// 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();

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));

代码样本从这里。

假设你有两个Date对象，你可以减去它们，以毫秒为单位得到差值:

var difference = date2 - date1;

从那里，您可以使用简单的算术来推导其他值。

我做了一个下面的函数来得到现在和“2021-02-26T21:50:42.123”之间的区别。

差值返回以毫秒为单位的答案，所以我使用以下公式进行转换:

(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"));

加上@paresh mayani的答案，像Facebook一样工作-显示了以秒/分钟/小时/周/月/年为单位流逝了多少时间

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);

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      
}