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

另一个功能,不是我做的,而是在网上找到的,并做了一些改进:

public static int GetAge(DateTime birthDate)
{
    DateTime n = DateTime.Now; // To avoid a race condition around midnight
    int age = n.Year - birthDate.Year;

    if (n.Month < birthDate.Month || (n.Month == birthDate.Month && n.Day < birthDate.Day))
        age--;

    return age;
}

我只想到了两件事:来自不使用公历的国家的人呢?DateTime。我认为现在是服务器特定的文化。我对实际使用亚洲日历一无所知,我不知道是否有一种简单的方法来转换日历之间的日期,但以防万一,你想知道4660年的中国人:-)

其他回答

还有一个答案:

public static int AgeInYears(DateTime birthday, DateTime today)
{
    return ((today.Year - birthday.Year) * 372 + (today.Month - birthday.Month) * 31 + (today.Day - birthday.Day)) / 372;
}

这已经过广泛的单元测试。它看起来确实有点“神奇”。数字372是如果每个月有31天,一年中会有多少天。

其工作原理的解释(此处省略)如下:

让我们设置Yn=DateTime.Now.Year,Yb=生日.Year,Mn=DateTime.Now.Month,Mb=生日.Month、Dn=DateTime.Now.Day,Db=生日.Day年龄=Yn-Yb+(31*(Mn-Mb)+(Dn-Db))/372我们知道,如果日期已经到达,我们需要的是Yn-Yb,如果日期尚未到达,则需要Yn-Yb-1。a) 如果Mn<Mb,我们有-341<=31*(Mn-Mb)<=-31和-30<=Dn-Db<=30-371<=31*(锰-Mb)+(Dn-Db)<=-1带整数除法(31*(Mn-Mb)+(Dn-Db))/372=-1b) 如果Mn=Mb和Dn<Db,则我们有31*(Mn-Mb)=0和-30<=Dn Db<=-1再次使用整数除法(31*(Mn-Mb)+(Dn-Db))/372=-1c) 如果Mn>Mb,我们有31<=31*(Mn-Mb)<=341和-30<=Dn-Db<=301<=31*(Mn-Mb)+(Dn-Db)<=371带整数除法(31*(Mn-Mb)+(Dn-Db))/372=0d) 如果Mn=Mb且Dn>Db,则我们有31*(Mn-Mb)=0且1<=Dn Db<=30再次使用整数除法(31*(Mn-Mb)+(Dn-Db))/372=0e) 如果Mn=Mb,Dn=Db,我们有31*(Mn-Mb)+Dn Db=0因此(31*(Mn-Mb)+(Dn-Db))/372=0

这个经典问题值得野田时间来解决。

static int GetAge(LocalDate dateOfBirth)
{
    Instant now = SystemClock.Instance.Now;

    // The target time zone is important.
    // It should align with the *current physical location* of the person
    // you are talking about.  When the whereabouts of that person are unknown,
    // then you use the time zone of the person who is *asking* for the age.
    // The time zone of birth is irrelevant!

    DateTimeZone zone = DateTimeZoneProviders.Tzdb["America/New_York"];

    LocalDate today = now.InZone(zone).Date;

    Period period = Period.Between(dateOfBirth, today, PeriodUnits.Years);

    return (int) period.Years;
}

用法:

LocalDate dateOfBirth = new LocalDate(1976, 8, 27);
int age = GetAge(dateOfBirth);

您可能还对以下改进感兴趣:

将时钟作为IClock传递,而不是使用SystemClock.Instance,将提高可测试性。目标时区可能会更改,因此您也需要DateTimeZone参数。

另请参阅我关于这个主题的博客文章:处理生日和其他周年纪念日

我花了一些时间研究这个问题,并用这个来计算某人的年龄,以年、月和日为单位。我已经针对2月29日的问题和闰年进行了测试,它似乎奏效了,我希望得到任何反馈:

public void LoopAge(DateTime myDOB, DateTime FutureDate)
{
    int years = 0;
    int months = 0;
    int days = 0;

    DateTime tmpMyDOB = new DateTime(myDOB.Year, myDOB.Month, 1);

    DateTime tmpFutureDate = new DateTime(FutureDate.Year, FutureDate.Month, 1);

    while (tmpMyDOB.AddYears(years).AddMonths(months) < tmpFutureDate)
    {
        months++;

        if (months > 12)
        {
            years++;
            months = months - 12;
        }
    }

    if (FutureDate.Day >= myDOB.Day)
    {
        days = days + FutureDate.Day - myDOB.Day;
    }
    else
    {
        months--;

        if (months < 0)
        {
            years--;
            months = months + 12;
        }

        days +=
            DateTime.DaysInMonth(
                FutureDate.AddMonths(-1).Year, FutureDate.AddMonths(-1).Month
            ) + FutureDate.Day - myDOB.Day;

    }

    //add an extra day if the dob is a leap day
    if (DateTime.IsLeapYear(myDOB.Year) && myDOB.Month == 2 && myDOB.Day == 29)
    {
        //but only if the future date is less than 1st March
        if (FutureDate >= new DateTime(FutureDate.Year, 3, 1))
            days++;
    }

}
        var EndDate = new DateTime(2022, 4, 21);

        var StartDate = new DateTime(1986, 4, 25);

        Int32 Months = EndDate.Month - StartDate.Month;

        Int32 Years = EndDate.Year - StartDate.Year;

        Int32 Days = EndDate.Day - StartDate.Day;

        if (Days < 0)
        {
            Months = Months - 1;
        }

        if (Months < 0)
        {
            Years = Years - 1;

            Months = Months + 12;
        }
        
        string Ages = Years.ToString() + " Year(s) " + Months.ToString() + " Month(s) ";

以下是使用DateTimeOffset和手动数学的答案:

var diff = DateTimeOffset.Now - dateOfBirth;
var sinceEpoch = DateTimeOffset.UnixEpoch + diff;

return sinceEpoch.Year - 1970;