public static int PayableMonthsInDuration(DateTime StartDate, DateTime EndDate)
{
    int sy = StartDate.Year; int sm = StartDate.Month; int count = 0;
    do
    {
        count++;if ((sy == EndDate.Year) && (sm >= EndDate.Month)) { break; }
        sm++;if (sm == 13) { sm = 1; sy++; }
    } while ((EndDate.Year >= sy) || (EndDate.Month >= sm));
    return (count);
}

这个解决方案是用于租金/订阅计算的，其中的差异并不意味着减法，它意味着这两个日期之间的跨度。

你可以这样做

if ( date1.AddMonths(x) > date2 )

Public Class ClassDateOperation
    Private prop_DifferenceInDay As Integer
    Private prop_DifferenceInMonth As Integer
    Private prop_DifferenceInYear As Integer


    Public Function DayMonthYearFromTwoDate(ByVal DateStart As Date, ByVal DateEnd As Date) As ClassDateOperation
        Dim differenceInDay As Integer
        Dim differenceInMonth As Integer
        Dim differenceInYear As Integer
        Dim myDate As Date

        DateEnd = DateEnd.AddDays(1)

        differenceInYear = DateEnd.Year - DateStart.Year

        If DateStart.Month <= DateEnd.Month Then
            differenceInMonth = DateEnd.Month - DateStart.Month
        Else
            differenceInYear -= 1
            differenceInMonth = (12 - DateStart.Month) + DateEnd.Month
        End If


        If DateStart.Day <= DateEnd.Day Then
            differenceInDay = DateEnd.Day - DateStart.Day
        Else

            myDate = CDate("01/" & DateStart.AddMonths(1).Month & "/" & DateStart.Year).AddDays(-1)
            If differenceInMonth <> 0 Then
                differenceInMonth -= 1
            Else
                differenceInMonth = 11
                differenceInYear -= 1
            End If

            differenceInDay = myDate.Day - DateStart.Day + DateEnd.Day

        End If

        prop_DifferenceInDay = differenceInDay
        prop_DifferenceInMonth = differenceInMonth
        prop_DifferenceInYear = differenceInYear

        Return Me
    End Function

    Public ReadOnly Property DifferenceInDay() As Integer
        Get
            Return prop_DifferenceInDay
        End Get
    End Property

    Public ReadOnly Property DifferenceInMonth As Integer
        Get
            Return prop_DifferenceInMonth
        End Get
    End Property

    Public ReadOnly Property DifferenceInYear As Integer
        Get
            Return prop_DifferenceInYear
        End Get
    End Property

End Class

似乎DateTimeSpan解决方案使许多人满意。我不知道。让我们考虑一下:

BeginDate = 1972/2/29销售= 1972/4/28。

基于DateTimeSpan的答案是:

1年(s)， 2个月(s)和0天(s)

我实现了一个方法，在此基础上，答案是:

1年、1个月及28天

显然没有两个月的时间。我想说的是，因为我们在开始日期的月末，剩下的实际上是整个3月加上结束日期(4月)的月份所经过的天数，所以1个月零28天。

如果你读到这里，你有兴趣，我把方法贴在下面。我在评论中解释了我所做的假设，因为有多少个月，月份的概念是一个不断变化的目标。多次测试，看看答案是否有意义。我通常选择相邻年份的考试日期，一旦我确认了答案，我就会前后移动一两天。到目前为止，它看起来不错，我相信你会发现一些bug:D。代码可能看起来有点粗糙，但我希望它足够清楚:

static void Main(string[] args) {
        DateTime EndDate = new DateTime(1973, 4, 28);
        DateTime BeginDate = new DateTime(1972, 2, 29);
        int years, months, days;
        GetYearsMonthsDays(EndDate, BeginDate, out years, out months, out days);
        Console.WriteLine($"{years} year(s), {months} month(s) and {days} day(s)");
    }

    /// <summary>
    /// Calculates how many years, months and days are between two dates.
    /// </summary>
    /// <remarks>
    /// The fundamental idea here is that most of the time all of us agree
    /// that a month has passed today since the same day of the previous month.
    /// A particular case is when both days are the last days of their respective months 
    /// when again we can say one month has passed.
    /// In the following cases the idea of a month is a moving target.
    /// - When only the beginning date is the last day of the month then we're left just with 
    /// a number of days from the next month equal to the day of the month that end date represent
    /// - When only the end date is the last day of its respective month we clearly have a 
    /// whole month plus a few days after the the day of the beginning date until the end of its
    /// respective months
    /// In all the other cases we'll check
    /// - beginingDay > endDay -> less then a month just daysToEndofBeginingMonth + dayofTheEndMonth
    /// - beginingDay < endDay -> full month + (endDay - beginingDay)
    /// - beginingDay == endDay -> one full month 0 days
    /// 
    /// </remarks>
    /// 
    private static void GetYearsMonthsDays(DateTime EndDate, DateTime BeginDate, out int years, out int months, out int days ) {
        var beginMonthDays = DateTime.DaysInMonth(BeginDate.Year, BeginDate.Month);
        var endMonthDays = DateTime.DaysInMonth(EndDate.Year, EndDate.Month);
        // get the full years
        years = EndDate.Year - BeginDate.Year - 1;
        // how many full months in the first year
        var firstYearMonths = 12 - BeginDate.Month;
        // how many full months in the last year
        var endYearMonths = EndDate.Month - 1;
        // full months
        months = firstYearMonths + endYearMonths;           
        days = 0;
        // Particular end of month cases
        if(beginMonthDays == BeginDate.Day && endMonthDays == EndDate.Day) {
            months++;
        }
        else if(beginMonthDays == BeginDate.Day) {
            days += EndDate.Day;
        }
        else if(endMonthDays == EndDate.Day) {
            days += beginMonthDays - BeginDate.Day;
        }
        // For all the other cases
        else if(EndDate.Day > BeginDate.Day) {
            months++;
            days += EndDate.Day - BeginDate.Day;
        }
        else if(EndDate.Day < BeginDate.Day) {                
            days += beginMonthDays - BeginDate.Day;
            days += EndDate.Day;
        }
        else {
            months++;
        }
        if(months >= 12) {
            years++;
            months = months - 12;
        }
    }

我写了一个函数来完成这个，因为其他的方法都不适合我。

public string getEndDate (DateTime startDate,decimal monthCount)
{
    int y = startDate.Year;
    int m = startDate.Month;

    for (decimal  i = monthCount; i > 1; i--)
    {
        m++;
        if (m == 12)
        { y++;
            m = 1;
        }
    }
    return string.Format("{0}-{1}-{2}", y.ToString(), m.ToString(), startDate.Day.ToString());
}

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 },
};