两个日期之间的月差

How to calculate the difference in months between two dates in C#?

如何在c#中计算两个日期之间的月差?

Is there is equivalent of VB's DateDiff() method in C#. I need to find difference in months between two dates that are years apart. The documentation says that I can use TimeSpan like:

在c#中是否有等价于VB的DateDiff()方法。我需要在相隔数年的两次约会中找出不同的月份。文档说明我可以使用TimeSpan:

TimeSpan ts = date1 - date2;

but this gives me data in Days. I don't want to divide this number by 30 because not every month is 30 days and since the two operand values are quite apart from each other, I am afraid dividing by 30 might give me a wrong value.

但这给了我几天的数据。我不想把这个数除以30因为不是每个月都是30天因为这两个操作数的值相差很大，我担心除以30会给我一个错误的值。

Any suggestions?

有什么建议吗?

32 个解决方案

#1

Assuming the day of the month is irrelevant (i.e. the diff between 2011.1.1 and 2010.12.31 is 1), with date1 > date2 giving a positive value and date2 > date1 a negative value

假设月份的日期无关(即2011.1.1 - 2010.12.31之间的差异为1)，date1 > date2为正值，date2 > date1为负值

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

Or, assuming you want an approximate number of 'average months' between the two dates, the following should work for all but very huge date differences.

或者，假设您想要两个日期之间的“平均月数”，下面的方法应该适用于几乎所有的日期差异。

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

Note, if you were to use the latter solution then your unit tests should state the widest date range which your application is designed to work with and validate the results of the calculation accordingly.

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用于处理的最宽的日期范围，并相应地验证计算结果。

---

Update (with thanks to Gary)

更新(感谢Gary)

If using the 'average months' method, a slightly more accurate number to use for the 'average number of days per year' is 365.2425.

如果使用“平均月份”方法，对“每年平均天数”使用的更准确的数字是365.2425。

#2

Here is a comprehensive solution to return a DateTimeSpan, similar to a TimeSpan, except that it includes all the date components in addition to the time components.

这里有一个全面的解决方案来返回DateTimeSpan，类似于TimeSpan，只是它除了time组件之外还包含所有的date组件。

Usage:

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);
}

Outputs:

输出:

Years: 1
Months: 5
Days: 27
Hours: 1
Minutes: 36
Seconds: 50
Milliseconds: 0

年:1个月:5天:27小时:1分钟:36秒:50毫秒:0

For convenience, I've lumped the logic into the DateTimeSpan struct, but you may move the method CompareDates wherever you see fit. Also note, it doesn't matter which date comes before the other.

为了方便起见，我将逻辑集中到DateTimeSpan结构中，但是您可以将方法CompareDates移动到您认为合适的地方。另外要注意的是，哪一个日期在另一个日期之前是不重要的。

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 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;
    }
}

#3

If you want the exact number of full months, always positive (2000-01-15, 2000-02-14 returns 0), considering a full month is when you reach the same day the next month (something like the age calculation)

如果你想要确切的月份数，总是正值(2000-01-15,2000-02-14返回0)，考虑到一个完整的月份是你在下个月到达同一天(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;
    }
}

Edit reason: the old code was not correct in some cases like :

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 },
};

#4

You could do

你可以做

if ( date1.AddMonths(x) > date2 )

#5

I checked the usage of this method in VB.NET via MSDN and it seems that it has a lot of usages. There is no such a built-in method in C#. (Even it's not a good idea) you can call VB's in C#.

我检查了这个方法在VB中的用法。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。

1. Add Microsoft.VisualBasic.dll to your project as a reference
2. 添加Microsoft.VisualBasic。dll到您的项目作为参考。
3. use Microsoft.VisualBasic.DateAndTime.DateDiff in your code
4. 使用Microsoft.VisualBasic.DateAndTime。DateDiff在代码中

#6

To get difference in months (both start and end inclusive), irrespective of dates:

以月为单位(包括开始和结束)，不论日期:

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);

#7

I just needed something simple to cater for e.g. employment dates where only the month/year is entered, so wanted distinct years and months worked in. This is what I use, here for usefullness only

我只是需要一些简单的东西来满足，例如，工作日期只输入月份/年，所以想要明确的年份和月份。这是我使用的，这里仅用于有用

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 Fiddle

net小提琴

#8

You can use the DateDiff class of the Time Period Library for .NET:

你可以使用。net的时间周期库的DateDiff类:

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

#9

Use Noda Time:

野田佳彦使用时间:

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

(example source)

(例子)

#10

This worked for what I needed it for. The day of month didn't matter in my case because it always happens to be the last day of the month.

这是我需要的。在我的情况下，一个月的一天并不重要，因为它总是恰好是一个月的最后一天。

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;
}

#11

The most precise way is this that return difference in months by fraction :

最精确的方法是将月差除以分数:

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;
}

#12

Here is a simple solution that works at least for me. It's probably not the fastest though because it uses the cool DateTime's AddMonth feature in a loop:

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cool DateTime的AddMonth特性:

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);
}

#13

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

#14

This is from my own library, will return the difference of months between two dates.

这是我自己的图书馆，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;
}

#15

You can have a function something like this.

你可以有一个像这样的函数。

For Example, from 2012/12/27 to 2012/12/29 becomes 3 days. Likewise, from 2012/12/15 to 2013/01/15 becomes 2 months, because up to 2013/01/14 it's 1 month. from 15th it's 2nd month started.

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

You can remove the "=" in the second if condition, if you do not want to include both days in the calculation. i.e, from 2012/12/15 to 2013/01/15 is 1 month.

如果不希望在计算中同时包含这两天，可以在第二个if条件中删除“=”。我。e，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;
}

#16

you can use the following extension: Code

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
}

Implementation !

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); 

#17

Here's a much more concise solution using VB.Net DateDiff for Year, Month, Day only. You can load the DateDiff library in C# as well.

这里有一个使用VB的更简洁的解决方案。净日期为年，月，天。您也可以在c#中加载DateDiff库。

date1 must be <= date2

date1必须是<= date2。

VB.NET

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#

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);

#18

Here is my contribution to get difference in Months that I've found to be accurate:

以下是我的贡献，在我发现是准确的几个月的差异:

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;
        }
    }
}

Usage:

用法:

Int32 months = DateTime.Now.DiffMonths( DateTime.Now.AddYears( 5 ) );

You can create another method called DiffYears and apply exactly the same logic as above and AddYears instead of AddMonths in the while loop.

您可以创建另一个名为DiffYears的方法，并与上面的逻辑完全相同，并在while循环中使用AddYears而不是AddMonths。

#19

This is in response to Kirk Woll's answer. I don't have enough reputation points to reply to a comment yet...

这是对柯克·沃尔的回答的回应。我还没有足够的声誉点来回复评论……

I liked Kirk's solution and was going to shamelessly rip it off and use it in my code, but when I looked through it I realized it's way too complicated. Unnecessary switching and looping, and a public constructor that is pointless to use.

我喜欢柯克的解决方案，打算无耻地把它撕下来，在我的代码中使用它，但当我查看它时，我发现它太复杂了。不必要的切换和循环，以及毫无意义的公共构造函数。

Here's my rewrite:

这是我重写:

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);
    }
}

Usage1, pretty much the same:

Usage1，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, similar:

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);
}

#20

In my case it is required to calculate the complete month from the start date to the day prior to this day in the next month or from start to end of month.

在我的情况下，需要计算从开始日期到下个月的前一天或从开始到月底的完整月份。

Ex: from 1/1/2018 to 31/1/2018 is a complete month
Ex2: from 5/1/2018 to 4/2/2018 is a complete month

2018年1月1日至2018年1月31日为完整月Ex2: 2018年5月1日至4月2日为完整月

so based on this here is my solution:

基于此，我的解决方案是:

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;
}

Usage:

用法:

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

Note: in my case it was required to calculate the remaining days after the complete months so if it's not your case you could ignore the days result or even you could change the method return from tuple to integer.

注意:在我的例子中，需要计算完整个月后的剩余天数，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值从tuple改为integer。

#21

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);
}

This solution is for Rental/subscription calculation, where difference doesn't means to be subtraction, it's meant to be the span in within those two dates.

这个解决方案用于租金/订阅计算，其中的差异并不意味着减法，而是指在这两个日期内的跨度。

#22

There's 3 cases: same year, previous year and other years.

有3个案例:同一年，前一年和其他年份。

If the day of the month does not matter...

如果这个月的哪一天不重要……

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;
    }
}

#23

I wrote a function to accomplish this, because the others ways weren't working for me.

我编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());
}

#24

My understanding of the total months difference between 2 dates has an integral and a fractional part (the date matters).

我对两个日期的月差的理解有一个积分和一个小数部分(日期很重要)。

The integral part is the full months difference.

完整的部分是完整的月差。

The fractional part, for me, is the difference of the % of the day (to the full days of month) between the starting and ending months.

对我来说，分数部分是开始月份和结束月份之间的日(到月满日)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;
    }
}

With this extension, those are the results:

通过这个扩展，这些就是结果:

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

#25

There are not a lot of clear answers on this because you are always assuming things.

关于这个问题没有很多明确的答案，因为你总是在假设。

This solution calculates between two dates the months between assuming you want to save the day of month for comparison, (meaning that the day of the month is considered in the calculation)

该解决方案计算两个日期之间的月份，假设您要保存月份的日期以便进行比较(这意味着在计算中考虑月份的日期)

Example, if you have a date of 30 Jan 2012, 29 Feb 2012 will not be a month but 01 March 2013 will.

例如，如果你的日期是2012年1月30日，2012年2月29日不是一个月，而是2013年3月1日。

It's been tested pretty thoroughly, probably will clean it up later as we use it, but here:

它经过了非常彻底的测试，以后我们使用的时候可能会把它清理干净，但是这里:

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
}

#26

To be able to calculate the difference between 2 dates in months is a perfectly logical thing to do, and is needed in many business applications. The several coders here who have provided comments such as - what's the difference in months between "May 1,2010" and "June 16,2010, what's the difference in months between 31 December 2010 and 1 Jan 2011? -- have failed to understand the very basics of business applications.

能够计算两个日期之间的月差是一件非常合理的事情，在许多业务应用程序中都需要这样做。这里的几位程序员发表了如下评论:“2010年5月1日”和“2010年6月16日”之间的月份差异是什么? 2010年12月31日和2011年1月1日之间的月份差异是什么?——未能理解业务应用程序的基本知识。

Here is the answer to the above 2 comments - The number of months between 1-may-2010 and 16-jun-2010 is 1 month, the number of months between 31-dec-2010 and 1-jan-2011 is 0. It would be very foolish to calculate them as 1.5 months and 1 second, as the coders above have suggested.

以下是以上两点评论的答案——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。就像上面的编码员所说的那样，将它们计算为1.5个月零1秒是非常愚蠢的。

People who have worked on credit card, mortgage processing, tax processing, rent processing, monthly interest calculations and a vast variety of other business solutions would agree.

那些曾从事过信用卡、抵押贷款处理、税务处理、租金处理、月度利息计算以及各种各样的商业解决方案的人会同意这一观点。

Problem is that such a function is not included in C# or VB.NET for that matter. Datediff only takes into account years or the month component, so is actually useless.

问题是c#或VB中不包含这样的函数。净。Datediff只考虑年份或月份，因此实际上是无用的。

Here are some real-life examples of where you need to and correctly can calculate months:

以下是一些现实生活中你需要并且正确计算月份的例子:

You lived in a short-term rental from 18-feb to 23-aug. How many months did you stay there? The answer is a simple - 6 months

从2月18日到8月23日，你住的是短期租房。你在那里呆了几个月?答案很简单——6个月

You have a bank acount where interest is calculated and paid at the end of every month. You deposit money on 10-jun and take it out 29-oct (same year). How many months do you get interest for? Very simple answer- 4 months (again the extra days do not matter)

你有一份银行帐单，利息在每个月底计算和支付。你把钱存到10月10日，10月29日(同年)取出。你的利息是多少个月?非常简单的答案- 4个月(额外的天数并不重要)

In business applications, most of the time, when you need to calculate months, it is because you need to know 'full' months based on how humans calculate time; not based on some abstract/irrelevant thoughts.

在业务应用程序中，大多数情况下，当您需要计算月份时，是因为您需要根据人类计算时间的方式知道“完整”月份;不是基于一些抽象的/不相关的想法。

#27

Expanded Kirks struct with ToString(format) and Duration(long ms)

用ToString(格式)和持续时间(long ms)扩展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;
    }
}

#28

  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));

#29

Here's how we approach this:

以下是我们的方法:

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;
    }
}

#30

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