如何在c#中计算两个日期之间的月差?

c#中是否有相当于VB的DateDiff()方法。我需要找出相隔数年的两个日期之间的月差。文档说我可以像这样使用TimeSpan:

TimeSpan ts = date1 - date2;

但这里的数据是以天为单位的。我不想把这个数字除以30,因为不是每个月都是30天,而且两个操作数的值相差很大,所以我担心除以30可能会得到错误的值。

有什么建议吗?


当前回答

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

其他回答

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);
}

这个解决方案是用于租金/订阅计算的,其中的差异并不意味着减法,它意味着这两个日期之间的跨度。

要获得月份的差异(包括开始和结束),而不考虑日期:

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);

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

(例子)

我对两个日期之间总月差的理解有一个整数部分和一个小数部分(日期很重要)。

积分部分是整个月的差额。

对我来说,小数部分是开始月份和结束月份之间一天的百分比(到一个月的全部天数)的差值。

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;
    }
}

有了这个扩展,这些是结果:

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

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);
 }