这里有一个简单的解决方案，至少对我来说是有效的。它可能不是最快的，因为它在循环中使用了很酷的DateTime的AddMonth功能:

public static int GetMonthsDiff(DateTime start, DateTime end)
{
    if (start > end)
        return GetMonthsDiff(end, start);

    int months = 0;
    do
    {
        start = start.AddMonths(1);
        if (start > end)
            return months;

        months++;
    }
    while (true);
}

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;
        }
    }

假设这个月的日期不相关(即2011.1.1和2010.12.31之间的差为1)，date1 > date2为正值，date2 > date1为负值

((date1.Year - date2.Year) * 12) + date1.Month - date2.Month

或者，假设你想要两个日期之间的“平均月”的大致数字，下面的方法应该适用于所有日期，但日期差异非常大。

date1.Subtract(date2).Days / (365.25 / 12)

注意，如果您要使用后一种解决方案，那么您的单元测试应该声明应用程序设计使用的最宽日期范围，并相应地验证计算结果。

更新(感谢Gary)

如果使用“平均月份”方法，“每年平均天数”的更准确数字是365.2425。

我对两个日期之间总月差的理解有一个整数部分和一个小数部分(日期很重要)。

积分部分是整个月的差额。

对我来说，小数部分是开始月份和结束月份之间一天的百分比(到一个月的全部天数)的差值。

public static class DateTimeExtensions
{
    public static double TotalMonthsDifference(this DateTime from, DateTime to)
    {
        //Compute full months difference between dates
        var fullMonthsDiff = (to.Year - from.Year)*12 + to.Month - from.Month;

        //Compute difference between the % of day to full days of each month
        var fractionMonthsDiff = ((double)(to.Day-1) / (DateTime.DaysInMonth(to.Year, to.Month)-1)) -
            ((double)(from.Day-1)/ (DateTime.DaysInMonth(from.Year, from.Month)-1));

        return fullMonthsDiff + fractionMonthsDiff;
    }
}

有了这个扩展，这些是结果:

2/29/2000 TotalMonthsDifference 2/28/2001 => 12
2/28/2000 TotalMonthsDifference 2/28/2001 => 12.035714285714286
01/01/2000 TotalMonthsDifference 01/16/2000 => 0.5
01/31/2000 TotalMonthsDifference 01/01/2000 => -1.0
01/31/2000 TotalMonthsDifference 02/29/2000 => 1.0
01/31/2000 TotalMonthsDifference 02/28/2000 => 0.9642857142857143
01/31/2001 TotalMonthsDifference 02/28/2001 => 1.0

一定是有人干的))

扩展方法返回给定日期之间的完整月数。无论以什么顺序接收日期，都会返回一个自然数。在“正确”答案中没有近似的计算。

    /// <summary>
    /// Returns the difference between dates in months.
    /// </summary>
    /// <param name="current">First considered date.</param>
    /// <param name="another">Second considered date.</param>
    /// <returns>The number of full months between the given dates.</returns>
    public static int DifferenceInMonths(this DateTime current, DateTime another)
    {
        DateTime previous, next;
        if (current > another)
        {
            previous = another;
            next     = current;
        }
        else
        {
            previous = current;
            next     = another;
        }

        return
            (next.Year - previous.Year) * 12     // multiply the difference in years by 12 months
          + next.Month - previous.Month          // add difference in months
          + (previous.Day <= next.Day ? 0 : -1); // if the day of the next date has not reached the day of the previous one, then the last month has not yet ended
    }

但如果你仍然想要得到月份的小数部分，你只需要在回报中再加一项:

+(下一个。Day - previous.Day) / DateTime.DaysInMonth(previous. Day)年,previous.Month)

如果你只关心月份和年份，想要触及两个日期(例如你想要从JAN/2021到AGO/2022)，你可以使用这个:

int numberOfMonths= (Year2 > Year1 ? ( Year2 - Year1 - 1) * 12 + (12 - Month1) + Month2 + 1 : Month2 - Month1 + 1); 

例子:

Year1/Month1: 2021/10   
Year2/Month2: 2022/08   
numberOfMonths = 11;

或者同年:

Year1/Month1: 2021/10   
Year2/Month2: 2021/12   
numberOfMonths = 3;

如果你只想触碰其中一个，就去掉两个+ 1。