抱歉，平毫秒计算不可靠 感谢所有的回复，但我尝试的功能很少失败 1. 一个接近今天的日期 2. 1970年或者 3.闰年的一天。

最适合我的方法，涵盖了所有场景，例如闰年，接近1970年的日期，2月29日等。

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.");

这就是如何在没有框架的情况下实现日期之间的差异。

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 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

像“差几天”这样的表达从来不像看起来那么简单。如果你有以下日期:

d1: 2011-10-15 23:59:00
d1: 2011-10-16 00:01:00

时间差2分钟，“天数差”应该是1还是0?类似的问题也出现在任何以月、年或其他形式表示的差异中，因为年、月和日的长度和时间不同(例如，夏令时开始的那一天比平时短1小时，比夏令时结束的那一天短2小时)。

这里是一个忽略时间的天数差函数，即对于上述日期，它返回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 );
}

这可以更简洁:

/* 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>'); });

            // 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();

基于javascript运行时原型实现，您可以使用简单的算术减去日期如下所示

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);

差值返回以毫秒为单位的答案，然后你必须通过除法转换它:

按1000换算成秒 通过1000×60转换为分钟 通过1000×60×60转换为小时 通过1000×60×60×24转换为日