下面是一个函数，它精确地提供了两个日期之间的月数。 默认行为只计算整个月份，例如3个月和1天将导致3个月的差异。您可以通过将roundUpFractionalMonths参数设置为true来防止这种情况，因此3个月和1天的差异将返回为4个月。

上面公认的答案(T.J.克劳德的答案)是不准确的，它有时会返回错误的值。

例如，monthDiff(new Date('Jul 01, 2015')， new Date('Aug 05, 2015'))返回0，这显然是错误的。正确的差值是1个月或2个月。

这是我写的函数:

function getMonthsBetween(date1,date2,roundUpFractionalMonths)
{
    //Months will be calculated between start and end dates.
    //Make sure start date is less than end date.
    //But remember if the difference should be negative.
    var startDate=date1;
    var endDate=date2;
    var inverse=false;
    if(date1>date2)
    {
        startDate=date2;
        endDate=date1;
        inverse=true;
    }

    //Calculate the differences between the start and end dates
    var yearsDifference=endDate.getFullYear()-startDate.getFullYear();
    var monthsDifference=endDate.getMonth()-startDate.getMonth();
    var daysDifference=endDate.getDate()-startDate.getDate();

    var monthCorrection=0;
    //If roundUpFractionalMonths is true, check if an extra month needs to be added from rounding up.
    //The difference is done by ceiling (round up), e.g. 3 months and 1 day will be 4 months.
    if(roundUpFractionalMonths===true && daysDifference>0)
    {
        monthCorrection=1;
    }
    //If the day difference between the 2 months is negative, the last month is not a whole month.
    else if(roundUpFractionalMonths!==true && daysDifference<0)
    {
        monthCorrection=-1;
    }

    return (inverse?-1:1)*(yearsDifference*12+monthsDifference+monthCorrection);
};

anyVar = (((DisplayTo.getFullYear() * 12) + DisplayTo.getMonth()) - ((DisplayFrom.getFullYear() * 12) + DisplayFrom.getMonth()));

有两种方法，数学的和快速的，但受制于日历的变幻莫测，或者迭代的和缓慢的，但处理所有奇怪的(或者至少委托处理它们到一个经过良好测试的库)。

If you iterate through the calendar, incrementing the start date by one month & seeing if we pass the end date. This delegates anomaly-handling to the built-in Date() classes, but could be slow IF you're doing this for a large number of dates. James' answer takes this approach. As much as I dislike the idea, I think this is the "safest" approach, and if you're only doing one calculation, the performance difference really is negligible. We tend to try to over-optimize tasks which will only be performed once.

现在，如果您在数据集中计算这个函数，您可能不希望在每一行上运行该函数(或者上帝禁止，每条记录多次)。在这种情况下，您几乎可以使用这里的任何其他答案，除了接受的答案，这是错误的(new Date()和new Date()之间的差异是-1)?

下面是我尝试的一种数学而快速的方法，它解释了不同的月份长度和闰年。你真的应该只使用这样的函数，如果你将应用它到一个数据集(做这个计算一遍又一遍)。如果只需要执行一次，可以使用上面James的迭代方法，因为您正在将所有(许多)异常的处理委托给Date()对象。

function diffInMonths(from, to){
    var months = to.getMonth() - from.getMonth() + (12 * (to.getFullYear() - from.getFullYear()));

    if(to.getDate() < from.getDate()){
        var newFrom = new Date(to.getFullYear(),to.getMonth(),from.getDate());
        if (to < newFrom  && to.getMonth() == newFrom.getMonth() && to.getYear() %4 != 0){
            months--;
        }
    }

    return months;
}

下面的逻辑将在几个月内取得差异

(endDate.getFullYear()*12+endDate.getMonth())-(startDate.getFullYear()*12+startDate.getMonth())

这是我能找到的最简单的解。这将直接返回月数。尽管，它总是给出一个绝对值。

new Date(new Date(d2) - new Date(d1)).getMonth();

对于非绝对值，您可以使用以下解决方案:

function diff_months(startDate, endDate) {
  let diff = new Date( new Date(endDate)  - new Date(startDate) ).getMonth();
  return endDate >= startDate ? diff : -diff;
}

下面的代码还将部分月份中的nr天考虑在内，从而返回两个日期之间的完整月份。

var monthDiff = function(d1, d2) {
  if( d2 < d1 ) { 
    var dTmp = d2;
    d2 = d1;
    d1 = dTmp;
  }

  var months = (d2.getFullYear() - d1.getFullYear()) * 12;
  months -= d1.getMonth() + 1;
  months += d2.getMonth();

  if( d1.getDate() <= d2.getDate() ) months += 1;

  return months;
}

monthDiff(new Date(2015, 01, 20), new Date(2015, 02, 20))
> 1

monthDiff(new Date(2015, 01, 20), new Date(2015, 02, 19))
> 0

monthDiff(new Date(2015, 01, 20), new Date(2015, 01, 22))
> 0