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;
    }
}

LINQ的解决方案,

DateTime ToDate = DateTime.Today;
DateTime FromDate = ToDate.Date.AddYears(-1).AddDays(1);

int monthCount = Enumerable.Range(0, 1 + ToDate.Subtract(FromDate).Days)
                    .Select(x => FromDate.AddDays(x))
                    .ToList<DateTime>()
                    .GroupBy(z => new { z.Year, z.Month })
                    .Count();

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

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

这个简单的静态函数计算两个Datetimes之间的月份分数。

1.1. 到31.1。= 1.0 1.4. 到15.4。= 0.5 16.4. 到30.4。= 0.5 1.3. 到1.4。= 1 + 1/30

该函数假设第一个日期比第二个日期小。要处理负时间间隔，可以通过在开始时引入符号和变量交换来轻松地修改函数。

public static double GetDeltaMonths(DateTime t0, DateTime t1)
{
     DateTime t = t0;
     double months = 0;
     while(t<=t1)
     {
         int daysInMonth = DateTime.DaysInMonth(t.Year, t.Month);
         DateTime endOfMonth = new DateTime(t.Year, t.Month, daysInMonth);
         int cutDay = endOfMonth <= t1 ? daysInMonth : t1.Day;
         months += (cutDay - t.Day + 1) / (double) daysInMonth;
         t = new DateTime(t.Year, t.Month, 1).AddMonths(1);
     }
     return Math.Round(months,2);
 }

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。