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"));

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.");

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>'); });

有很多方法可以做到这一点。 是的，你可以使用普通的旧JS。试试:

let dt1 = new Date()
let dt2 = new Date()

让我们使用Date.prototype.setMinutes模拟通道，并确保我们在范围内。

dt1.setMinutes(7)
dt2.setMinutes(42)
console.log('Elapsed seconds:',(dt2-dt1)/1000)

或者你也可以使用一些像js-joda这样的库，在那里你可以很容易地做这样的事情(直接从文档中):

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

有更多的ofc库，但js-joda还有一个额外的好处，它也可以在Java中使用，在Java中已经进行了广泛的测试。所有这些测试都已迁移到js-joda，它也是不可变的。

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

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;