你可以使用野田时间https://nodatime.org/

LocalDate start = new LocalDate(2010, 1, 5);
LocalDate end = new LocalDate(2012, 6, 1);
Period period = Period.Between(start, end, PeriodUnits.Months);
Console.WriteLine(period.Months);

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;
    }

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。

这是对Kirk Woll的回答的回应。我还没有足够的声望点来回复评论……

我喜欢Kirk的解决方案，并打算无耻地窃取它并在我的代码中使用它，但当我仔细查看它时，我意识到它太复杂了。不必要的切换和循环，以及使用毫无意义的公共构造函数。

以下是我的改写:

public class DateTimeSpan {
    private DateTime _date1;
    private DateTime _date2;
    private int _years;
    private int _months;
    private int _days;
    private int _hours;
    private int _minutes;
    private int _seconds;
    private int _milliseconds;

    public int Years { get { return _years; } }
    public int Months { get { return _months; } }
    public int Days { get { return _days; } }
    public int Hours { get { return _hours; } }
    public int Minutes { get { return _minutes; } }
    public int Seconds { get { return _seconds; } }
    public int Milliseconds { get { return _milliseconds; } }

    public DateTimeSpan(DateTime date1, DateTime date2) {
        _date1 = (date1 > date2) ? date1 : date2;
        _date2 = (date2 < date1) ? date2 : date1;

        _years = _date1.Year - _date2.Year;
        _months = (_years * 12) + _date1.Month - _date2.Month;
        TimeSpan t = (_date2 - _date1);
        _days = t.Days;
        _hours = t.Hours;
        _minutes = t.Minutes;
        _seconds = t.Seconds;
        _milliseconds = t.Milliseconds;

    }

    public static DateTimeSpan CompareDates(DateTime date1, DateTime date2) {
        return new DateTimeSpan(date1, date2);
    }
}

用法1，基本相同:

void Main()
{
    DateTime compareTo = DateTime.Parse("8/13/2010 8:33:21 AM");
    DateTime now = DateTime.Parse("2/9/2012 10:10:11 AM");
    var dateSpan = new DateTimeSpan(compareTo, now);
    Console.WriteLine("Years: " + dateSpan.Years);
    Console.WriteLine("Months: " + dateSpan.Months);
    Console.WriteLine("Days: " + dateSpan.Days);
    Console.WriteLine("Hours: " + dateSpan.Hours);
    Console.WriteLine("Minutes: " + dateSpan.Minutes);
    Console.WriteLine("Seconds: " + dateSpan.Seconds);
    Console.WriteLine("Milliseconds: " + dateSpan.Milliseconds);
}

Usage2类似:

void Main()
{
    DateTime compareTo = DateTime.Parse("8/13/2010 8:33:21 AM");
    DateTime now = DateTime.Parse("2/9/2012 10:10:11 AM");
    Console.WriteLine("Years: " + DateTimeSpan.CompareDates(compareTo, now).Years);
    Console.WriteLine("Months: " + DateTimeSpan.CompareDates(compareTo, now).Months);
    Console.WriteLine("Days: " + DateTimeSpan.CompareDates(compareTo, now).Days);
    Console.WriteLine("Hours: " + DateTimeSpan.CompareDates(compareTo, now).Hours);
    Console.WriteLine("Minutes: " + DateTimeSpan.CompareDates(compareTo, now).Minutes);
    Console.WriteLine("Seconds: " + DateTimeSpan.CompareDates(compareTo, now).Seconds);
    Console.WriteLine("Milliseconds: " + DateTimeSpan.CompareDates(compareTo, now).Milliseconds);
}

我写了一个函数来完成这个，因为其他的方法都不适合我。

public string getEndDate (DateTime startDate,decimal monthCount)
{
    int y = startDate.Year;
    int m = startDate.Month;

    for (decimal  i = monthCount; i > 1; i--)
    {
        m++;
        if (m == 12)
        { y++;
            m = 1;
        }
    }
    return string.Format("{0}-{1}-{2}", y.ToString(), m.ToString(), startDate.Day.ToString());
}

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