How to calculate date difference in JavaScript?

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I want to calculate date difference in days, hours, minutes, seconds, milliseconds, nanoseconds. How can I do it?

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18条回答

灰色年华

2020-11-22 03:58

<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>

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迷失自我

2020-11-22 04:05

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.

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梦谈多话

2020-11-22 04:05

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">

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野性不改

2020-11-22 04:05

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;

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隐瞒了意图╮

2020-11-22 04:06
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.");
```
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逝去的感伤

2020-11-22 04:08
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
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