在这个问题上没有很多明确的答案，因为你总是在假设事情。

这个解决方案在两个日期之间进行计算，假设您想保存一个月中的某一天进行比较，(这意味着在计算中考虑了这个月中的某一天)

例如，如果你的日期是2012年1月30日，2012年2月29日就不是一个月，但2013年3月1日就不是一个月。

它经过了相当彻底的测试，可能稍后我们会在使用时清理它，但这里:

private static int TotalMonthDifference(DateTime dtThis, DateTime dtOther)
{
    int intReturn = 0;
    bool sameMonth = false;

    if (dtOther.Date < dtThis.Date) //used for an error catch in program, returns -1
        intReturn--;

    int dayOfMonth = dtThis.Day; //captures the month of day for when it adds a month and doesn't have that many days
    int daysinMonth = 0; //used to caputre how many days are in the month

    while (dtOther.Date > dtThis.Date) //while Other date is still under the other
    {
        dtThis = dtThis.AddMonths(1); //as we loop, we just keep adding a month for testing
        daysinMonth = DateTime.DaysInMonth(dtThis.Year, dtThis.Month); //grabs the days in the current tested month

        if (dtThis.Day != dayOfMonth) //Example 30 Jan 2013 will go to 28 Feb when a month is added, so when it goes to march it will be 28th and not 30th
        {
            if (daysinMonth < dayOfMonth) // uses day in month max if can't set back to day of month
                dtThis.AddDays(daysinMonth - dtThis.Day);
            else
                dtThis.AddDays(dayOfMonth - dtThis.Day);
        }
        if (((dtOther.Year == dtThis.Year) && (dtOther.Month == dtThis.Month))) //If the loop puts it in the same month and year
        {
            if (dtOther.Day >= dayOfMonth) //check to see if it is the same day or later to add one to month
                intReturn++;
            sameMonth = true; //sets this to cancel out of the normal counting of month
        }
        if ((!sameMonth)&&(dtOther.Date > dtThis.Date))//so as long as it didn't reach the same month (or if i started in the same month, one month ahead, add a month)
            intReturn++;
    }
    return intReturn; //return month
}

简单的修复。工作的100%

        var exactmonth = (date1.Year - date2.Year) * 12 + date1.Month - 
        date2.Month +  (date1.Day >= date2.Day ? 0 : -1);
        Console.WriteLine(exactmonth);

以下是我对获得Months差异的贡献，我发现这是准确的:

namespace System
{
     public static class DateTimeExtensions
     {
         public static Int32 DiffMonths( this DateTime start, DateTime end )
         {
             Int32 months = 0;
             DateTime tmp = start;

             while ( tmp < end )
             {
                 months++;
                 tmp = tmp.AddMonths( 1 );
             }

             return months;
        }
    }
}

用法:

Int32 months = DateTime.Now.DiffMonths( DateTime.Now.AddYears( 5 ) );

您可以创建另一个名为DiffYears的方法，并应用与上面完全相同的逻辑，并在while循环中使用AddYears而不是AddMonths。

你可以在。net中使用Time Period Library的DateDiff类:

// ----------------------------------------------------------------------
public void DateDiffSample()
{
  DateTime date1 = new DateTime( 2009, 11, 8, 7, 13, 59 );
  DateTime date2 = new DateTime( 2011, 3, 20, 19, 55, 28 );
  DateDiff dateDiff = new DateDiff( date1, date2 );

  // differences
  Console.WriteLine( "DateDiff.Months: {0}", dateDiff.Months );
  // > DateDiff.Months: 16

  // elapsed
  Console.WriteLine( "DateDiff.ElapsedMonths: {0}", dateDiff.ElapsedMonths );
  // > DateDiff.ElapsedMonths: 4

  // description
  Console.WriteLine( "DateDiff.GetDescription(6): {0}", dateDiff.GetDescription( 6 ) );
  // > DateDiff.GetDescription(6): 1 Year 4 Months 12 Days 12 Hours 41 Mins 29 Secs
} // DateDiffSample

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)

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;
        }
    }