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);

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)

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);
}

下面是一个使用VB的更简洁的解决方案。只适用于年、月、日的净日期。你也可以在c#中加载DateDiff库。

Date1必须<= date2

VB。网

Dim date1 = Now.AddDays(-2000)
Dim date2 = Now
Dim diffYears = DateDiff(DateInterval.Year, date1, date2) - If(date1.DayOfYear > date2.DayOfYear, 1, 0)
Dim diffMonths = DateDiff(DateInterval.Month, date1, date2) - diffYears * 12 - If(date1.Day > date2.Day, 1, 0)
Dim diffDays = If(date2.Day >= date1.Day, date2.Day - date1.Day, date2.Day + (Date.DaysInMonth(date1.Year, date1.Month) - date1.Day))

C#

DateTime date1 = Now.AddDays(-2000);
DateTime date2 = Now;
int diffYears = DateDiff(DateInterval.Year, date1, date2) - date1.DayOfYear > date2.DayOfYear ? 1 : 0;
int diffMonths = DateDiff(DateInterval.Month, date1, date2) - diffYears * 12 - date1.Day > date2.Day ? 1 : 0;
int diffDays = date2.Day >= date1.Day ? date2.Day - date1.Day : date2.Day + (System.DateTime.DaysInMonth(date1.Year, date1.Month) - date1.Day);

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);

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.
            //

        }