给定代表某人生日的DateTime,我如何计算他们的年龄(以年为单位)?


当前回答

这不是一个直接的答案,但更多的是从准科学的角度对当前问题进行哲学推理。

我认为,这个问题并没有具体说明衡量年龄的单位或文化,大多数答案似乎都假设了一个整数年表示。时间的国际单位制单位是秒,因此正确的通用答案应该是(当然,假设标准化日期时间,不考虑相对论效应):

var lifeInSeconds = (DateTime.Now.Ticks - then.Ticks)/TickFactor;

在基督教以年计算年龄的方法中:

var then = ... // Then, in this case the birthday
var now = DateTime.UtcNow;
int age = now.Year - then.Year;
if (now.AddYears(-age) < then) age--;

在金融领域,当计算通常被称为日计数分数(Day Count Fraction)的东西时,也存在类似的问题,该分数大致是给定时期的年数。年龄问题确实是一个衡量时间的问题。

实际/实际(正确计算所有天数)惯例示例:

DateTime start, end = .... // Whatever, assume start is before end

double startYearContribution = 1 - (double) start.DayOfYear / (double) (DateTime.IsLeapYear(start.Year) ? 366 : 365);
double endYearContribution = (double)end.DayOfYear / (double)(DateTime.IsLeapYear(end.Year) ? 366 : 365);
double middleContribution = (double) (end.Year - start.Year - 1);

double DCF = startYearContribution + endYearContribution + middleContribution;

另一种很常见的衡量时间的方法通常是“序列化”(命名这一日期惯例的家伙一定是认真的“trippin”):

DateTime start, end = .... // Whatever, assume start is before end
int days = (end - start).Days;

我想知道,在相对论年龄(以秒为单位)变得比迄今为止地球围绕太阳周期的粗略近似更有用之前,我们还需要多长时间:)或者换句话说,当一个周期必须给定一个位置或一个表示其自身运动的函数才能有效时:)

其他回答

我认为TimeSpan包含了我们所需要的一切,而不必求助于365.25(或任何其他近似值)。扩展Aug的示例:

DateTime myBD = new DateTime(1980, 10, 10);
TimeSpan difference = DateTime.Now.Subtract(myBD);

textBox1.Text = difference.Years + " years " + difference.Months + " Months " + difference.Days + " days";

我使用ScArcher2的解决方案来精确计算人的年龄,但我需要进一步计算他们的月和日以及年。

    public static Dictionary<string,int> CurrentAgeInYearsMonthsDays(DateTime? ndtBirthDate, DateTime? ndtReferralDate)
    {
        //----------------------------------------------------------------------
        // Can't determine age if we don't have a dates.
        //----------------------------------------------------------------------
        if (ndtBirthDate == null) return null;
        if (ndtReferralDate == null) return null;

        DateTime dtBirthDate = Convert.ToDateTime(ndtBirthDate);
        DateTime dtReferralDate = Convert.ToDateTime(ndtReferralDate);

        //----------------------------------------------------------------------
        // Create our Variables
        //----------------------------------------------------------------------
        Dictionary<string, int> dYMD = new Dictionary<string,int>();
        int iNowDate, iBirthDate, iYears, iMonths, iDays;
        string sDif = "";

        //----------------------------------------------------------------------
        // Store off current date/time and DOB into local variables
        //---------------------------------------------------------------------- 
        iNowDate = int.Parse(dtReferralDate.ToString("yyyyMMdd"));
        iBirthDate = int.Parse(dtBirthDate.ToString("yyyyMMdd"));

        //----------------------------------------------------------------------
        // Calculate Years
        //----------------------------------------------------------------------
        sDif = (iNowDate - iBirthDate).ToString();
        iYears = int.Parse(sDif.Substring(0, sDif.Length - 4));

        //----------------------------------------------------------------------
        // Store Years in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Years", iYears);

        //----------------------------------------------------------------------
        // Calculate Months
        //----------------------------------------------------------------------
        if (dtBirthDate.Month > dtReferralDate.Month)
            iMonths = 12 - dtBirthDate.Month + dtReferralDate.Month - 1;
        else
            iMonths = dtBirthDate.Month - dtReferralDate.Month;

        //----------------------------------------------------------------------
        // Store Months in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Months", iMonths);

        //----------------------------------------------------------------------
        // Calculate Remaining Days
        //----------------------------------------------------------------------
        if (dtBirthDate.Day > dtReferralDate.Day)
            //Logic: Figure out the days in month previous to the current month, or the admitted month.
            //       Subtract the birthday from the total days which will give us how many days the person has lived since their birthdate day the previous month.
            //       then take the referral date and simply add the number of days the person has lived this month.

            //If referral date is january, we need to go back to the following year's December to get the days in that month.
            if (dtReferralDate.Month == 1)
                iDays = DateTime.DaysInMonth(dtReferralDate.Year - 1, 12) - dtBirthDate.Day + dtReferralDate.Day;       
            else
                iDays = DateTime.DaysInMonth(dtReferralDate.Year, dtReferralDate.Month - 1) - dtBirthDate.Day + dtReferralDate.Day;       
        else
            iDays = dtReferralDate.Day - dtBirthDate.Day;             

        //----------------------------------------------------------------------
        // Store Days in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Days", iDays);

        return dYMD;
}

对此的简单答案是应用AddYears,如下所示,因为这是唯一一种将年份添加到闰年2月29日的本地方法,并获得普通年份2月28日的正确结果。

有些人认为3月1日是勒普林斯的生日,但.Net和任何官方规则都不支持这一点,也没有常见的逻辑解释为什么一些出生在2月的人应该在另一个月拥有75%的生日。

此外,Age方法可以作为DateTime的扩展添加。由此,您可以以最简单的方式获得年龄:

列表项目

int age=出生日期.age();

public static class DateTimeExtensions
{
    /// <summary>
    /// Calculates the age in years of the current System.DateTime object today.
    /// </summary>
    /// <param name="birthDate">The date of birth</param>
    /// <returns>Age in years today. 0 is returned for a future date of birth.</returns>
    public static int Age(this DateTime birthDate)
    {
        return Age(birthDate, DateTime.Today);
    }

    /// <summary>
    /// Calculates the age in years of the current System.DateTime object on a later date.
    /// </summary>
    /// <param name="birthDate">The date of birth</param>
    /// <param name="laterDate">The date on which to calculate the age.</param>
    /// <returns>Age in years on a later day. 0 is returned as minimum.</returns>
    public static int Age(this DateTime birthDate, DateTime laterDate)
    {
        int age;
        age = laterDate.Year - birthDate.Year;

        if (age > 0)
        {
            age -= Convert.ToInt32(laterDate.Date < birthDate.Date.AddYears(age));
        }
        else
        {
            age = 0;
        }

        return age;
    }
}

现在,运行此测试:

class Program
{
    static void Main(string[] args)
    {
        RunTest();
    }

    private static void RunTest()
    {
        DateTime birthDate = new DateTime(2000, 2, 28);
        DateTime laterDate = new DateTime(2011, 2, 27);
        string iso = "yyyy-MM-dd";

        for (int i = 0; i < 3; i++)
        {
            for (int j = 0; j < 3; j++)
            {
                Console.WriteLine("Birth date: " + birthDate.AddDays(i).ToString(iso) + "  Later date: " + laterDate.AddDays(j).ToString(iso) + "  Age: " + birthDate.AddDays(i).Age(laterDate.AddDays(j)).ToString());
            }
        }

        Console.ReadKey();
    }
}

关键日期示例如下:

出生日期:2000-02-29出生日期:2011-02-28年龄:11

输出:

{
    Birth date: 2000-02-28  Later date: 2011-02-27  Age: 10
    Birth date: 2000-02-28  Later date: 2011-02-28  Age: 11
    Birth date: 2000-02-28  Later date: 2011-03-01  Age: 11
    Birth date: 2000-02-29  Later date: 2011-02-27  Age: 10
    Birth date: 2000-02-29  Later date: 2011-02-28  Age: 11
    Birth date: 2000-02-29  Later date: 2011-03-01  Age: 11
    Birth date: 2000-03-01  Later date: 2011-02-27  Age: 10
    Birth date: 2000-03-01  Later date: 2011-02-28  Age: 10
    Birth date: 2000-03-01  Later date: 2011-03-01  Age: 11
}

2012年2月28日晚些时候:

{
    Birth date: 2000-02-28  Later date: 2012-02-28  Age: 12
    Birth date: 2000-02-28  Later date: 2012-02-29  Age: 12
    Birth date: 2000-02-28  Later date: 2012-03-01  Age: 12
    Birth date: 2000-02-29  Later date: 2012-02-28  Age: 11
    Birth date: 2000-02-29  Later date: 2012-02-29  Age: 12
    Birth date: 2000-02-29  Later date: 2012-03-01  Age: 12
    Birth date: 2000-03-01  Later date: 2012-02-28  Age: 11
    Birth date: 2000-03-01  Later date: 2012-02-29  Age: 11
    Birth date: 2000-03-01  Later date: 2012-03-01  Age: 12
}

我想添加希伯来文日历计算(或其他系统。全球化日历可以以相同的方式使用),使用此线程中的重写函数:

Public Shared Function CalculateAge(BirthDate As DateTime) As Integer
    Dim HebCal As New System.Globalization.HebrewCalendar ()
    Dim now = DateTime.Now()
    Dim iAge = HebCal.GetYear(now) - HebCal.GetYear(BirthDate)
    Dim iNowMonth = HebCal.GetMonth(now), iBirthMonth = HebCal.GetMonth(BirthDate)
    If iNowMonth < iBirthMonth Or (iNowMonth = iBirthMonth AndAlso HebCal.GetDayOfMonth(now) < HebCal.GetDayOfMonth(BirthDate)) Then iAge -= 1
    Return iAge
End Function

这不是一个直接的答案,但更多的是从准科学的角度对当前问题进行哲学推理。

我认为,这个问题并没有具体说明衡量年龄的单位或文化,大多数答案似乎都假设了一个整数年表示。时间的国际单位制单位是秒,因此正确的通用答案应该是(当然,假设标准化日期时间,不考虑相对论效应):

var lifeInSeconds = (DateTime.Now.Ticks - then.Ticks)/TickFactor;

在基督教以年计算年龄的方法中:

var then = ... // Then, in this case the birthday
var now = DateTime.UtcNow;
int age = now.Year - then.Year;
if (now.AddYears(-age) < then) age--;

在金融领域,当计算通常被称为日计数分数(Day Count Fraction)的东西时,也存在类似的问题,该分数大致是给定时期的年数。年龄问题确实是一个衡量时间的问题。

实际/实际(正确计算所有天数)惯例示例:

DateTime start, end = .... // Whatever, assume start is before end

double startYearContribution = 1 - (double) start.DayOfYear / (double) (DateTime.IsLeapYear(start.Year) ? 366 : 365);
double endYearContribution = (double)end.DayOfYear / (double)(DateTime.IsLeapYear(end.Year) ? 366 : 365);
double middleContribution = (double) (end.Year - start.Year - 1);

double DCF = startYearContribution + endYearContribution + middleContribution;

另一种很常见的衡量时间的方法通常是“序列化”(命名这一日期惯例的家伙一定是认真的“trippin”):

DateTime start, end = .... // Whatever, assume start is before end
int days = (end - start).Days;

我想知道,在相对论年龄(以秒为单位)变得比迄今为止地球围绕太阳周期的粗略近似更有用之前,我们还需要多长时间:)或者换句话说,当一个周期必须给定一个位置或一个表示其自身运动的函数才能有效时:)