我们是这样做的:

public static int MonthDiff(DateTime date1, DateTime date2)
{
    if (date1.Month < date2.Month)
    {
        return (date2.Year - date1.Year) * 12 + date2.Month - date1.Month;
    }
    else
    {
        return (date2.Year - date1.Year - 1) * 12 + date2.Month - date1.Month + 12;
    }
}

如果您想要完整月份的确切数目，总是正的(2000-01-15,2000-02-14返回0)，则考虑完整月份是当您到达下个月的同一天时(类似于年龄计算)

public static int GetMonthsBetween(DateTime from, DateTime to)
{
    if (from > to) return GetMonthsBetween(to, from);

    var monthDiff = Math.Abs((to.Year * 12 + (to.Month - 1)) - (from.Year * 12 + (from.Month - 1)));

    if (from.AddMonths(monthDiff) > to || to.Day < from.Day)
    {
        return monthDiff - 1;
    }
    else
    {
        return monthDiff;
    }
}

编辑原因:旧代码在某些情况下不正确，如:

new { From = new DateTime(1900, 8, 31), To = new DateTime(1901, 8, 30), Result = 11 },

Test cases I used to test the function:

var tests = new[]
{
    new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 1, 1), Result = 0 },
    new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 1, 2), Result = 0 },
    new { From = new DateTime(1900, 1, 2), To = new DateTime(1900, 1, 1), Result = 0 },
    new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 2, 1), Result = 1 },
    new { From = new DateTime(1900, 2, 1), To = new DateTime(1900, 1, 1), Result = 1 },
    new { From = new DateTime(1900, 1, 31), To = new DateTime(1900, 2, 1), Result = 0 },
    new { From = new DateTime(1900, 8, 31), To = new DateTime(1900, 9, 30), Result = 0 },
    new { From = new DateTime(1900, 8, 31), To = new DateTime(1900, 10, 1), Result = 1 },
    new { From = new DateTime(1900, 1, 1), To = new DateTime(1901, 1, 1), Result = 12 },
    new { From = new DateTime(1900, 1, 1), To = new DateTime(1911, 1, 1), Result = 132 },
    new { From = new DateTime(1900, 8, 31), To = new DateTime(1901, 8, 30), Result = 11 },
};

有3种情况:同一年，前一年和其他年份。

如果日期不重要的话……

public int GetTotalNumberOfMonths(DateTime start, DateTime end)
{
    // work with dates in the right order
    if (start > end)
    {
        var swapper = start;
        start = end;
        end = swapper;
    }

    switch (end.Year - start.Year)
    {
        case 0: // Same year
            return end.Month - start.Month;

        case 1: // last year
            return (12 - start.Month) + end.Month;

        default:
            return 12 * (3 - (end.Year - start.Year)) + (12 - start.Month) + end.Month;
    }
}

这是我所需要的。对我来说，一个月的哪一天并不重要，因为它总是碰巧是一个月的最后一天。

public static int MonthDiff(DateTime d1, DateTime d2){
    int retVal = 0;

    if (d1.Month<d2.Month)
    {
        retVal = (d1.Month + 12) - d2.Month;
        retVal += ((d1.Year - 1) - d2.Year)*12;
    }
    else
    {
        retVal = d1.Month - d2.Month;
        retVal += (d1.Year - d2.Year)*12;
    }
    //// Calculate the number of years represented and multiply by 12
    //// Substract the month number from the total
    //// Substract the difference of the second month and 12 from the total
    //retVal = (d1.Year - d2.Year) * 12;
    //retVal = retVal - d1.Month;
    //retVal = retVal - (12 - d2.Month);

    return retVal;
}

你可以使用野田时间https://nodatime.org/

LocalDate start = new LocalDate(2010, 1, 5);
LocalDate end = new LocalDate(2012, 6, 1);
Period period = Period.Between(start, end, PeriodUnits.Months);
Console.WriteLine(period.Months);

我只是需要一些简单的东西来满足例如，只输入月份/年的就业日期，所以希望工作的年份和月份不同。这就是我所使用的，只是为了实用

public static YearsMonths YearMonthDiff(DateTime startDate, DateTime endDate) {
    int monthDiff = ((endDate.Year * 12) + endDate.Month) - ((startDate.Year * 12) + startDate.Month) + 1;
    int years = (int)Math.Floor((decimal) (monthDiff / 12));
    int months = monthDiff % 12;
    return new YearsMonths {
        TotalMonths = monthDiff,
            Years = years,
            Months = months
    };
}

net小提琴