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

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));
}

其他回答

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)

假设这个月的日期不相关(即2011.1.1和2010.12.31之间的差为1),date1 > date2为正值,date2 > date1为负值

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

或者,假设你想要两个日期之间的“平均月”的大致数字,下面的方法应该适用于所有日期,但日期差异非常大。

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

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“平均月份”方法,“每年平均天数”的更准确数字是365.2425。

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

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

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

这是我所需要的。对我来说,一个月的哪一天并不重要,因为它总是碰巧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;
}