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

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);
 }

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。

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。

  var dt1 = (DateTime.Now.Year * 12) + DateTime.Now.Month;
  var dt2 = (DateTime.Now.AddMonths(-13).Year * 12) + DateTime.Now.AddMonths(-13).Month;
  Console.WriteLine(dt1);
  Console.WriteLine(dt2);
  Console.WriteLine((dt1 - dt2));

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 },
};