDate = date.addDays(day - dateDay);
break;
}
}
if (!lastUpdateDate)
lastUpdateDate = date.addDays(updateDaysReverse[0] - (7 + dateDay));
lastUpdateDate = new Date(Date.UTC(1900 + lastUpdateDate.getYear(), lastUpdateDate.getMonth(), lastUpdateDate.getDate(), updateTime, 0, 0, 0));
var secs = Math.trunc((lastUpdateDate - firstUpdateDate));
if (secs < 0) return 0;
var dayDiffs = [];
for(day of updateDays)
dayDiffs.push(day - updateDays[0]);
var weeks = Math.trunc(secs / 604800000);
var days = Math.trunc((secs % 604800000) / 86400000);
var apples = weeks * updateDays.length;
for(diff of dayDiffs)
{
if (diff <= days)
apples++;
else
break;
}
return apples * applesPerUpdate;
}
// important, day and month is zero-based
var startDate = new Date(Date.UTC(2016, 01, 07, 0, 0, 0, 0));
var updateDays = [2, 4];// 0 = sunday , have to be in order and must be 0 <= x < 7
var updateTime = 9;
console.log(getApples(startDate, new Date(), updateDays, updateTime, 2));
Was more coplicated than I thaught, and i havn't testet it much, so there may be bugs.
Here is a plunker to play with the values.

How to exclude weekends between two dates using Moment.js

I am trying to exclude weekends in my JavaScript code. I use moment.js and having difficulty choosing the right variable for 'days'.
So far I have thought that I need to exclude day 6 (saturday) and day 0 (sunday) by changing the weekday variable to count from day 1 to day 5 only. But not sure how it changes.
My jsfiddle is shown here: FIDDLE
HTML:
<div id="myContent">
<input type="radio" value="types" class="syncTypes" name="syncTypes"> <td><label for="xshipping.xshipping1">Free Shipping: (<span id="fsv1" value="5">5</span> to <span id="fsv2" value="10">10</span> working days)</label> </td><br>
<div id="contacts" style="display:none;border:1px #666 solid;padding:3px;top:15px;position:relative;margin-bottom:25px;">
Contacts
</div>
<input type="radio" value="groups" class="syncTypes" name="syncTypes"> <td><label for="xshipping.xshipping2">Express Shipping: (<span id="esv1" value="3">3</span> to <span id="esv2" value="4">4</span> working days)</label> </td>
<div id="groups" style="display:none;border:1px #666 solid;padding:3px;top:15px;position:relative">
Groups
</div>
</div>
JavaScript:
var a = 5; //Free shipping between a
var b = 10;//and b
var c = 3;//Express shipping between c
var d = 4;//and d
var now = moment();
var f = "Your item will be delivered between " + now.add("days",a).format("Do MMMM") + " and " + now.add("days",b).format("Do MMMM");
var g = "Your item will be delivered between " + now.add("days".c).format("Do MMMM") + " and " + now.add("days",d).format("Do MMMM");
var h = document.getElementById('contacts');
h.innerHTML = g
var i = document.getElementById('groups');
i.innerHTML = f
$(function() {
$types = $('.syncTypes');
$contacts = $('#contacts');
$groups = $('#groups');
$types.change(function() {
$this = $(this).val();
if ($this == "types") {
$groups.slideUp(300);
$contacts.delay(200).slideDown(300);
}
else if ($this == "groups") {
$contacts.slideUp(300);
$groups.delay(200).slideDown(300);
}
});
});
Here you go!
function addWeekdays(date, days) {
date = moment(date); // use a clone
while (days > 0) {
date = date.add(1, 'days');
// decrease "days" only if it's a weekday.
if (date.isoWeekday() !== 6 && date.isoWeekday() !== 7) {
days -= 1;
}
}
return date;
}
You call it like this
var date = addWeekdays(moment(), 5);
I used .isoWeekday instead of .weekday because it doesn't depend on the locale (.weekday(0) can be either Monday or Sunday).
Don't subtract weekdays, i.e addWeekdays(moment(), -3) otherwise this simple function will loop forever!
Updated JSFiddle http://jsfiddle.net/Xt2e6/39/ (using different momentjs cdn)
Those iteration looped solutions would not fit my needs.
They were too slow for large numbers.
So I made my own version:
https://github.com/leonardosantos/momentjs-business
Hope you find it useful.
https://github.com/andruhon/moment-weekday-calc plugin for momentJS might be helpful for similar tasks
It does not solves the exact problem, but it is able to calculate specific weekdays in the range.
Usage:
moment().isoWeekdayCalc({
rangeStart: '1 Apr 2015',
rangeEnd: '31 Mar 2016',
weekdays: [1,2,3,4,5], //weekdays Mon to Fri
exclusions: ['6 Apr 2015','7 Apr 2015'] //public holidays
}) //returns 260 (260 workdays excluding two public holidays)
If you want a pure JavaScript version (not relying on Moment.js) try this...
function addWeekdays(date, days) {
date.setDate(date.getDate());
var counter = 0;
if(days > 0 ){
while (counter < days) {
date.setDate(date.getDate() + 1 ); // Add a day to get the date tomorrow
var check = date.getDay(); // turns the date into a number (0 to 6)
if (check == 0 || check == 6) {
// Do nothing it's the weekend (0=Sun & 6=Sat)
}
else{
counter++; // It's a weekday so increase the counter
}
}
}
return date;
}
You call it like this...
var date = addWeekdays(new Date(), 3);
This function checks each next day to see if it falls on a Saturday (day 6) or Sunday (day 0). If true, the counter is not increased yet the date is increased.
This script is fine for small date increments like a month or less.
I would suggest adding a function to the moment prototype.
Something like this maybe? (untested)
nextWeekday : function () {
var day = this.clone(this);
day = day.add('days', 1);
while(day.weekday() == 0 || day.weekday() == 6){
day = day.add("days", 1);
}
return day;
},
nthWeekday : function (n) {
var day = this.clone(this);
for (var i=0;i<n;i++) {
day = day.nextWeekday();
}
return day;
},
And when you're done and written some tests, send in a pull request for bonus points.
d1 and d2 are moment dates passed as an argument to calculateBusinessDays
calculateBusinessDays(d1, d2) {
const days = d2.diff(d1, "days") + 1;
let newDay: any = d1.toDate(),
workingDays: number = 0,
sundays: number = 0,
saturdays: number = 0;
for (let i = 0; i < days; i++) {
const day = newDay.getDay();
newDay = d1.add(1, "days").toDate();
const isWeekend = ((day % 6) === 0);
if (!isWeekend) {
workingDays++;
}
else {
if (day === 6) saturdays++;
if (day === 0) sundays++;
}
}
console.log("Total Days:", days, "workingDays", workingDays, "saturdays", saturdays, "sundays", sundays);
return workingDays;
}
If you want a version of #acorio's code sample which is performant (using #Isantos's optimisation) and can deal with negative numbers use this:
moment.fn.addWorkdays = function (days) {
// Getting negative / positive increment
var increment = days / Math.abs(days);
// Looping weeks for each full 5 workdays
var date = this.clone().add(Math.floor(Math.abs(days) / 5) * 7 * increment, 'days');
// Account for starting on Saturdays and Sundays
if(date.isoWeekday() === 6) { date.add(-increment, 'days'); }
else if(date.isoWeekday() === 7) { date.add(-2 * increment, 'days'); }
// Adding / removing remaining days in a short loop, jumping over weekends
var remaining = days % 5;
while(remaining != 0) {
date.add(increment, 'days');
if(date.isoWeekday() !== 6 && date.isoWeekday() !== 7)
remaining -= increment;
}
return date;
};
See Fiddle here: http://jsfiddle.net/dain/5xrr79h0/
Edit: now fixed issue adding 5 days to a day initially on a weekend
I know this question was posted long ago, but in case somebody bump on this, here is optimized solution using moment.js:
function getBusinessDays(startDate, endDate){
var startDateMoment = moment(startDate);
var endDateMoment = moment(endDate)
var days = Math.round(startDateMoment.diff(endDateMoment, 'days') - startDateMoment .diff(endDateMoment, 'days') / 7 * 2);
if (endDateMoment.day() === 6) {
days--;
}
if (startDateMoment.day() === 7) {
days--;
}
return days;
}
const calcBusinessDays = (d1, d2) => {
// Calc all days used including last day ( the +1 )
const days = d2.diff(d1, 'days') + 1;
console.log('Days:', days);
// how many full weekends occured in this time span
const weekends = Math.floor( days / 7 );
console.log('Full Weekends:', weekends);
// Subtract all the weekend days
let businessDays = days - ( weekends * 2);
// Special case for weeks less than 7
if( weekends === 0 ){
const cur = d1.clone();
for( let i =0; i < days; i++ ){
if( cur.day() === 0 || cur.day() === 6 ){
businessDays--;
}
cur.add(1, 'days')
}
} else {
// If the last day is a saturday we need to account for it
if (d2.day() === 6 ) {
console.log('Extra weekend day (Saturday)');
businessDays--;
}
// If the first day is a sunday we need to account for it
if (d1.day() === 0) {
console.log('Extra weekend day (Sunday)');
businessDays--;
}
}
console.log('Business days:', businessDays);
return businessDays;
}
This can be done without looping between all dates in between.
// get nb of weekend days
var startDateMonday = startDate.clone().startOf('isoWeek');
var endDateMonday = endDate.clone().startOf('isoWeek');
var nbWeekEndDays = 2 * endDateMonday.diff(startDateMonday, 'days') / 7;
var isoDayStart = startDate.isoWeekday();
if (isoDayStart > 5) // starts during the weekend
{
nbWeekEndDays -= (8 - isoDayStart);
}
var isoDayEnd = endDate.isoWeekday();
if (isoDayEnd > 5) // ends during the weekend
{
nbWeekEndDays += (8 - isoDayEnd);
}
// if we want to also exlcude holidays
var startOfStartDate = startDate.clone().startOf('day');
var nbHolidays = holidays.filter(h => {
return h.isSameOrAfter(startOfStartDate) && h.isSameOrBefore(endDate);
}).length;
var duration = moment.duration(endDate.diff(startDate));
duration = duration.subtract({ days: nbWeekEndDays + nbHolidays });
var nbWorkingDays = Math.floor(duration.asDays()); // get only nb of complete days
I am iterating from start date to end date and only counting days which are weekdays.
const calculateBusinessDays = (start_date, end_date) => {
const d1 = start_date.clone();
let num_days = 0;
while(end_date.diff(d1.add(1, 'days')) > 0) {
if ([0, 6].includes(d1.day())) {
// Don't count the days
} else {
num_days++;
}
}
return num_days;
}

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