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Home/ Questions/Q 823029
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T02:53:14+00:00 2026-05-15T02:53:14+00:00

This is what I’ve coded it up, and it appears to work. window.onload =

  • 0

This is what I’ve coded it up, and it appears to work.

window.onload = function() {
   var currentSpan = document.getElementById('current');

   var minute = 60000,
       hour = minute * 60,
       day = hour * 24,
       week = day * 7,
       month = week * 4,
       year = day * 365;
       var start = new Date(2009, 6, 1);

   setInterval(function() {

       var now = new Date();



       var difference = now - start;


       var years = Math.floor(difference / year),
           months = Math.floor((difference - (years * year)) / month),
           weeks = Math.floor((difference - (months * month + years * year)) / week),
           days = Math.floor((difference - (weeks * week + months * month + years * year)) / day);



   currentSpan.innerHTML = 'Since has passed: ' + years + ' years, ' + months + ' months, ' + weeks + ' weeks and ' + days + ' days';

   }, 500);

};

This seems to update my span fine, and all the numbers look correct.

However, the code looks quite ugly. Do I really need to set up faux constants like that, and then do all that math to calculate what I want?

It’s been a while since I’ve worked with the Date object.

Is this the best way to do this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T02:53:15+00:00Added an answer on May 15, 2026 at 2:53 am

    You could put those constants in an array and then just iterate through it:

    function tdiff(utc) {
      var diff = new Date() - new Date(utc);
      var units = [
        1000 * 60 * 60 * 24 * 365,
        1000 * 60 * 60 * 24 * 28,
        1000 * 60 * 60 * 24 * 7,
        1000 * 60 * 60 * 24,
        1000 * 60 * 60,
        1000 * 60,
        1000
      ];
    
      var rv = [];
      for (var i = 0; i < units.length; ++i) {
        rv.push(Math.floor(diff / units[i]));
        diff = diff % units[i];
      }
      return rv;
    }
    

    Of course since months and years aren’t always the same length, this isn’t really that accurate, but I figure you realize that 🙂

    Also see this: http://timeago.yarp.com/ it’s kind-of cool

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