用年份格式化 TimeSpan

Question

我有一个具有 2 个日期属性的类：FirstDay 和 LastDay。 LastDay 可以为空。我想生成一个"x year(s) y day(s)" 格式的字符串。如果总年份小于 1，我想省略年份部分。如果总天数小于 1，我想省略天部分。如果年或日为 0，则应分别表示“日/年”，而不是“日/年”。

示例：
2.2 年：             “2 年 73 天”
1.002738 年：“1 年 1 天”
0.2 年：             “73 天”
2 年：                “2 年”

我有什么作品，但是很长：

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;
    }
}

有没有更简洁的方法来做到这一点（但仍然可读）？

Answer 1

TimeSpan 没有合理的“年”概念，因为它取决于起点和终点。 （月份类似 - 29 天有多少个月？嗯，这取决于...）

为了给一个无耻的插件，我的Noda Time 项目让这变得非常简单：

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);
    }
}

输出：

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

Answer 2

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);
}

Answer 3

我不会用TimeSpan 这样做。一旦超过天数，日期数学就会变得棘手，因为一个月中的天数和一年中的天数不再是恒定的。这可能是TimeSpan 不包含Years 和Months 的属性的原因。相反，我会确定两个 DateTime 值之间的年数/月数/天数等，并相应地显示结果。

Answer 4

我认为这应该可行：

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;
}

Answer 5

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

Answer 6

我需要为 Core 3 执行此操作。NodaTime 似乎依赖于 Framework 4.7.2。我编写了以下方法，该方法似乎可以将时间跨度格式化为年、月和日，省略了不需要的部分。

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();
    }