我也有同样的问题，但如果你在SQL查询上做的话会更好:

DateDiff(DAY, StartValue,GETDATE()) AS CountDays

查询将自动生成一个列CountDays

当我想在两个日期上做一些计算时，我发现了这个问题，但是日期有小时和分钟的值，我修改了@michael-liu的答案来满足我的要求，它通过了我的测试。

差异日期2012-12-31 23:00和2013-01-01 01:00应该等于1。(2小时) 差异日期2012-12-31 01:00和2013-01-01 23:00应该等于1。(46个小时)

function treatAsUTC(date) {
    var result = new Date(date);
    result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
    return result;
}

var millisecondsPerDay = 24 * 60 * 60 * 1000;
function diffDays(startDate, endDate) {
    return Math.floor(treatAsUTC(endDate) / millisecondsPerDay) - Math.floor(treatAsUTC(startDate) / millisecondsPerDay);
}

我只有两个以毫秒为单位的时间戳，所以我必须用moment.js做一些额外的步骤来获得天数。

const getDaysDiff = (fromTimestamp, toTimestamp) => {
    // set timezone offset with utcOffset if needed
    let fromDate = moment(fromTimestamp).utcOffset(8);
    let toDate = moment(toTimestamp).utcOffset(8);
    // get the start moment of the day
    fromDate.set({'hour':0, 'minute': 0, 'second': 0, 'millisecond': 0});
    toDate.set({'hour':0, 'minute': 0, 'second': 0, 'millisecond': 0});
    let diffDays = toDate.diff(fromDate, 'days');

    return diffDays;
}

getDaysDiff(1528889400000, 1528944180000)// 1 

夏令时问题使这里的许多答案无效。我将使用一个helper函数来获得给定日期的唯一天数——通过使用UTC方法:

const dayNumber = a => Date.UTC(a.getFullYear(), a.getMonth(), a.getDate()) / (24*60*60*1000); const daysBetween = (a, b) => dayNumber(b) - dayNumber(a); // Testing it const start = new Date(1000, 0, 1); // 1 January 1000 const end = new Date(3000, 0, 1); // 1 January 3000 let current = new Date(start); for (let days = 0; current < end; days++) { const diff = daysBetween(start, current); if (diff !== days) throw "test failed"; current.setDate(current.getDate() + 1); // move current date one day forward } console.log("tests succeeded");

在撰写本文时，其他答案中只有一个正确处理DST(夏令时)转换。以下是位于加州的一个系统的结果:

                                        1/1/2013- 3/10/2013- 11/3/2013-
User       Formula                      2/1/2013  3/11/2013  11/4/2013  Result
---------  ---------------------------  --------  ---------  ---------  ---------
Miles                   (d2 - d1) / N   31        0.9583333  1.0416666  Incorrect
some         Math.floor((d2 - d1) / N)  31        0          1          Incorrect
fuentesjr    Math.round((d2 - d1) / N)  31        1          1          Correct
toloco     Math.ceiling((d2 - d1) / N)  31        1          2          Incorrect

N = 86400000

虽然数学。round返回正确的结果，我认为它有点笨拙。相反，当DST开始或结束时，通过显式计算UTC偏移量的变化，我们可以使用精确的算术:

function treatAsUTC(date) {
    var result = new Date(date);
    result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
    return result;
}

function daysBetween(startDate, endDate) {
    var millisecondsPerDay = 24 * 60 * 60 * 1000;
    return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}

alert(daysBetween($('#first').val(), $('#second').val()));

解释

JavaScript的日期计算很棘手，因为date对象内部存储的时间是UTC，而不是本地时间。例如，3/10/2013太平洋标准时间12:00 AM (UTC-08:00)存储为3/10/2013上午8:00 UTC, 3/11/2013太平洋夏令时12:00 AM (UTC-07:00)存储为3/11/2013上午7:00 UTC。在这一天，从午夜到午夜，当地时间在UTC只有23小时!

虽然本地时间中的一天可以大于或小于24小时，但国际标准时间中的一天总是24小时上面所示的daysBetween方法利用了这一事实，它首先调用treatAsUTC将本地时间调整为午夜UTC，然后再进行减法和除法。

1. JavaScript忽略闰秒。

function formatDate(seconds, dictionary) {
    var foo = new Date;
    var unixtime_ms = foo.getTime();
    var unixtime = parseInt(unixtime_ms / 1000);
    var diff = unixtime - seconds;
    var display_date;
    if (diff <= 0) {
        display_date = dictionary.now;
    } else if (diff < 60) {
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.second;
        } else {
            display_date = diff + ' ' + dictionary.seconds;
        }
    } else if (diff < 3540) {
        diff = Math.round(diff / 60);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.minute;
        } else {
            display_date = diff + ' ' + dictionary.minutes;
        }
    } else if (diff < 82800) {
        diff = Math.round(diff / 3600);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.hour;
        } else {
            display_date = diff + ' ' + dictionary.hours;
        }
    } else {
        diff = Math.round(diff / 86400);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.day;
        } else {
            display_date = diff + ' ' + dictionary.days;
        }
    }
    return display_date;
}