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

为了计算一个人的年龄,

DateTime dateOfBirth;

int ageInYears = DateTime.Now.Year - dateOfBirth.Year;

if (dateOfBirth > today.AddYears(-ageInYears )) ageInYears --;

其他回答

我使用这个:

public static class DateTimeExtensions
{
    public static int Age(this DateTime birthDate)
    {
        return Age(birthDate, DateTime.Now);
    }

    public static int Age(this DateTime birthDate, DateTime offsetDate)
    {
        int result=0;
        result = offsetDate.Year - birthDate.Year;

        if (offsetDate.DayOfYear < birthDate.DayOfYear)
        {
              result--;
        }

        return result;
    }
}

我已经创建了一个SQL Server用户定义函数来计算某人的年龄,给定他们的出生日期。当您需要它作为查询的一部分时,这很有用:

using System;
using System.Data;
using System.Data.Sql;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;

public partial class UserDefinedFunctions
{
    [SqlFunction(DataAccess = DataAccessKind.Read)]
    public static SqlInt32 CalculateAge(string strBirthDate)
    {
        DateTime dtBirthDate = new DateTime();
        dtBirthDate = Convert.ToDateTime(strBirthDate);
        DateTime dtToday = DateTime.Now;

        // get the difference in years
        int years = dtToday.Year - dtBirthDate.Year;

        // subtract another year if we're before the
        // birth day in the current year
        if (dtToday.Month < dtBirthDate.Month || (dtToday.Month == dtBirthDate.Month && dtToday.Day < dtBirthDate.Day))
            years=years-1;

        int intCustomerAge = years;
        return intCustomerAge;
    }
};

这里有一个小的C#代码示例,我总结了一下,请注意边缘情况,特别是闰年,并不是所有上述解决方案都考虑到这些情况。将答案推出来作为DateTime可能会导致问题,因为你可能会在一个特定的月份中投入太多的时间,例如2月的30天。

public string 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 = 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++;
    }

    return "Years: " + years + " Months: " + months + " Days: " + days;
}

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

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