两个日期之间的月份差异

假设月份中的某一天无关紧要（即 2011.1.1 和 2010.12.31 之间的差值为 1），其中 date1 > date2 给出一个正值，date2 > date1 给出一个负值

((date1.Year - date2.Year) * 12) + date1.Month - date2.Month

或者，假设您想要两个日期之间的“平均月数”的大致数量，那么除了非常大的日期差异之外，以下内容应该适用于所有日期。

date1.Subtract(date2).Days / (365.25 / 12)

请注意，如果要使用后一种解决方案，则单元测试应说明应用程序设计用于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“平均月数”方法，则用于“每年平均天数”的更准确的数字是 365.2425。

我通过MSDN检查了此方法在 VB.NET 中的用法，似乎它有很多用法。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。

1. 搭 您的项目作为参考Microsoft.VisualBasic.dll
2. 在代码中使用Microsoft.VisualBasic.DateAndTime.DateDiff

Public Class ClassDateOperation
    Private prop_DifferenceInDay As Integer
    Private prop_DifferenceInMonth As Integer
    Private prop_DifferenceInYear As Integer


    Public Function DayMonthYearFromTwoDate(ByVal DateStart As Date, ByVal DateEnd As Date) As ClassDateOperation
        Dim differenceInDay As Integer
        Dim differenceInMonth As Integer
        Dim differenceInYear As Integer
        Dim myDate As Date

        DateEnd = DateEnd.AddDays(1)

        differenceInYear = DateEnd.Year - DateStart.Year

        If DateStart.Month <= DateEnd.Month Then
            differenceInMonth = DateEnd.Month - DateStart.Month
        Else
            differenceInYear -= 1
            differenceInMonth = (12 - DateStart.Month) + DateEnd.Month
        End If


        If DateStart.Day <= DateEnd.Day Then
            differenceInDay = DateEnd.Day - DateStart.Day
        Else

            myDate = CDate("01/" & DateStart.AddMonths(1).Month & "/" & DateStart.Year).AddDays(-1)
            If differenceInMonth <> 0 Then
                differenceInMonth -= 1
            Else
                differenceInMonth = 11
                differenceInYear -= 1
            End If

            differenceInDay = myDate.Day - DateStart.Day + DateEnd.Day

        End If

        prop_DifferenceInDay = differenceInDay
        prop_DifferenceInMonth = differenceInMonth
        prop_DifferenceInYear = differenceInYear

        Return Me
    End Function

    Public ReadOnly Property DifferenceInDay() As Integer
        Get
            Return prop_DifferenceInDay
        End Get
    End Property

    Public ReadOnly Property DifferenceInMonth As Integer
        Get
            Return prop_DifferenceInMonth
        End Get
    End Property

    Public ReadOnly Property DifferenceInYear As Integer
        Get
            Return prop_DifferenceInYear
        End Get
    End Property

End Class

你可以做

if ( date1.AddMonths(x) > date2 )

这是一个返回 a 的综合解决方案，类似于 ，只是除了时间部分之外，它还包括所有日期组件。DateTimeSpanTimeSpan

用法：

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 = DateTimeSpan.CompareDates(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);
}

输出：

年：1 个月：5
天：27
小时：1

分钟：36
秒：50
毫秒：0

为方便起见，我已将逻辑归入结构中，但您可以将该方法移动到您认为合适的任何位置。另请注意，哪个日期在另一个日期之前并不重要。DateTimeSpanCompareDates

public struct DateTimeSpan
{
    public int Years { get; }
    public int Months { get; }
    public int Days { get; }
    public int Hours { get; }
    public int Minutes { get; }
    public int Seconds { get; }
    public int Milliseconds { get; }

    public DateTimeSpan(int years, int months, int days, int hours, int minutes, int seconds, int milliseconds)
    {
        Years = years;
        Months = months;
        Days = days;
        Hours = hours;
        Minutes = minutes;
        Seconds = seconds;
        Milliseconds = milliseconds;
    }

    enum Phase { Years, Months, Days, Done }

    public static DateTimeSpan CompareDates(DateTime date1, DateTime date2)
    {
        if (date2 < date1)
        {
            var sub = date1;
            date1 = date2;
            date2 = sub;
        }

        DateTime current = date1;
        int years = 0;
        int months = 0;
        int days = 0;

        Phase phase = Phase.Years;
        DateTimeSpan span = new DateTimeSpan();
        int officialDay = current.Day;

        while (phase != Phase.Done)
        {
            switch (phase)
            {
                case Phase.Years:
                    if (current.AddYears(years + 1) > date2)
                    {
                        phase = Phase.Months;
                        current = current.AddYears(years);
                    }
                    else
                    {
                        years++;
                    }
                    break;
                case Phase.Months:
                    if (current.AddMonths(months + 1) > date2)
                    {
                        phase = Phase.Days;
                        current = current.AddMonths(months);
                        if (current.Day < officialDay && officialDay <= DateTime.DaysInMonth(current.Year, current.Month))
                            current = current.AddDays(officialDay - current.Day);
                    }
                    else
                    {
                        months++;
                    }
                    break;
                case Phase.Days:
                    if (current.AddDays(days + 1) > date2)
                    {
                        current = current.AddDays(days);
                        var timespan = date2 - current;
                        span = new DateTimeSpan(years, months, days, timespan.Hours, timespan.Minutes, timespan.Seconds, timespan.Milliseconds);
                        phase = Phase.Done;
                    }
                    else
                    {
                        days++;
                    }
                    break;
            }
        }

        return span;
    }
}

如果你想要确切的整月数，总是正数（2000-01-15,2000-02-14返回0），考虑到一个完整的月份是你到达下个月的同一天（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，例如：

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 },
};

这是来自我自己的库，将返回两个日期之间的月差。

public static int MonthDiff(DateTime d1, DateTime d2)
{
    int retVal = 0;

    // Calculate the number of years represented and multiply by 12
    // Substract the month number from the total
    // Substract the difference of the second month and 12 from the total
    retVal = (d1.Year - d2.Year) * 12;
    retVal = retVal - d1.Month;
    retVal = retVal - (12 - d2.Month);

    return retVal;
}

能够以月为单位计算 2 个日期之间的差异是一件完全合乎逻辑的事情，并且在许多业务应用程序中都是必需的。这里的几位编码员提供了诸如 - “2010 年 5 月 1 日”和“2010 年 6 月 16 日”之间有什么月份区别，2010 年 12 月 31 日和 2011 年 1 月 1 日之间有什么月份区别？-- 未能理解业务应用程序的基础知识。

以下是以上 2 条评论的答案 - 2010 年 5 月 1 日至 2010 年 6 月 16 日之间的月数为 1 个月，2010 年 12 月 31 日至 2011 年 1 月 1 日之间的月数为 0。正如上面的编码人员所建议的那样，将它们计算为 1.5 个月零 1 秒是非常愚蠢的。

从事过信用卡、抵押贷款处理、税务处理、租金处理、每月利息计算和各种其他业务解决方案的人会同意。

问题是这样的函数不包含在 C# 或 VB.NET 中。Datediff 只考虑年份或月份部分，因此实际上毫无用处。

以下是一些现实生活中的例子，说明您需要并正确计算月份：

您从2月18日至8月23日住在短期出租屋中。你在那里住了几个月？答案很简单 - 6个月

您有一个银行账户，每个月底都会计算和支付利息。您在 6 月 10 日存钱，并在 10 月 29 日（同年）取出。您获得多少个月的利息？非常简单的答案 - 4 个月（同样，额外的天数无关紧要）

在业务应用程序中，大多数时候，当你需要计算月份时，这是因为你需要根据人类计算时间的方式来了解“完整”的月份;不是基于一些抽象/无关紧要的想法。

你可以有一个这样的函数。

例如，从 2012/12/27 到 2012/12/29 变为 3 天。同样，从 2012/12/15 到 2013/01/15 变成了 2 个月，因为到 2013/01/14 是 1 个月。从15日开始，这是第二个月开始。

如果不想在计算中包括这两天，则可以删除第二个 if 条件中的“=”。即，从 2012/12/15 到 2013/01/15 是 1 个月。

public int GetMonths(DateTime startDate, DateTime endDate)
{
    if (startDate > endDate)
    {
        throw new Exception("Start Date is greater than the End Date");
    }

    int months = ((endDate.Year * 12) + endDate.Month) - ((startDate.Year * 12) + startDate.Month);

    if (endDate.Day >= startDate.Day)
    {
        months++;
    }

    return months;
}

public static int PayableMonthsInDuration(DateTime StartDate, DateTime EndDate)
{
    int sy = StartDate.Year; int sm = StartDate.Month; int count = 0;
    do
    {
        count++;if ((sy == EndDate.Year) && (sm >= EndDate.Month)) { break; }
        sm++;if (sm == 13) { sm = 1; sy++; }
    } while ((EndDate.Year >= sy) || (EndDate.Month >= sm));
    return (count);
}

此解决方案用于租赁/订阅计算，其中差异并不意味着减法，而是这两个日期内的跨度。

这适用于我需要它。就我而言，月份的某一天并不重要，因为它总是恰好是该月的最后一天。

public static int MonthDiff(DateTime d1, DateTime d2){
    int retVal = 0;

    if (d1.Month<d2.Month)
    {
        retVal = (d1.Month + 12) - d2.Month;
        retVal += ((d1.Year - 1) - d2.Year)*12;
    }
    else
    {
        retVal = d1.Month - d2.Month;
        retVal += (d1.Year - d2.Year)*12;
    }
    //// Calculate the number of years represented and multiply by 12
    //// Substract the month number from the total
    //// Substract the difference of the second month and 12 from the total
    //retVal = (d1.Year - d2.Year) * 12;
    //retVal = retVal - d1.Month;
    //retVal = retVal - (12 - d2.Month);

    return retVal;
}

我写了一个函数来实现这一点，因为其他方法对我不起作用。

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());
}

有 3 种情况：同年、上一年和其他年份。

如果这个月的日期无关紧要......

public int GetTotalNumberOfMonths(DateTime start, DateTime end)
{
    // work with dates in the right order
    if (start > end)
    {
        var swapper = start;
        start = end;
        end = swapper;
    }

    switch (end.Year - start.Year)
    {
        case 0: // Same year
            return end.Month - start.Month;

        case 1: // last year
            return (12 - start.Month) + end.Month;

        default:
            return 12 * (3 - (end.Year - start.Year)) + (12 - start.Month) + end.Month;
    }
}

使用 ToString（format） 和 Duration（long ms） 扩展了 Kirks 结构

 public struct DateTimeSpan
{
    private readonly int years;
    private readonly int months;
    private readonly int days;
    private readonly int hours;
    private readonly int minutes;
    private readonly int seconds;
    private readonly int milliseconds;

    public DateTimeSpan(int years, int months, int days, int hours, int minutes, int seconds, int milliseconds)
    {
        this.years = years;
        this.months = months;
        this.days = days;
        this.hours = hours;
        this.minutes = minutes;
        this.seconds = seconds;
        this.milliseconds = 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; } }

    enum Phase { Years, Months, Days, Done }


    public string ToString(string format)
    {
        format = format.Replace("YYYY", Years.ToString());
        format = format.Replace("MM", Months.ToString());
        format = format.Replace("DD", Days.ToString());
        format = format.Replace("hh", Hours.ToString());
        format = format.Replace("mm", Minutes.ToString());
        format = format.Replace("ss", Seconds.ToString());
        format = format.Replace("ms", Milliseconds.ToString());
        return format;
    }


    public static DateTimeSpan Duration(long ms)
    {
        DateTime dt = new DateTime();
        return CompareDates(dt, dt.AddMilliseconds(ms));
    }


    public static DateTimeSpan CompareDates(DateTime date1, DateTime date2)
    {
        if (date2 < date1)
        {
            var sub = date1;
            date1 = date2;
            date2 = sub;
        }

        DateTime current = date1;
        int years = 0;
        int months = 0;
        int days = 0;

        Phase phase = Phase.Years;
        DateTimeSpan span = new DateTimeSpan();

        while (phase != Phase.Done)
        {
            switch (phase)
            {
                case Phase.Years:
                    if (current.AddYears(years + 1) > date2)
                    {
                        phase = Phase.Months;
                        current = current.AddYears(years);
                    }
                    else
                    {
                        years++;
                    }
                    break;
                case Phase.Months:
                    if (current.AddMonths(months + 1) > date2)
                    {
                        phase = Phase.Days;
                        current = current.AddMonths(months);
                    }
                    else
                    {
                        months++;
                    }
                    break;
                case Phase.Days:
                    if (current.AddDays(days + 1) > date2)
                    {
                        current = current.AddDays(days);
                        var timespan = date2 - current;
                        span = new DateTimeSpan(years, months, days, timespan.Hours, timespan.Minutes, timespan.Seconds, timespan.Milliseconds);
                        phase = Phase.Done;
                    }
                    else
                    {
                        days++;
                    }
                    break;
            }
        }

        return span;
    }
}

关于这个问题没有太多明确的答案，因为你总是在假设事情。

此解决方案在两个日期之间计算假设您要保存月份中的某一天进行比较之间的月份（这意味着在计算中考虑了该月的某一天）

例如，如果您的日期是 2012 年 1 月 30 日，则 2012 年 2 月 29 日将不是一个月，而是 2013 年 3 月 1 日。

它已经过非常彻底的测试，可能会在我们以后使用它时清理它，但在这里：

private static int TotalMonthDifference(DateTime dtThis, DateTime dtOther)
{
    int intReturn = 0;
    bool sameMonth = false;

    if (dtOther.Date < dtThis.Date) //used for an error catch in program, returns -1
        intReturn--;

    int dayOfMonth = dtThis.Day; //captures the month of day for when it adds a month and doesn't have that many days
    int daysinMonth = 0; //used to caputre how many days are in the month

    while (dtOther.Date > dtThis.Date) //while Other date is still under the other
    {
        dtThis = dtThis.AddMonths(1); //as we loop, we just keep adding a month for testing
        daysinMonth = DateTime.DaysInMonth(dtThis.Year, dtThis.Month); //grabs the days in the current tested month

        if (dtThis.Day != dayOfMonth) //Example 30 Jan 2013 will go to 28 Feb when a month is added, so when it goes to march it will be 28th and not 30th
        {
            if (daysinMonth < dayOfMonth) // uses day in month max if can't set back to day of month
                dtThis.AddDays(daysinMonth - dtThis.Day);
            else
                dtThis.AddDays(dayOfMonth - dtThis.Day);
        }
        if (((dtOther.Year == dtThis.Year) && (dtOther.Month == dtThis.Month))) //If the loop puts it in the same month and year
        {
            if (dtOther.Day >= dayOfMonth) //check to see if it is the same day or later to add one to month
                intReturn++;
            sameMonth = true; //sets this to cancel out of the normal counting of month
        }
        if ((!sameMonth)&&(dtOther.Date > dtThis.Date))//so as long as it didn't reach the same month (or if i started in the same month, one month ahead, add a month)
            intReturn++;
    }
    return intReturn; //return month
}

  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));

要获得月份（包括开始和结束）的差异，无论日期如何：

DateTime start = new DateTime(2013, 1, 1);
DateTime end = new DateTime(2014, 2, 1);
var diffMonths = (end.Month + end.Year * 12) - (start.Month + start.Year * 12);

可以将 .NET 的时间段库的 DateDiff 类用于：

// ----------------------------------------------------------------------
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

我只是需要一些简单的东西来满足，例如只输入月/年的就业日期，所以想要不同的工作年份和月份。这就是我使用的，这里只是为了有用

public static YearsMonths YearMonthDiff(DateTime startDate, DateTime endDate) {
    int monthDiff = ((endDate.Year * 12) + endDate.Month) - ((startDate.Year * 12) + startDate.Month) + 1;
    int years = (int)Math.Floor((decimal) (monthDiff / 12));
    int months = monthDiff % 12;
    return new YearsMonths {
        TotalMonths = monthDiff,
            Years = years,
            Months = months
    };
}

.NET 小提琴

以下是我们处理此问题的方法：

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;
    }
}

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

使用野田时间：

LocalDate start = new LocalDate(2013, 1, 5);
LocalDate end = new LocalDate(2014, 6, 1);
Period period = Period.Between(start, end, PeriodUnits.Months);
Console.WriteLine(period.Months); // 16

（示例来源）

我对 2 个日期之间的总月差的理解有一个整体和小数部分（日期很重要）。

不可分割的部分是完整的月差。

对我来说，小数部分是开始月份和结束月份之间一天（到一个月的整天）的百分比之差。

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;
    }
}

使用此扩展，结果是：

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

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); 

最精确的方法是，以月为单位的分数返回差异：

private double ReturnDiffereceBetweenTwoDatesInMonths(DateTime startDateTime, DateTime endDateTime)
{
    double result = 0;
    double days = 0;
    DateTime currentDateTime = startDateTime;
    while (endDateTime > currentDateTime.AddMonths(1))
    {
        result ++;

        currentDateTime = currentDateTime.AddMonths(1);
    }

    if (endDateTime > currentDateTime)
    {
        days = endDateTime.Subtract(currentDateTime).TotalDays;

    }
    return result + days/endDateTime.GetMonthDays;
}

这是一个更简洁的解决方案，仅对年、月、日使用 VB.Net DateDiff。您也可以在 C# 中加载 DateDiff 库。

date1 必须是 <= date2

VB.NET

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);

LINQ 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();

这是一个简单的解决方案，至少对我有用。不过，它可能不是最快的，因为它在循环中使用了很酷的 DateTime 的 AddMonth 功能：

public static int GetMonthsDiff(DateTime start, DateTime end)
{
    if (start > end)
        return GetMonthsDiff(end, start);

    int months = 0;
    do
    {
        start = start.AddMonths(1);
        if (start > end)
            return months;

        months++;
    }
    while (true);
}

以下是我发现准确的月份差异的贡献：

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，并在 while 循环中应用与上述和 AddYears 完全相同的逻辑，而不是 AddMonths。

这是对柯克·沃尔（Kirk Woll）回答的回应。我还没有足够的声望点数来回复评论......

我喜欢 Kirk 的解决方案，并打算无耻地将其撕掉并在我的代码中使用它，但当我浏览它时，我意识到它太复杂了。不必要的切换和循环，以及使用起来毫无意义的公共构造函数。

这是我的重写：

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，几乎相同：

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);
}

用法2，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);
}

就我而言，需要计算从下个月开始日期到此日期前一天或从月初到月底的完整月份。

例：从 1/1/2018 到 31/1/2018 是一个完整的月份
Ex2：从 5/1/2018 到 4/2/2018 是一个完整的月

因此，基于此，这是我的解决方案：

public static DateTime GetMonthEnd(DateTime StartDate, int MonthsCount = 1)
{
    return StartDate.AddMonths(MonthsCount).AddDays(-1);
}
public static Tuple<int, int> CalcPeriod(DateTime StartDate, DateTime EndDate)
{
    int MonthsCount = 0;
    Tuple<int, int> Period;
    while (true)
    {
        if (GetMonthEnd(StartDate) > EndDate)
            break;
        else
        {
            MonthsCount += 1;
            StartDate = StartDate.AddMonths(1);
        }
    }
    int RemainingDays = (EndDate - StartDate).Days + 1;
    Period = new Tuple<int, int>(MonthsCount, RemainingDays);
    return Period;
}

用法：

Tuple<int, int> Period = CalcPeriod(FromDate, ToDate);

注意：就我而言，需要计算完整月份后的剩余天数，因此，如果不是您的情况，您可以忽略天数结果，甚至可以将方法返回从元组更改为整数。

简单修复。100%工作

        var exactmonth = (date1.Year - date2.Year) * 12 + date1.Month - 
        date2.Month +  (date1.Day >= date2.Day ? 0 : -1);
        Console.WriteLine(exactmonth);

public partial class Form1 : Form
{
    public Form1()
    {
        InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {

        label3.Text = new DateDifference(Convert.ToDateTime("2018-09-13"), Convert.ToDateTime("2018-11-15")).ToString();
        label2.Text = new DateDifference(Convert.ToDateTime("2018-10-12"), Convert.ToDateTime("2018-11-15")).ToString();

        DateDifference oDateDifference = new DateDifference(Convert.ToDateTime("2018-11-12"));
       label1.Text  =   oDateDifference.ToString();

    }
}




public class DateDifference
{
    public DateTime start { get; set; }
    public DateTime currentDAte { get; set; }
    public DateTime origstart { get; set; }
    public DateTime origCurrentDAte { get; set; }

    int days { get; set; }
    int months { get; set; }
    int years { get; set; }

    public DateDifference(DateTime postedDate, DateTime currentDAte)
    {
        this.start = this.removeTime(postedDate);
        this.currentDAte = this.removeTime(currentDAte);
        this.origstart = postedDate;
        this.origCurrentDAte = currentDAte;

    }

    public DateDifference(DateTime postedDate)
    {
        DateTime currentDate_ = DateTime.Now;
        this.start = this.removeTime(postedDate);
        this.currentDAte = this.removeTime(currentDate_);
        this.origstart = postedDate;
        this.origCurrentDAte = currentDate_;
        if (start > this.currentDAte)
        {
            throw new Exception("Current date is greater than date posted");
        }
        this.compute();
    }

    void compute()
    {
        while (this.start.Year <= this.currentDAte.Year)
        {
            if (this.start.Year <= this.currentDAte.Year && (this.start.AddMonths(1) <= this.currentDAte))
            {
                ++this.months;
                this.start = this.start.AddMonths(1);
            }

            if ((this.start.Year == this.currentDAte.Year) && (this.start >= this.currentDAte.AddMonths(-1) && this.start <= this.currentDAte))
            {
                break;
            }
        }

        while (this.start.DayOfYear < this.currentDAte.DayOfYear)
        {
            ++this.days;
            this.start = start.AddDays(1);
        }

        if (this.months > 11)
        {
            while (this.months > 11)
            {
                ++this.years;
                this.months = months - 12;
            }
        }

    }


    public override string ToString()
    {
        if (this.start > this.currentDAte)
        {
            throw new Exception("Current date is greater than date posted");
        }
        String ret = this.ComposeTostring();
        this.reset();
        return ret;
    }

    private String ComposeTostring()
    {
        this.compute();
        if (this.years > 0)
        {
            if (this.months > 0)
            {
                if (this.days > 0)
                {
                    return String.Format("{0} year{1}, {2} month{3} && {4} Day{5} ago", this.years, plural(this.years), this.months, plural(this.months), this.days, plural(this.days));
                }
                return String.Format("{0} year{1}, {2} month{3} ago", this.years, plural(this.years), this.months, plural(this.months));
            }
            else
            {
                if (this.days > 0)
                {
                    return String.Format("{0} year{1},{2} day{3} ago", this.years, plural(this.years), this.days, plural(this.days));
                }

                return String.Format("{0} year{1} ago", this.years, plural(this.years));

            }
        }

        if (this.months > 0)
        {
            if (this.days > 0)
            {
                return String.Format("{0} month{1}, {2} day{3} ago", this.months, plural(this.months), this.days, plural(this.days));
            }
            else
            {
                return String.Format("{0} month{1} ago", this.months, plural(this.months));
            }
        }

        if ((this.origCurrentDAte - this.origstart).Days > 0)
        {
            int daysDiff = (this.origCurrentDAte - this.origstart).Days;
            this.origstart = this.origstart.AddDays(daysDiff);
            int HoursDiff = (this.origCurrentDAte - this.origstart).Hours;
            return String.Format("{0} day{1}, {2} hour{3} ago", daysDiff, plural(daysDiff), HoursDiff, plural(HoursDiff));

        }
        else if ((this.origCurrentDAte - this.origstart).Hours > 0)
        {
            int HoursDiff = (this.origCurrentDAte - this.origstart).Hours;
            this.origstart = this.origstart.AddHours(HoursDiff);
            int MinDiff = (this.origCurrentDAte - this.origstart).Minutes;
            return String.Format("{0} hour{1}, {2} minute{3} ago", HoursDiff, plural(HoursDiff), MinDiff, plural(MinDiff));
        }
        else if ((this.origCurrentDAte - this.origstart).Minutes > 0)
        {

            int MinDiff = (this.origCurrentDAte - this.origstart).Minutes;
            this.origstart = this.origstart.AddMinutes(MinDiff);
            int SecDiff = (this.origCurrentDAte - this.origstart).Seconds;
            return String.Format("{0} minute{1}, {2} second{3} ago", MinDiff, plural(MinDiff), SecDiff, plural(SecDiff));
        }
        else if ((this.origCurrentDAte - this.origstart).Seconds > 0)
        {
            int sec = (this.origCurrentDAte - this.origstart).Seconds;
            return String.Format("{0} second{1}", sec, plural(sec));
        }

        return "";
    }

    String plural(int val)
    {
        return (val > 1 ? "s" : String.Empty);
    }

    DateTime removeTime(DateTime dtime)
    {
        dtime = dtime.AddHours(-dtime.Hour);
        dtime = dtime.AddMinutes(-dtime.Minute);
        dtime = dtime.AddSeconds(-dtime.Second);
        return dtime;
    }

    public void reset()
    {

        this.days = 0;
        this.months = 0;
        this.years = 0;
        this.start = DateTime.MinValue;
        this.currentDAte = DateTime.MinValue;
        this.origstart = DateTime.MinValue;
        this.origCurrentDAte = DateTime.MinValue;
    }
}

基于上面完成的出色 DateTimeSpan 工作，我对代码进行了一些规范化;这似乎效果很好：

public class DateTimeSpan
{
  private DateTimeSpan() { }

  private DateTimeSpan(int years, int months, int days, int hours, int minutes, int seconds, int milliseconds)
  {
    Years = years;
    Months = months;
    Days = days;
    Hours = hours;
    Minutes = minutes;
    Seconds = seconds;
    Milliseconds = milliseconds;
  }

  public int Years { get; private set; } = 0;
  public int Months { get; private set; } = 0;
  public int Days { get; private set; } = 0;
  public int Hours { get; private set; } = 0;
  public int Minutes { get; private set; } = 0;
  public int Seconds { get; private set; } = 0;
  public int Milliseconds { get; private set; } = 0;

  public static DateTimeSpan CompareDates(DateTime StartDate, DateTime EndDate)
  {
    if (StartDate.Equals(EndDate)) return new DateTimeSpan();
    DateTimeSpan R = new DateTimeSpan();
    bool Later;
    if (Later = StartDate > EndDate)
    {
      DateTime D = StartDate;
      StartDate = EndDate;
      EndDate = D;
    }

    // Calculate Date Stuff
    for (DateTime D = StartDate.AddYears(1); D < EndDate; D = D.AddYears(1), R.Years++) ;
    if (R.Years > 0) StartDate = StartDate.AddYears(R.Years);
    for (DateTime D = StartDate.AddMonths(1); D < EndDate; D = D.AddMonths(1), R.Months++) ;
    if (R.Months > 0) StartDate = StartDate.AddMonths(R.Months);
    for (DateTime D = StartDate.AddDays(1); D < EndDate; D = D.AddDays(1), R.Days++) ;
    if (R.Days > 0) StartDate = StartDate.AddDays(R.Days);

    // Calculate Time Stuff
    TimeSpan T1 = EndDate - StartDate;
    R.Hours = T1.Hours;
    R.Minutes = T1.Minutes;
    R.Seconds = T1.Seconds;
    R.Milliseconds = T1.Milliseconds;

    // Return answer. Negate values if the Start Date was later than the End Date
    if (Later)
      return new DateTimeSpan(-R.Years, -R.Months, -R.Days, -R.Hours, -R.Minutes, -R.Seconds, -R.Milliseconds);
    return R;
  }
}

简单快速的解决方案来计算 2 个日期之间的总月数。 如果您只想获得不同的月份，而不计算从日期中的月份 - 只需从代码中删除 +1 即可。

public static int GetTotalMonths(DateTime From, DateTime Till)
        {
            int MonthDiff = 0;

            for (int i = 0; i < 12; i++)
            {
                if (From.AddMonths(i).Month == Till.Month)
                {
                    MonthDiff = i + 1;
                    break;
                }
            }

            return MonthDiff;
        }

我的问题通过这个解决方案解决了：

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.
            //

        }

似乎 DateTimeSpan 解决方案让很多人感到高兴。我不知道。让我们考虑一下：

开始日期 = 1972/2/29 结束日期 = 1972/4/28。

基于 DateTimeSpan 的答案是：

1年2个月0天

我实现了一个方法，基于此，答案是：

1年1个月28天

显然，那里没有整整 2 个月。我想说的是，因为我们处于开始日期的月底，所以剩下的实际上是 3 月的完整月份加上结束日期月份（4 月）经过的天数，所以 1 个月零 28 天。

如果您到目前为止阅读并且很感兴趣，我在下面发布了该方法。我在评论中解释了我所做的假设，因为有多少个月，几个月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。我通常会选择相邻年份的考试日期，一旦我验证了答案，我就会来回移动一两天。到目前为止，它看起来不错，我相信你会:D发现一些错误。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;
        }
    }

这个简单的静态函数计算两个日期时间之间的月数，例如

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

该函数假定第一个日期小于第二个日期。为了处理负时间间隔，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);
 }

您可以使用Noda Time 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);

除了所有给出的答案之外，我发现这段代码非常简单。AS DateTime.MinValue 是 1/1/1，我们必须从月、年和日中减去 1。

var timespan = endDate.Subtract(startDate);
var tempdate = DateTime.MinValue + timespan;

var totalMonths = (tempdate.Year - 1) * 12 + tempdate.Month - 1;

var totalDays = tempdate.Day - 1;
if (totalDays > 0)
{
    totalMonths = totalMonths + 1;
}

疯狂的方法，计算所有的日子，如此超级精确

助手类：

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;
    }

如果您只关心月份和年份并想触摸这两个日期（例如，您想从 2021 年 1 月到 2022 年 AGO），您可以使用以下命令：

int numberOfMonths= (Year2 > Year1 ? ( Year2 - Year1 - 1) * 12 + (12 - Month1) + Month2 + 1 : Month2 - Month1 + 1); 

例：

Year1/Month1: 2021/10   
Year2/Month2: 2022/08   
numberOfMonths = 11;

或同年：

Year1/Month1: 2021/10   
Year2/Month2: 2021/12   
numberOfMonths = 3;

如果您只想触摸其中之一，请删除两个 + 1。

单线解决方案

首先，检查两个日期是否都在当前年份，如果不是，则获取整年的月份，然后从开始和结束年份添加月份。

DateTime dateFrom = new DateTime(2019, 2, 1);
DateTime dateTo = new DateTime(2021, 5, 25);

随着第一个月

var monthCount = dateFrom.Year != dateTo.Year ? ((dateTo.Year - dateFrom.Year - 1) * 12) + (13 - dateFrom.Month + dateTo.Month) : dateTo.Month - dateFrom.Month + 1;

结果 = 28

没有第一个月

monthCount = dateFrom.Year != dateTo.Year ? ((dateTo.Year - dateFrom.Year - 1) * 12) + (12 - dateFrom.Month + dateTo.Month) : dateTo.Month - dateFrom.Month;

结果 = 27

我正在做一个只用了几年和几个月的项目。

/// <summary>
/// Get the total months between two date.  This will count whole months and not care about the day.
/// </summary>
/// <param name="firstDate">First date.</param>
/// <param name="lastDate">Last date.</param>
/// <returns>Number of month apart.</returns>
private static int GetTotalMonths(DateOnly firstDate, DateOnly lastDate)
{
    int yearsApart = lastDate.Year - firstDate.Year;
    int monthsApart = lastDate.Month - firstDate.Month;
    return (yearsApart * 12) + monthsApart;
}

private static int GetTotalMonths(DateTime firstDate, DateTime lastDate)
{
    return GetTotalMonths(DateOnly.FromDateTime(firstDate), DateOnly.FromDateTime(lastDate));
}

一定是有人干的））

extension 方法返回给定日期之间的整月数。无论以何种顺序接收日期，都将始终返回一个自然数。没有像“正确”答案那样的近似计算。

    /// <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
    }

但是，如果您仍然想获得月份的小数部分，则只需在返回值中再添加一个项：

+ (next.Day - previous.Day) / DateTime.DaysInMonth(previous.Year, previous.Month)