用年份格式化时间跨度

TimeSpan没有"年"的合理概念，因为它取决于起点和终点。(月也是一样——29天里有几个月?嗯，这要看……)

为了给一个无耻的宣传，我的野田时间项目让它变得非常简单:

using System;
using NodaTime;
public class Test
{
    static void Main(string[] args)
    {
        LocalDate start = new LocalDate(2010, 6, 19);
        LocalDate end = new LocalDate(2013, 4, 11);
        Period period = Period.Between(start, end,
                                       PeriodUnits.Years | PeriodUnits.Days);
        Console.WriteLine("Between {0} and {1} are {2} years and {3} days",
                          start, end, period.Years, period.Days);
    }
}

Between 19 June 2010 and 11 April 2013 are 2 years and 296 days

public string GetAgeText(DateTime birthDate)
{
        const double ApproxDaysPerMonth = 30.4375;
        const double ApproxDaysPerYear = 365.25;
        /*
        The above are the average days per month/year over a normal 4 year period
        We use these approximations as they are more accurate for the next century or so
        After that you may want to switch over to these 400 year approximations
           ApproxDaysPerMonth = 30.436875
           ApproxDaysPerYear  = 365.2425 
          How to get theese numbers:
            The are 365 days in a year, unless it is a leepyear.
            Leepyear is every forth year if Year % 4 = 0
            unless year % 100 == 1
            unless if year % 400 == 0 then it is a leep year.
            This gives us 97 leep years in 400 years. 
            So 400 * 365 + 97 = 146097 days.
            146097 / 400      = 365.2425
            146097 / 400 / 12 = 30,436875
        Due to the nature of the leap year calculation, on this side of the year 2100
        you can assume every 4th year is a leap year and use the other approximatiotions
        */
    //Calculate the span in days
    int iDays = (DateTime.Now - birthDate).Days;
    //Calculate years as an integer division
    int iYear = (int)(iDays / ApproxDaysPerYear);
    //Decrease remaing days
    iDays -= (int)(iYear * ApproxDaysPerYear);
    //Calculate months as an integer division
    int iMonths = (int)(iDays / ApproxDaysPerMonth);
    //Decrease remaing days
    iDays -= (int)(iMonths * ApproxDaysPerMonth);
    //Return the result as an string   
    return string.Format("{0} years, {1} months, {2} days", iYear, iMonths, iDays);
}

我不会这样做与TimeSpan。一旦超过天数，日期计算就会变得棘手，因为一个月和一年的天数不再是恒定的。这可能就是为什么TimeSpan不包含Years和Months的属性。相反，我将确定两个DateTime值之间的年/月/日等数，并相应地显示结果。

我想这应该行得通:

public static int DiffYears(DateTime dateValue1, DateTime dateValue2)
{
    var intToCompare1 = Convert.ToInt32(dateValue1.ToString("yyyyMMdd"));
    var intToCompare2 = Convert.ToInt32(dateValue2.ToString("yyyyMMdd"));
    return (intToCompare2 - intToCompare1) / 10000;
}

Public Function TimeYMDBetween(StartDate As DateTime, EndDate As DateTime) As String
    Dim Years As Integer = EndDate.Year - StartDate.Year
    Dim Months As Integer = EndDate.Month - StartDate.Month
    Dim Days As Integer = EndDate.Day - StartDate.Day
    Dim DaysLastMonth As Integer
    'figure out how many days were in last month
    If EndDate.Month = 1 Then
        DaysLastMonth = DateTime.DaysInMonth(EndDate.Year - 1, 12)
    Else
        DaysLastMonth = DateTime.DaysInMonth(EndDate.Year, EndDate.Month - 1)
    End If
    'adjust for negative days
    If Days < 0 Then
        Months = Months - 1
        Days = Days + DaysLastMonth 'borrowing from last month
    End If
    'adjust for negative Months
    If Months < 0 Then 'startdate hasn't happend this year yet
        Years = Years - 1
        Months = Months + 12
    End If
    Return Years.ToString() + " Years, " + Months.ToString() + " Months and " + Days.ToString() + " Days"
End Function

我需要在Core 3中这样做。NodaTime似乎依赖于框架4.7.2。我编写了下面的方法，它似乎可以将时间范围格式化为年、月和日，省略不需要的部分。

public static string ToYearsMonthsAndDays(this TimeSpan span)
    {
        var result = string.Empty;
        var totalYears = span.Days / 364.25;
        var fullYears = Math.Floor(totalYears);
        var totalMonths = (span.Days - (365.24 * fullYears)) / 30;
        var fullMonths = Math.Floor(totalMonths);
        var totalDays = (span.Days - (365.24 * totalYears) - (30 * fullMonths)) / 30;
        var fullDays = Math.Floor(totalDays);
        var sb = new StringBuilder();
        if (fullYears > 0)
        {
            if (sb.Length > 0)
                sb.Append(", ");
            sb.Append(fullYears + "y");
        }
        if (fullMonths > 0)
        {
            if (sb.Length > 0)
                sb.Append(", ");
            sb.Append(fullMonths + "m");
        }
        if (fullDays > 0)
        {
            if (sb.Length > 0)
                sb.Append(", ");
            sb.Append(fullDays + "d");
        }
        return sb.ToString();
    }