两个日期之间的月差

最精确的方法是以月为单位的分数返回差值:

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;
}

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必须为<= 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);

LINQ的解决方案,

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();

这里有一个简单的解决方案，至少对我来说是有效的。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);
}

以下是我对获得Months差异的贡献，我发现这是准确的:

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，并应用与上面完全相同的逻辑，并在while循环中使用AddYears而不是AddMonths。

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);
}

在我的情况下，需要计算从开始日期到下个月这一天的前一天或从月初到月底的完整月份。

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

基于此，我的解决方案如下:

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:在我的情况下，需要计算完整月份之后的剩余天数，所以如果不是你的情况，你可以忽略天数结果，甚至可以将方法返回值从元组更改为整数。

简单的修复。工作的100%

        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;
}
}

基于上面出色的DateTimeSpan工作，我将代码规范化了一些;这似乎很有效:

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;
}
}

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.
//


}

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 EndDate = 1972/4/28。

基于DateTimeSpan的答案是:

1年、2个月及0天

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;
}
}

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);
}

你可以使用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，我发现这段代码非常简单。DateTime。MinValue是1/1/1，我们必须从月，年和日减去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，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;
}

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。

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，则获取全年的月份，然后从年初和年末添加月份。

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

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

我当时在做一个项目，只需要几年和几个月。

/// <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))

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。没有近似计算，如“正确”;的答案。

    /// <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:

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