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);

似乎DateTimeSpan解决方案使许多人满意。我不知道。让我们考虑一下:

BeginDate = 1972/2/29销售= 1972/4/28。

基于DateTimeSpan的答案是:

1年(s)， 2个月(s)和0天(s)

我实现了一个方法，在此基础上，答案是:

1年、1个月及28天

显然没有两个月的时间。我想说的是，因为我们在开始日期的月末，剩下的实际上是整个3月加上结束日期(4月)的月份所经过的天数，所以1个月零28天。

如果你读到这里，你有兴趣，我把方法贴在下面。我在评论中解释了我所做的假设，因为有多少个月，月份的概念是一个不断变化的目标。多次测试，看看答案是否有意义。我通常选择相邻年份的考试日期，一旦我确认了答案，我就会前后移动一两天。到目前为止，它看起来不错，我相信你会发现一些bug:D。代码可能看起来有点粗糙，但我希望它足够清楚:

static void Main(string[] args) {
        DateTime EndDate = new DateTime(1973, 4, 28);
        DateTime BeginDate = new DateTime(1972, 2, 29);
        int years, months, days;
        GetYearsMonthsDays(EndDate, BeginDate, out years, out months, out days);
        Console.WriteLine($"{years} year(s), {months} month(s) and {days} day(s)");
    }

    /// <summary>
    /// Calculates how many years, months and days are between two dates.
    /// </summary>
    /// <remarks>
    /// The fundamental idea here is that most of the time all of us agree
    /// that a month has passed today since the same day of the previous month.
    /// A particular case is when both days are the last days of their respective months 
    /// when again we can say one month has passed.
    /// In the following cases the idea of a month is a moving target.
    /// - When only the beginning date is the last day of the month then we're left just with 
    /// a number of days from the next month equal to the day of the month that end date represent
    /// - When only the end date is the last day of its respective month we clearly have a 
    /// whole month plus a few days after the the day of the beginning date until the end of its
    /// respective months
    /// In all the other cases we'll check
    /// - beginingDay > endDay -> less then a month just daysToEndofBeginingMonth + dayofTheEndMonth
    /// - beginingDay < endDay -> full month + (endDay - beginingDay)
    /// - beginingDay == endDay -> one full month 0 days
    /// 
    /// </remarks>
    /// 
    private static void GetYearsMonthsDays(DateTime EndDate, DateTime BeginDate, out int years, out int months, out int days ) {
        var beginMonthDays = DateTime.DaysInMonth(BeginDate.Year, BeginDate.Month);
        var endMonthDays = DateTime.DaysInMonth(EndDate.Year, EndDate.Month);
        // get the full years
        years = EndDate.Year - BeginDate.Year - 1;
        // how many full months in the first year
        var firstYearMonths = 12 - BeginDate.Month;
        // how many full months in the last year
        var endYearMonths = EndDate.Month - 1;
        // full months
        months = firstYearMonths + endYearMonths;           
        days = 0;
        // Particular end of month cases
        if(beginMonthDays == BeginDate.Day && endMonthDays == EndDate.Day) {
            months++;
        }
        else if(beginMonthDays == BeginDate.Day) {
            days += EndDate.Day;
        }
        else if(endMonthDays == EndDate.Day) {
            days += beginMonthDays - BeginDate.Day;
        }
        // For all the other cases
        else if(EndDate.Day > BeginDate.Day) {
            months++;
            days += EndDate.Day - BeginDate.Day;
        }
        else if(EndDate.Day < BeginDate.Day) {                
            days += beginMonthDays - BeginDate.Day;
            days += EndDate.Day;
        }
        else {
            months++;
        }
        if(months >= 12) {
            years++;
            months = months - 12;
        }
    }

我在VB中检查了这个方法的用法。NET通过MSDN，它似乎有很多用途。c#中没有这样的内置方法。(即使这不是一个好主意)你可以在c#中调用VB。

将Microsoft.VisualBasic.dll添加到 你的项目作为参考 使用 Microsoft.VisualBasic.DateAndTime.DateDiff 在代码中

您可以使用以下扩展: 代码

public static class Ext
{
    #region Public Methods

    public static int GetAge(this DateTime @this)
    {
        var today = DateTime.Today;
        return ((((today.Year - @this.Year) * 100) + (today.Month - @this.Month)) * 100 + today.Day - @this.Day) / 10000;
    }

    public static int DiffMonths(this DateTime @from, DateTime @to)
    {
        return (((((@to.Year - @from.Year) * 12) + (@to.Month - @from.Month)) * 100 + @to.Day - @from.Day) / 100);
    }

    public static int DiffYears(this DateTime @from, DateTime @to)
    {
        return ((((@to.Year - @from.Year) * 100) + (@to.Month - @from.Month)) * 100 + @to.Day - @from.Day) / 10000;
    }

    #endregion Public Methods
}

实现!

int Age;
int years;
int Months;
//Replace your own date
var d1 = new DateTime(2000, 10, 22);
var d2 = new DateTime(2003, 10, 20);
//Age
Age = d1.GetAge();
Age = d2.GetAge();
//positive
years = d1.DiffYears(d2);
Months = d1.DiffMonths(d2);
//negative
years = d2.DiffYears(d1);
Months = d2.DiffMonths(d1);
//Or
Months = Ext.DiffMonths(d1, d2);
years = Ext.DiffYears(d1, d2); 

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 },
};

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