我的问题用这个方法解决了:

static void Main(string[] args)
        {
            var date1 = new DateTime(2018, 12, 05);
            var date2 = new DateTime(2019, 03, 01);

            int CountNumberOfMonths() => (date2.Month - date1.Month) + 12 * (date2.Year - date1.Year);

            var numberOfMonths = CountNumberOfMonths();

            Console.WriteLine("Number of months between {0} and {1}: {2} months.", date1.ToString(), date2.ToString(), numberOfMonths.ToString());

            Console.ReadKey();

            //
            // *** Console Output:
            // Number of months between 05/12/2018 00:00:00 and 01/03/2019 00:00:00: 3 months.
            //

        }

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。

int nMonths = 0;
if (FDate.ToDateTime().Year == TDate.ToDateTime().Year)
     nMonths = TDate.ToDateTime().Month - FDate.ToDateTime().Month;                         
else
nMonths = (12 - FDate.Month) + TDate.Month;                          

疯狂的方法，计算所有的日子，超级精确

Helper类:

public class DaysInMonth
{
    public int Days { get; set; }
    public int Month { get; set; }
    public int Year { get; set; }
    public bool Full { get; set; }
}

功能:

    public static List<DaysInMonth> MonthsDelta(DateTime start, DateTime end)
    {
        
        var dates = Enumerable.Range(0, 1 + end.Subtract(start).Days)
          .Select(offset => start.AddDays(offset))
          .ToArray();

        DateTime? prev = null;
        int days = 0;

        List < DaysInMonth > list = new List<DaysInMonth>();

        foreach (DateTime date in dates)
        {
            if (prev != null)
            {
                if(date.Month!=prev.GetValueOrDefault().Month)
                {
                    DaysInMonth daysInMonth = new DaysInMonth();
                    daysInMonth.Days = days;
                    daysInMonth.Month = prev.GetValueOrDefault().Month;
                    daysInMonth.Year = prev.GetValueOrDefault().Year;
                    daysInMonth.Full = DateTime.DaysInMonth(daysInMonth.Year, daysInMonth.Month) == daysInMonth.Days;
                    list.Add(daysInMonth);
                    days = 0;
                }
            }
            days++;
            prev = date;
        }

        //------------------ add last
        if (days > 0)
        {
            DaysInMonth daysInMonth = new DaysInMonth();
            daysInMonth.Days = days;
            daysInMonth.Month = prev.GetValueOrDefault().Month;
            daysInMonth.Year = prev.GetValueOrDefault().Year;
            daysInMonth.Full = DateTime.DaysInMonth(daysInMonth.Year, daysInMonth.Month) == daysInMonth.Days;
            list.Add(daysInMonth);
        }

        return list;
    }

你可以有一个这样的函数。

例如，从2012/12/27到2012/12/29变成3天。同样，从2012/12/15到2013/01/15变成了2个月，因为到2013/01/14是1个月。从15号开始是第二个月。

如果您不想在计算中包括这两天，则可以删除第二个if条件中的“=”。即从2012/12/15到2013/01/15为1个月。

public int GetMonths(DateTime startDate, DateTime endDate)
{
    if (startDate > endDate)
    {
        throw new Exception("Start Date is greater than the End Date");
    }

    int months = ((endDate.Year * 12) + endDate.Month) - ((startDate.Year * 12) + startDate.Month);

    if (endDate.Day >= startDate.Day)
    {
        months++;
    }

    return months;
}

简单快速的解决方案，计算2个日期之间的总月份。 如果你只想得到不同的月份，而不计算From date中的月份-只需从代码中删除+1。

public static int GetTotalMonths(DateTime From, DateTime Till)
        {
            int MonthDiff = 0;

            for (int i = 0; i < 12; i++)
            {
                if (From.AddMonths(i).Month == Till.Month)
                {
                    MonthDiff = i + 1;
                    break;
                }
            }

            return MonthDiff;
        }