Is there a way to check if a date is less than 1 hour ago like this?
// old date
var olddate = new Date("February 9, 2012, 12:15");
// current date
var currentdate = new Date();
if (olddate >= currentdate - 1 hour) {
alert("newer than 1 hour");
else {
alert("older than 1 hour");
}
Also, different question - is there a way to add hours to a date like this?
var olddate = new Date("February 9, 2012, 12:15") + 15 HOURS; // output: February 10, 2012, 3:15
Define
var ONE_HOUR = 60 * 60 * 1000; /* ms */
then you can do
((new Date) - myDate) < ONE_HOUR
To get one hour from a date, try
new Date(myDate.getTime() + ONE_HOUR)
Using some ES6 syntax:
const lessThanOneHourAgo = (date) => {
const HOUR = 1000 * 60 * 60;
const anHourAgo = Date.now() - HOUR;
return date > anHourAgo;
}
Using the Moment library:
const lessThanOneHourAgo = (date) => {
return moment(date).isAfter(moment().subtract(1, 'hours'));
}
Shorthand syntax with Moment:
const lessThanOneHourAgo = (date) => moment(date).isAfter(moment().subtract(1, 'hours'));
the moment library can really help express this. The trick is to take the date, add time, and see if it's before or after now:
lastSeenAgoLabel: function() {
var d = this.lastLogin();
if (! moment(d).isValid()) return 'danger'; // danger if not a date.
if (moment(d).add(10, 'minutes').isBefore(/*now*/)) return 'danger'; // danger if older than 10 mins
if (moment(d).add(5, 'minutes').isBefore(/*now*/)) return 'warning'; // warning if older than 5mins
return 'success'; // Looks good!
},
Using moment will be much easier in this case, You could try this:
let hours = moment().diff(moment(yourDateString), 'hours');
It will give you integer value like 1,2,5,0etc so you can easily use condition check like:
if(hours < 1) {
Also, one more thing is you can get more accurate result of the time difference (in decimals like 1.2,1.5,0.7etc) to get this kind of result use this syntax:
let hours = moment().diff(moment(yourDateString), 'hours', true);
Let me know if you have any further query
//for adding hours to a date
Date.prototype.addHours= function(hrs){
this.setHours(this.getHours()+hrs);
return this;
}
Call function like this:
//test alert(new Date().addHours(4));
You can do it as follows:
First find difference of two dates i-e in milliseconds
Convert milliseconds into minutes
If minutes are less than 60, then it means date is within hour else not within hour.
var date = new Date("2020-07-12 11:30:10");
var now = new Date();
var diffInMS = now - date;
var msInHour = Math.floor(diffInMS/1000/60);
if (msInHour < 60) {
console.log('Within hour');
} else {
console.log('Not within the hour');
}
Plain JavaScript solution with in 12 days and 12 days ago option
const timeAgo = ( inputDate ) => {
const date = ( inputDate instanceof Date) ? inputDate : new Date(inputDate);
const FORMATTER = new Intl.RelativeTimeFormat('en');
const RANGES = {
years : 3600 * 24 * 365,
months : 3600 * 24 * 30,
weeks : 3600 * 24 * 7,
days : 3600 * 24,
hours : 3600,
minutes : 60,
seconds : 1
};
const secondsElapsed = (date.getTime() - Date.now()) / 1000;
for (let key in RANGES) {
if ( RANGES[key] < Math.abs(secondsElapsed) ) {
const delta = secondsElapsed / RANGES[key];
return FORMATTER.format(Math.round(delta), key);
}
}
}
// OUTPUTS
console.log( timeAgo('2040-12-24') )
console.log( timeAgo('6 Sept, 2012') );
console.log( timeAgo('2022-05-27T17:45:01+0000') );
let d = new Date()
console.log( "Date will change: ", timeAgo( d.setHours(24,0,0,0) ) );
// d.setDate( d.getDate() - 0 );
d.setHours(-24,0,0,0); // (H,M,S,MS) | 24 hours format
console.log("Day started: " , timeAgo( d ) );
//try this:
// to compare two date's:
<Script Language=Javascript>
function CompareDates()
{
var str1 = document.getElementById("Fromdate").value;
var str2 = document.getElementById("Todate").value;
var dt1 = parseInt(str1.substring(0,2),10);
var mon1 = parseInt(str1.substring(3,5),10);
var yr1 = parseInt(str1.substring(6,10),10);
var dt2 = parseInt(str2.substring(0,2),10);
var mon2 = parseInt(str2.substring(3,5),10);
var yr2 = parseInt(str2.substring(6,10),10);
var date1 = new Date(yr1, mon1, dt1);
var date2 = new Date(yr2, mon2, dt2);
if(date2 < date1)
{
alert("To date cannot be greater than from date");
return false;
}
else
{
alert("Submitting ...");
}
}
</Script>
Hope it will work 4 u...
Related
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).
This question already has answers here:
How to calculate number of days between two dates?
(42 answers)
Closed 3 months ago.
I want to calculate date difference in days, hours, minutes, seconds, milliseconds, nanoseconds. How can I do it?
Assuming you have two Date objects, you can just subtract them to get the difference in milliseconds:
var difference = date2 - date1;
From there, you can use simple arithmetic to derive the other values.
var DateDiff = {
inDays: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return Math.floor((t2-t1)/(24*3600*1000));
},
inWeeks: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000*7));
},
inMonths: function(d1, d2) {
var d1Y = d1.getFullYear();
var d2Y = d2.getFullYear();
var d1M = d1.getMonth();
var d2M = d2.getMonth();
return (d2M+12*d2Y)-(d1M+12*d1Y);
},
inYears: function(d1, d2) {
return d2.getFullYear()-d1.getFullYear();
}
}
var dString = "May, 20, 1984";
var d1 = new Date(dString);
var d2 = new Date();
document.write("<br />Number of <b>days</b> since "+dString+": "+DateDiff.inDays(d1, d2));
document.write("<br />Number of <b>weeks</b> since "+dString+": "+DateDiff.inWeeks(d1, d2));
document.write("<br />Number of <b>months</b> since "+dString+": "+DateDiff.inMonths(d1, d2));
document.write("<br />Number of <b>years</b> since "+dString+": "+DateDiff.inYears(d1, d2));
Code sample taken from here.
Another solution is convert difference to a new Date object and get that date's year(diff from 1970), month, day etc.
var date1 = new Date(2010, 6, 17);
var date2 = new Date(2013, 12, 18);
var diff = new Date(date2.getTime() - date1.getTime());
// diff is: Thu Jul 05 1973 04:00:00 GMT+0300 (EEST)
console.log(diff.getUTCFullYear() - 1970); // Gives difference as year
// 3
console.log(diff.getUTCMonth()); // Gives month count of difference
// 6
console.log(diff.getUTCDate() - 1); // Gives day count of difference
// 4
So difference is like "3 years and 6 months and 4 days". If you want to take difference in a human readable style, that can help you.
Expressions like "difference in days" are never as simple as they seem. If you have the following dates:
d1: 2011-10-15 23:59:00
d1: 2011-10-16 00:01:00
the difference in time is 2 minutes, should the "difference in days" be 1 or 0? Similar issues arise for any expression of the difference in months, years or whatever since years, months and days are of different lengths and different times (e.g. the day that daylight saving starts is 1 hour shorter than usual and two hours shorter than the day that it ends).
Here is a function for a difference in days that ignores the time, i.e. for the above dates it returns 1.
/*
Get the number of days between two dates - not inclusive.
"between" does not include the start date, so days
between Thursday and Friday is one, Thursday to Saturday
is two, and so on. Between Friday and the following Friday is 7.
e.g. getDaysBetweenDates( 22-Jul-2011, 29-jul-2011) => 7.
If want inclusive dates (e.g. leave from 1/1/2011 to 30/1/2011),
use date prior to start date (i.e. 31/12/2010 to 30/1/2011).
Only calculates whole days.
Assumes d0 <= d1
*/
function getDaysBetweenDates(d0, d1) {
var msPerDay = 8.64e7;
// Copy dates so don't mess them up
var x0 = new Date(d0);
var x1 = new Date(d1);
// Set to noon - avoid DST errors
x0.setHours(12,0,0);
x1.setHours(12,0,0);
// Round to remove daylight saving errors
return Math.round( (x1 - x0) / msPerDay );
}
This can be more concise:
/* Return number of days between d0 and d1.
** Returns positive if d0 < d1, otherwise negative.
**
** e.g. between 2000-02-28 and 2001-02-28 there are 366 days
** between 2015-12-28 and 2015-12-29 there is 1 day
** between 2015-12-28 23:59:59 and 2015-12-29 00:00:01 there is 1 day
** between 2015-12-28 00:00:01 and 2015-12-28 23:59:59 there are 0 days
**
** #param {Date} d0 - start date
** #param {Date} d1 - end date
** #returns {number} - whole number of days between d0 and d1
**
*/
function daysDifference(d0, d1) {
var diff = new Date(+d1).setHours(12) - new Date(+d0).setHours(12);
return Math.round(diff/8.64e7);
}
// Simple formatter
function formatDate(date){
return [date.getFullYear(),('0'+(date.getMonth()+1)).slice(-2),('0'+date.getDate()).slice(-2)].join('-');
}
// Examples
[[new Date(2000,1,28), new Date(2001,1,28)], // Leap year
[new Date(2001,1,28), new Date(2002,1,28)], // Not leap year
[new Date(2017,0,1), new Date(2017,1,1)]
].forEach(function(dates) {
document.write('From ' + formatDate(dates[0]) + ' to ' + formatDate(dates[1]) +
' is ' + daysDifference(dates[0],dates[1]) + ' days<br>');
});
<html lang="en">
<head>
<script>
function getDateDiff(time1, time2) {
var str1= time1.split('/');
var str2= time2.split('/');
// yyyy , mm , dd
var t1 = new Date(str1[2], str1[0]-1, str1[1]);
var t2 = new Date(str2[2], str2[0]-1, str2[1]);
var diffMS = t1 - t2;
console.log(diffMS + ' ms');
var diffS = diffMS / 1000;
console.log(diffS + ' ');
var diffM = diffS / 60;
console.log(diffM + ' minutes');
var diffH = diffM / 60;
console.log(diffH + ' hours');
var diffD = diffH / 24;
console.log(diffD + ' days');
alert(diffD);
}
//alert(getDateDiff('10/18/2013','10/14/2013'));
</script>
</head>
<body>
<input type="button"
onclick="getDateDiff('10/18/2013','10/14/2013')"
value="clickHere()" />
</body>
</html>
use Moment.js for all your JavaScript related date-time calculation
Answer to your question is:
var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b) // 86400000
Complete details can be found here
adding to #paresh mayani 's answer, to work like Facebook - showing how much time has passed in sec/min/hours/weeks/months/years
var DateDiff = {
inSeconds: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/1000);
},
inMinutes: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/60000);
},
inHours: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/3600000);
},
inDays: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000));
},
inWeeks: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000*7));
},
inMonths: function(d1, d2) {
var d1Y = d1.getFullYear();
var d2Y = d2.getFullYear();
var d1M = d1.getMonth();
var d2M = d2.getMonth();
return (d2M+12*d2Y)-(d1M+12*d1Y);
},
inYears: function(d1, d2) {
return d2.getFullYear()-d1.getFullYear();
}
}
var dString = "May, 20, 1984"; //will also get (Y-m-d H:i:s)
var d1 = new Date(dString);
var d2 = new Date();
var timeLaps = DateDiff.inSeconds(d1, d2);
var dateOutput = "";
if (timeLaps<60)
{
dateOutput = timeLaps+" seconds";
}
else
{
timeLaps = DateDiff.inMinutes(d1, d2);
if (timeLaps<60)
{
dateOutput = timeLaps+" minutes";
}
else
{
timeLaps = DateDiff.inHours(d1, d2);
if (timeLaps<24)
{
dateOutput = timeLaps+" hours";
}
else
{
timeLaps = DateDiff.inDays(d1, d2);
if (timeLaps<7)
{
dateOutput = timeLaps+" days";
}
else
{
timeLaps = DateDiff.inWeeks(d1, d2);
if (timeLaps<4)
{
dateOutput = timeLaps+" weeks";
}
else
{
timeLaps = DateDiff.inMonths(d1, d2);
if (timeLaps<12)
{
dateOutput = timeLaps+" months";
}
else
{
timeLaps = DateDiff.inYears(d1, d2);
dateOutput = timeLaps+" years";
}
}
}
}
}
}
alert (dateOutput);
With momentjs it's simple:
moment("2016-04-08").fromNow();
function DateDiff(date1, date2) {
date1.setHours(0);
date1.setMinutes(0, 0, 0);
date2.setHours(0);
date2.setMinutes(0, 0, 0);
var datediff = Math.abs(date1.getTime() - date2.getTime()); // difference
return parseInt(datediff / (24 * 60 * 60 * 1000), 10); //Convert values days and return value
}
var d1=new Date(2011,0,1); // jan,1 2011
var d2=new Date(); // now
var diff=d2-d1,sign=diff<0?-1:1,milliseconds,seconds,minutes,hours,days;
diff/=sign; // or diff=Math.abs(diff);
diff=(diff-(milliseconds=diff%1000))/1000;
diff=(diff-(seconds=diff%60))/60;
diff=(diff-(minutes=diff%60))/60;
days=(diff-(hours=diff%24))/24;
console.info(sign===1?"Elapsed: ":"Remains: ",
days+" days, ",
hours+" hours, ",
minutes+" minutes, ",
seconds+" seconds, ",
milliseconds+" milliseconds.");
I think this should do it.
let today = new Date();
let form_date=new Date('2019-10-23')
let difference=form_date>today ? form_date-today : today-form_date
let diff_days=Math.floor(difference/(1000*3600*24))
based on javascript runtime prototype implementation you can use simple arithmetic to subtract dates as in bellow
var sep = new Date(2020, 07, 31, 23, 59, 59);
var today = new Date();
var diffD = Math.floor((sep - today) / (1000 * 60 * 60 * 24));
console.log('Day Diff: '+diffD);
the difference return answer as milliseconds, then you have to convert it by division:
by 1000 to convert to second
by 1000×60 convert to minute
by 1000×60×60 convert to hour
by 1000×60×60×24 convert to day
function DateDiff(b, e)
{
let
endYear = e.getFullYear(),
endMonth = e.getMonth(),
years = endYear - b.getFullYear(),
months = endMonth - b.getMonth(),
days = e.getDate() - b.getDate();
if (months < 0)
{
years--;
months += 12;
}
if (days < 0)
{
months--;
days += new Date(endYear, endMonth, 0).getDate();
}
return [years, months, days];
}
[years, months, days] = DateDiff(
new Date("October 21, 1980"),
new Date("July 11, 2017")); // 36 8 20
Sorry but flat millisecond calculation is not reliable
Thanks for all the responses, but few of the functions I tried are failing either on
1. A date near today's date
2. A date in 1970 or
3. A date in a leap year.
Approach that best worked for me and covers all scenario e.g. leap year, near date in 1970, feb 29 etc.
var someday = new Date("8/1/1985");
var today = new Date();
var years = today.getFullYear() - someday.getFullYear();
// Reset someday to the current year.
someday.setFullYear(today.getFullYear());
// Depending on when that day falls for this year, subtract 1.
if (today < someday)
{
years--;
}
document.write("Its been " + years + " full years.");
This code will return the difference between two dates in days:
const previous_date = new Date("2019-12-23");
const current_date = new Date();
const current_year = current_date.getFullYear();
const previous_date_year =
previous_date.getFullYear();
const difference_in_years = current_year -
previous_date_year;
let months = current_date.getMonth();
months = months + 1; // for making the indexing
// of months from 1
for(let i = 0; i < difference_in_years; i++){
months = months + 12;
}
let days = current_date.getDate();
days = days + (months * 30.417);
console.log(`The days between ${current_date} and
${previous_date} are : ${days} (approximately)`);
If you are using moment.js then it is pretty simple to find date difference.
var now = "04/09/2013 15:00:00";
var then = "04/09/2013 14:20:30";
moment.utc(moment(now,"DD/MM/YYYY HH:mm:ss").diff(moment(then,"DD/MM/YYYY HH:mm:ss"))).format("HH:mm:ss")
This is how you can implement difference between dates without a framework.
function getDateDiff(dateOne, dateTwo) {
if(dateOne.charAt(2)=='-' & dateTwo.charAt(2)=='-'){
dateOne = new Date(formatDate(dateOne));
dateTwo = new Date(formatDate(dateTwo));
}
else{
dateOne = new Date(dateOne);
dateTwo = new Date(dateTwo);
}
let timeDiff = Math.abs(dateOne.getTime() - dateTwo.getTime());
let diffDays = Math.ceil(timeDiff / (1000 * 3600 * 24));
let diffMonths = Math.ceil(diffDays/31);
let diffYears = Math.ceil(diffMonths/12);
let message = "Difference in Days: " + diffDays + " " +
"Difference in Months: " + diffMonths+ " " +
"Difference in Years: " + diffYears;
return message;
}
function formatDate(date) {
return date.split('-').reverse().join('-');
}
console.log(getDateDiff("23-04-2017", "23-04-2018"));
function daysInMonth (month, year) {
return new Date(year, month, 0).getDate();
}
function getduration(){
let A= document.getElementById("date1_id").value
let B= document.getElementById("date2_id").value
let C=Number(A.substring(3,5))
let D=Number(B.substring(3,5))
let dif=D-C
let arr=[];
let sum=0;
for (let i=0;i<dif+1;i++){
sum+=Number(daysInMonth(i+C,2019))
}
let sum_alter=0;
for (let i=0;i<dif;i++){
sum_alter+=Number(daysInMonth(i+C,2019))
}
let no_of_month=(Number(B.substring(3,5)) - Number(A.substring(3,5)))
let days=[];
if ((Number(B.substring(3,5)) - Number(A.substring(3,5)))>0||Number(B.substring(0,2)) - Number(A.substring(0,2))<0){
days=Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter
}
if ((Number(B.substring(3,5)) == Number(A.substring(3,5)))){
console.log(Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter)
}
time_1=[]; time_2=[]; let hour=[];
time_1=document.getElementById("time1_id").value
time_2=document.getElementById("time2_id").value
if (time_1.substring(0,2)=="12"){
time_1="00:00:00 PM"
}
if (time_1.substring(9,11)==time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))
}
if (time_1.substring(9,11)!=time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))+12
}
let min=Math.abs(Number(time_1.substring(3,5))-Number(time_2.substring(3,5)))
document.getElementById("duration_id").value=days +" days "+ hour+" hour " + min+" min "
}
<input type="text" id="date1_id" placeholder="28/05/2019">
<input type="text" id="date2_id" placeholder="29/06/2019">
<br><br>
<input type="text" id="time1_id" placeholder="08:01:00 AM">
<input type="text" id="time2_id" placeholder="00:00:00 PM">
<br><br>
<button class="text" onClick="getduration()">Submit </button>
<br><br>
<input type="text" id="duration_id" placeholder="days hour min">
var date1 = new Date("06/30/2019");
var date2 = new Date("07/30/2019");
// To calculate the time difference of two dates
var Difference_In_Time = date2.getTime() - date1.getTime();
// To calculate the no. of days between two dates
var Difference_In_Days = Difference_In_Time / (1000 * 3600 * 24);
//To display the final no. of days (result)
document.write("Total number of days between dates <br>"
+ date1 + "<br> and <br>"
+ date2 + " is: <br> "
+ Difference_In_Days);
this should work just fine if you just need to show what time left, since JavaScript uses frames for its time you'll have get your End Time - The Time RN after that we can divide it by 1000 since apparently 1000 frames = 1 seconds, after that you can use the basic math of time, but there's still a problem to this code, since the calculation is static, it can't compensate for the different day total in a year (360/365/366), the bunch of IF after the calculation is to make it null if the time is lower than 0, hope this helps even though it's not exactly what you're asking :)
var now = new Date();
var end = new Date("End Time");
var total = (end - now) ;
var totalD = Math.abs(Math.floor(total/1000));
var years = Math.floor(totalD / (365*60*60*24));
var months = Math.floor((totalD - years*365*60*60*24) / (30*60*60*24));
var days = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24)/ (60*60*24));
var hours = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24)/ (60*60));
var minutes = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60)/ (60));
var seconds = Math.floor(totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60 - minutes*60);
var Y = years < 1 ? "" : years + " Years ";
var M = months < 1 ? "" : months + " Months ";
var D = days < 1 ? "" : days + " Days ";
var H = hours < 1 ? "" : hours + " Hours ";
var I = minutes < 1 ? "" : minutes + " Minutes ";
var S = seconds < 1 ? "" : seconds + " Seconds ";
var A = years == 0 && months == 0 && days == 0 && hours == 0 && minutes == 0 && seconds == 0 ? "Sending" : " Remaining";
document.getElementById('txt').innerHTML = Y + M + D + H + I + S + A;
Ok, there are a bunch of ways you can do that.
Yes, you can use plain old JS. Just try:
let dt1 = new Date()
let dt2 = new Date()
Let's emulate passage using Date.prototype.setMinutes and make sure we are in range.
dt1.setMinutes(7)
dt2.setMinutes(42)
console.log('Elapsed seconds:',(dt2-dt1)/1000)
Alternatively you could use some library like js-joda, where you can easily do things like this (directly from docs):
var dt1 = LocalDateTime.parse("2016-02-26T23:55:42.123");
var dt2 = dt1
.plusYears(6)
.plusMonths(12)
.plusHours(2)
.plusMinutes(42)
.plusSeconds(12);
// obtain the duration between the two dates
dt1.until(dt2, ChronoUnit.YEARS); // 7
dt1.until(dt2, ChronoUnit.MONTHS); // 84
dt1.until(dt2, ChronoUnit.WEEKS); // 356
dt1.until(dt2, ChronoUnit.DAYS); // 2557
dt1.until(dt2, ChronoUnit.HOURS); // 61370
dt1.until(dt2, ChronoUnit.MINUTES); // 3682242
dt1.until(dt2, ChronoUnit.SECONDS); // 220934532
There are plenty more libraries ofc, but js-joda has an added bonus of being available also in Java, where it has been extensively tested. All those tests have been migrated to js-joda, it's also immutable.
I made a below function to get the difference between now and "2021-02-26T21:50:42.123".
The difference return answer as milliseconds, so I convert it by using this formula:
(1000 * 3600 * 24).
function getDiff(dateAcquired) {
let calDiff = Math.floor(
(new Date() - new Date(dateAcquired)) / (1000 * 3600 * 24)
);
return calDiff;
}
console.log(getDiff("2021-02-26T21:50:42.123"));
Can be useful :
const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/24 * 60 * 60 * 1000)
or
const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/86400000)
where 24 * 60 * 60 * 1000 is (day * minutes * seconds * milliseconds) = 86400000 milliseconds in one day
Thank you
// the idea is to get time left for new year.
// Not considering milliseconds as of now, but that
// can be done
var newYear = '1 Jan 2023';
const secondsInAMin = 60;
const secondsInAnHour = 60 * secondsInAMin;
const secondsInADay = 24 * secondsInAnHour;
function DateDiffJs() {
var newYearDate = new Date(newYear);
var currDate = new Date();
var remainingSecondsInDateDiff = (newYearDate - currDate) / 1000;
var days = Math.floor(remainingSecondsInDateDiff / secondsInADay);
var remainingSecondsAfterDays = remainingSecondsInDateDiff - (days * secondsInADay);
var hours = Math.floor(remainingSecondsAfterDays / secondsInAnHour);
var remainingSecondsAfterhours = remainingSecondsAfterDays - (hours * secondsInAnHour);
var mins = Math.floor(remainingSecondsAfterhours / secondsInAMin);
var seconds = Math.floor(remainingSecondsAfterhours - (mins * secondsInAMin));
console.log(`days :: ${days}`)
console.log(`hours :: ${hours}`)
console.log(`mins :: ${mins}`)
console.log(`seconds :: ${seconds}`)
}
DateDiffJs();
I have a datetimepicker where the user picks up a date, and my requirement is I need 7 days difference between his selected date.
For eg,
if user has selected 2017-03-01 so i need last 7 days from 2017-03-01 and NOT the current date
All answers i checked here were based on days difference from today.
Can anyone help me out here ?
$("#dateTimePickerIdWhereUserSelectsHisDate").val() - (7 * 24 * 60 * 60 * 1000);
this was on one of the answers but didn't work.
How can I achieve this ?
Try This
SelectDateTime will give you selected date
604800000 is 7 days in miliseconds
prevDate will give you last 7 days Date
$("#startDate").on("dp.change", function(e) {
if (e.oldDate != null) {
if (e.date.format('D') != e.oldDate.format('D')) {
var selectDateTime = e.date["_d"].getTime();
var prevDateTImeMili = selectDateTime - 604800000;
var prevDate = msToDateTime(prevDateTImeMili)
$('#startDate').data("DateTimePicker").hide();
}
}
});
msToDateTime is a function which converts milliseconds to DateTime
function msToDateTime(s) {
Number.prototype.padLeft = function(base,chr){
var len = (String(base || 10).length - String(this).length)+1;
return len > 0? new Array(len).join(chr || '0')+this : this;
}
if(s != null){
s = new Date(s);
// var d = new Date(s);
// var d = new Date(s.getTime()+s.getTimezoneOffset()*60*1000+timeConversionToMilliseconds(sessionStorage.getItem("accounttimezone").split('+')[1]+':00'))
var d = new Date(s.getTime()+(s.getTimezoneOffset()*60*1000)+ (330 *60*1000));
dformat = [ d.getFullYear(),
(d.getMonth()+1).padLeft(),
d.getDate().padLeft()].join('-')+
' ' +
[ d.getHours().padLeft(),
d.getMinutes().padLeft(),
d.getSeconds().padLeft()].join(':');
return dformat;
}else{
return " ";
}
}
function getNDaysBefore(dateString, numberOfDaysBefore) {
let startingDate = new Date(dateString).getTime();
let datesArray = [],
daysCounter = 0,
day = 1000 * 60 * 60 * 24;
while (daysCounter < numberOfDaysBefore + 1) {
let newDateBeforeStaring = startingDate - day * daysCounter;
datesArray.push(new Date(newDateBeforeStaring));
daysCounter++;
}
return datesArray;
}
var dateString = "2016-03-01";
alert(getNDaysBefore(dateString,7));
With that kind of a function you can get any N days before the given date as an array of Date objects
I'm passing my calendar selected date of birth to following JS function for calculating Age:
var DOBmdy = date.split("-");
Bdate = new Date(DOBmdy[2],DOBmdy[0]-1,DOBmdy[1]);
BDateArr = (''+Bdate).split(' ');
//document.getElementById('DOW').value = BDateArr[0];
Cdate = new Date;
CDateArr = (''+Cdate).split(" ");
Age = CDateArr[3] - BDateArr[3];
Now, lets say, input age is: 2nd Aug 1983 and age count comes: 28, while as August month has not been passed yet, i want to show the current age of 27 and not 28
Any idea, how can i write that logic, to count age 27 perfectly with my JS function.
Thanks !
Let birth date be august 2nd 1983, then the difference in milliseconds between now an that date is:
var diff = new Date - new Date('1983-08-02');
The difference in days is (1 second = 1000 ms, 1 hour = 60*60 seconds, 1 day = 24 * 1 hour)
var diffdays = diff / 1000 / (60 * 60 * 24);
The difference in years (so, the age) becomes (.25 to account for leapyears):
var age = Math.floor(diffdays / 365.25);
Now try it with
diff = new Date('2011-08-01') - new Date('1983-08-02'); //=> 27
diff = new Date('2011-08-02') - new Date('1983-08-02'); //=> 28
diff = new Date('2012-08-02') - new Date('1983-08-02'); //=> 29
So, your javascript could be rewritten as:
var Bdate = new Date(date.split("-").reverse().join('-')),
age = Math.floor( ( (Cdate - Bdate) / 1000 / (60 * 60 * 24) ) / 365.25 );
[edit] Didn't pay enough attention. date.split('-') gives the array [dd,mm,yyyy], so reversing it results in[yyyy,mm,dd]. Now joining that again using '-', the result is the string 'yyyy-mm-dd', which is valid input for a new Date.
(new Date() - new Date('08-02-1983')) / 1000 / 60 / 60 / 24 / 365.25
That will get you the difference in years, you will occasionally run into off-by-one-day issues using this.
May be this works:
var today = new Date();
var d = document.getElementById("dob").value;
if (!/\d{4}\-\d{2}\-\d{2}/.test(d)) { // check valid format
return false;
}
d = d.split("-");
var byr = parseInt(d[0]);
var nowyear = today.getFullYear();
if (byr >= nowyear || byr < 1900) { // check valid year
return false;
}
var bmth = parseInt(d[1],10)-1;
if (bmth<0 || bmth>11) { // check valid month 0-11
return false;
}
var bdy = parseInt(d[2],10);
if (bdy<1 || bdy>31) { // check valid date according to month
return false;
}
var age = nowyear - byr;
var nowmonth = today.getMonth();
var nowday = today.getDate();
if (bmth > nowmonth) {age = age - 1} // next birthday not yet reached
else if (bmth == nowmonth && nowday < bdy) {age = age - 1}
alert('You are ' + age + ' years old');
I just had to write a function to do this and thought'd I'd share.
This is accurate from a human point of view! None of that crazy 365.2425 stuff.
var ageCheck = function(yy, mm, dd) {
// validate input
yy = parseInt(yy,10);
mm = parseInt(mm,10);
dd = parseInt(dd,10);
if(isNaN(dd) || isNaN(mm) || isNaN(yy)) { return 0; }
if((dd < 1 || dd > 31) || (mm < 1 || mm > 12)) { return 0; }
// change human inputted month to javascript equivalent
mm = mm - 1;
// get today's date
var today = new Date();
var t_dd = today.getDate();
var t_mm = today.getMonth();
var t_yy = today.getFullYear();
// We are using last two digits, so make a guess of the century
if(yy == 0) { yy = "00"; }
else if(yy < 9) { yy = "0"+yy; }
yy = (today.getFullYear() < "20"+yy ? "19"+yy : "20"+yy);
// Work out the age!
var age = t_yy - yy - 1; // Starting point
if( mm < t_mm ) { age++;} // If it's past their birth month
if( mm == t_mm && dd <= t_dd) { age++; } // If it's past their birth day
return age;
}
How to find the difference between two dates?
By using the Date object and its milliseconds value, differences can be calculated:
var a = new Date(); // Current date now.
var b = new Date(2010, 0, 1, 0, 0, 0, 0); // Start of 2010.
var d = (b-a); // Difference in milliseconds.
You can get the number of seconds (as a integer/whole number) by dividing the milliseconds by 1000 to convert it to seconds then converting the result to an integer (this removes the fractional part representing the milliseconds):
var seconds = parseInt((b-a)/1000);
You could then get whole minutes by dividing seconds by 60 and converting it to an integer, then hours by dividing minutes by 60 and converting it to an integer, then longer time units in the same way. From this, a function to get the maximum whole amount of a time unit in the value of a lower unit and the remainder lower unit can be created:
function get_whole_values(base_value, time_fractions) {
time_data = [base_value];
for (i = 0; i < time_fractions.length; i++) {
time_data.push(parseInt(time_data[i]/time_fractions[i]));
time_data[i] = time_data[i] % time_fractions[i];
}; return time_data;
};
// Input parameters below: base value of 72000 milliseconds, time fractions are
// 1000 (amount of milliseconds in a second) and 60 (amount of seconds in a minute).
console.log(get_whole_values(72000, [1000, 60]));
// -> [0,12,1] # 0 whole milliseconds, 12 whole seconds, 1 whole minute.
If you're wondering what the input parameters provided above for the second Date object are, see their names below:
new Date(<year>, <month>, <day>, <hours>, <minutes>, <seconds>, <milliseconds>);
As noted in the comments of this solution, you don't necessarily need to provide all these values unless they're necessary for the date you wish to represent.
I have found this and it works fine for me:
Calculating the Difference between Two Known Dates
Unfortunately, calculating a date interval such as days, weeks, or months between two known dates is not as easy because you can't just add Date objects together. In order to use a Date object in any sort of calculation, we must first retrieve the Date's internal millisecond value, which is stored as a large integer. The function to do that is Date.getTime(). Once both Dates have been converted, subtracting the later one from the earlier one returns the difference in milliseconds. The desired interval can then be determined by dividing that number by the corresponding number of milliseconds. For instance, to obtain the number of days for a given number of milliseconds, we would divide by 86,400,000, the number of milliseconds in a day (1000 x 60 seconds x 60 minutes x 24 hours):
Date.daysBetween = function( date1, date2 ) {
//Get 1 day in milliseconds
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 = date2_ms - date1_ms;
// Convert back to days and return
return Math.round(difference_ms/one_day);
}
//Set the two dates
var y2k = new Date(2000, 0, 1);
var Jan1st2010 = new Date(y2k.getFullYear() + 10, y2k.getMonth(), y2k.getDate());
var today= new Date();
//displays 726
console.log( 'Days since '
+ Jan1st2010.toLocaleDateString() + ': '
+ Date.daysBetween(Jan1st2010, today));
The rounding is optional, depending on whether you want partial days or not.
Reference
If you are looking for a difference expressed as a combination of years, months, and days, I would suggest this function:
function interval(date1, date2) {
if (date1 > date2) { // swap
var result = interval(date2, date1);
result.years = -result.years;
result.months = -result.months;
result.days = -result.days;
result.hours = -result.hours;
return result;
}
result = {
years: date2.getYear() - date1.getYear(),
months: date2.getMonth() - date1.getMonth(),
days: date2.getDate() - date1.getDate(),
hours: date2.getHours() - date1.getHours()
};
if (result.hours < 0) {
result.days--;
result.hours += 24;
}
if (result.days < 0) {
result.months--;
// days = days left in date1's month,
// plus days that have passed in date2's month
var copy1 = new Date(date1.getTime());
copy1.setDate(32);
result.days = 32-date1.getDate()-copy1.getDate()+date2.getDate();
}
if (result.months < 0) {
result.years--;
result.months+=12;
}
return result;
}
// Be aware that the month argument is zero-based (January = 0)
var date1 = new Date(2015, 4-1, 6);
var date2 = new Date(2015, 5-1, 9);
document.write(JSON.stringify(interval(date1, date2)));
This solution will treat leap years (29 February) and month length differences in a way we would naturally do (I think).
So for example, the interval between 28 February 2015 and 28 March 2015 will be considered exactly one month, not 28 days. If both those days are in 2016, the difference will still be exactly one month, not 29 days.
Dates with exactly the same month and day, but different year, will always have a difference of an exact number of years. So the difference between 2015-03-01 and 2016-03-01 will be exactly 1 year, not 1 year and 1 day (because of counting 365 days as 1 year).
// This is for first date
first = new Date(2010, 03, 08, 15, 30, 10); // Get the first date epoch object
document.write((first.getTime())/1000); // get the actual epoch values
second = new Date(2012, 03, 08, 15, 30, 10); // Get the second date epoch object
document.write((second.getTime())/1000); // get the actual epoch values
diff= second - first ;
one_day_epoch = 24*60*60 ; // calculating one epoch
if ( diff/ one_day_epoch > 365 ) // check if it is exceeding regular calendar year
{
alert( 'date is exceeding one year');
}
This answer, based on another one (link at end), is about the difference between two dates.
You can see how it works because it's simple, also it includes splitting the difference into
units of time (a function that I made) and converting to UTC to stop time zone problems.
function date_units_diff(a, b, unit_amounts) {
var split_to_whole_units = function (milliseconds, unit_amounts) {
// unit_amounts = list/array of amounts of milliseconds in a
// second, seconds in a minute, etc., for example "[1000, 60]".
time_data = [milliseconds];
for (i = 0; i < unit_amounts.length; i++) {
time_data.push(parseInt(time_data[i] / unit_amounts[i]));
time_data[i] = time_data[i] % unit_amounts[i];
}; return time_data.reverse();
}; if (unit_amounts == undefined) {
unit_amounts = [1000, 60, 60, 24];
};
var utc_a = new Date(a.toUTCString());
var utc_b = new Date(b.toUTCString());
var diff = (utc_b - utc_a);
return split_to_whole_units(diff, unit_amounts);
}
// Example of use:
var d = date_units_diff(new Date(2010, 0, 1, 0, 0, 0, 0), new Date()).slice(0,-2);
document.write("In difference: 0 days, 1 hours, 2 minutes.".replace(
/0|1|2/g, function (x) {return String( d[Number(x)] );} ));
How my code above works
A date/time difference, as milliseconds, can be calculated using the Date object:
var a = new Date(); // Current date now.
var b = new Date(2010, 0, 1, 0, 0, 0, 0); // Start of 2010.
var utc_a = new Date(a.toUTCString());
var utc_b = new Date(b.toUTCString());
var diff = (utc_b - utc_a); // The difference as milliseconds.
Then to work out the number of seconds in that difference, divide it by 1000 to convert
milliseconds to seconds, then change the result to an integer (whole number) to remove
the milliseconds (fraction part of that decimal): var seconds = parseInt(diff/1000).
Also, I could get longer units of time using the same process, for example:
- (whole) minutes, dividing seconds by 60 and changing the result to an integer,
- hours, dividing minutes by 60 and changing the result to an integer.
I created a function for doing that process of splitting the difference into
whole units of time, named split_to_whole_units, with this demo:
console.log(split_to_whole_units(72000, [1000, 60]));
// -> [1,12,0] # 1 (whole) minute, 12 seconds, 0 milliseconds.
This answer is based on this other one.
You can also use it
export function diffDateAndToString(small: Date, big: Date) {
// To calculate the time difference of two dates
const Difference_In_Time = big.getTime() - small.getTime()
// To calculate the no. of days between two dates
const Days = Difference_In_Time / (1000 * 3600 * 24)
const Mins = Difference_In_Time / (60 * 1000)
const Hours = Mins / 60
const diffDate = new Date(Difference_In_Time)
console.log({ date: small, now: big, diffDate, Difference_In_Days: Days, Difference_In_Mins: Mins, Difference_In_Hours: Hours })
var result = ''
if (Mins < 60) {
result = Mins + 'm'
} else if (Hours < 24) result = diffDate.getMinutes() + 'h'
else result = Days + 'd'
return { result, Days, Mins, Hours }
}
results in { result: '30d', Days: 30, Mins: 43200, Hours: 720 }
Date.prototype.addDays = function(days) {
var dat = new Date(this.valueOf())
dat.setDate(dat.getDate() + days);
return dat;
}
function getDates(startDate, stopDate) {
var dateArray = new Array();
var currentDate = startDate;
while (currentDate <= stopDate) {
dateArray.push(currentDate);
currentDate = currentDate.addDays(1);
}
return dateArray;
}
var dateArray = getDates(new Date(), (new Date().addDays(7)));
for (i = 0; i < dateArray.length; i ++ ) {
// alert (dateArray[i]);
date=('0'+dateArray[i].getDate()).slice(-2);
month=('0' +(dateArray[i].getMonth()+1)).slice(-2);
year=dateArray[i].getFullYear();
alert(date+"-"+month+"-"+year );
}
var DateDiff = function(type, start, end) {
let // or var
years = end.getFullYear() - start.getFullYear(),
monthsStart = start.getMonth(),
monthsEnd = end.getMonth()
;
var returns = -1;
switch(type){
case 'm': case 'mm': case 'month': case 'months':
returns = ( ( ( years * 12 ) - ( 12 - monthsEnd ) ) + ( 12 - monthsStart ) );
break;
case 'y': case 'yy': case 'year': case 'years':
returns = years;
break;
case 'd': case 'dd': case 'day': case 'days':
returns = ( ( end - start ) / ( 1000 * 60 * 60 * 24 ) );
break;
}
return returns;
}
Usage
var qtMonths = DateDiff('mm', new Date('2015-05-05'), new Date());
var qtYears = DateDiff('yy', new Date('2015-05-05'), new Date());
var qtDays = DateDiff('dd', new Date('2015-05-05'), new Date());
OR
var qtMonths = DateDiff('m', new Date('2015-05-05'), new Date()); // m || y || d
var qtMonths = DateDiff('month', new Date('2015-05-05'), new Date()); // month || year || day
var qtMonths = DateDiff('months', new Date('2015-05-05'), new Date()); // months || years || days
...
var DateDiff = function (type, start, end) {
let // or var
years = end.getFullYear() - start.getFullYear(),
monthsStart = start.getMonth(),
monthsEnd = end.getMonth()
;
if(['m', 'mm', 'month', 'months'].includes(type)/*ES6*/)
return ( ( ( years * 12 ) - ( 12 - monthsEnd ) ) + ( 12 - monthsStart ) );
else if(['y', 'yy', 'year', 'years'].includes(type))
return years;
else if (['d', 'dd', 'day', 'days'].indexOf(type) !== -1/*EARLIER JAVASCRIPT VERSIONS*/)
return ( ( end - start ) / ( 1000 * 60 * 60 * 24 ) );
else
return -1;
}