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年でTimeSpanをフォーマットする

2つの日付プロパティFirstDayとLastDayのクラスがあります。 LastDayはNULL可能です。 "x year(s) y day(s)"の形式で文字列を生成したいと思います。合計年数が1未満の場合、年のセクションは省略します。合計日数が1日未満の場合は、日セクションを省略します。年数または日数のいずれかが0の場合、それぞれ「日/年」ではなく「日/年」と言う必要があります。

例：
2.2年： "2年73日"
1.002738年：「1年1日」
0.2年： "73日"
2年：「2年」

私が持っているものは機能しますが、それは長いです：

private const decimal DaysInAYear = 365.242M;

public string LengthInYearsAndDays
{
    get
    {
        var lastDay = this.LastDay ?? DateTime.Today;
        var lengthValue = lastDay - this.FirstDay;

        var builder = new StringBuilder();

        var totalDays = (decimal)lengthValue.TotalDays;
        var totalYears = totalDays / DaysInAYear;
        var years = (int)Math.Floor(totalYears);

        totalDays -= (years * DaysInAYear);
        var days = (int)Math.Floor(totalDays);

        Func<int, string> sIfPlural = value =>
            value > 1 ? "s" : string.Empty;

        if (years > 0)
        {
            builder.AppendFormat(
                CultureInfo.InvariantCulture,
                "{0} year{1}",
                years,
                sIfPlural(years));

            if (days > 0)
            {
                builder.Append(" ");
            }
        }

        if (days > 0)
        {
            builder.AppendFormat(
                CultureInfo.InvariantCulture,
                "{0} day{1}",
                days,
                sIfPlural(days));
        }

        var length = builder.ToString();
        return length;
    }
}

これを行うためのより簡潔な方法はありますか？

c#.netformatting.net-4.5timespan
21
Dan

TimeSpanは、開始点と終了点に依存するため、「年」という賢明な概念はありません。 （月も同様です-29日で何ヶ月ありますか？まあ、それは状況によります...）

恥知らずなプラグインを与えるために、私の Noda Time プロジェクトはこれを本当に簡単にします：

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);
    }
}

出力：

Between 19 June 2010 and 11 April 2013 are 2 years and 296 days
45
Jon Skeet
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);
}
7
Jens Borrisholt

TimeSpanを使用してこれを行うことはありません。 1か月の日数と1年の日数が一定ではなくなるため、日付の計算は、日数を超えるとすぐに難しくなります。 TimeSpanにYearsとMonthsのプロパティが含まれていないのは、おそらくそのためです。代わりに、2つのDateTime値の間の年/月/日などの数を決定し、それに応じて結果を表示します。

1
D Stanley

私はこれがうまくいくと思います：

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;
}
0
user3469676
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
0
Mark