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
private int GetAge(int _year, int _month, int _day
{
DateTime yourBirthDate= new DateTime(_year, _month, _day);
DateTime todaysDateTime = DateTime.Today;
int noOfYears = todaysDateTime.Year - yourBirthDate.Year;
if (DateTime.Now.Month < yourBirthDate.Month ||
(DateTime.Now.Month == yourBirthDate.Month && DateTime.Now.Day < yourBirthDate.Day))
{
noOfYears--;
}
return noOfYears;
}
其他回答
因为闰年和所有事情,我知道的最好的方法是:
DateTime birthDate = new DateTime(2000,3,1);
int age = (int)Math.Floor((DateTime.Now - birthDate).TotalDays / 365.25D);
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++;
}
}
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
保持简单(可能是愚蠢的:)。
DateTime birth = new DateTime(1975, 09, 27, 01, 00, 00, 00);
TimeSpan ts = DateTime.Now - birth;
Console.WriteLine("You are approximately " + ts.TotalSeconds.ToString() + " seconds old.");
我对DateTime一无所知,但我能做的就是:
using System;
public class Program
{
public static int getAge(int month, int day, int year) {
DateTime today = DateTime.Today;
int currentDay = today.Day;
int currentYear = today.Year;
int currentMonth = today.Month;
int age = 0;
if (currentMonth < month) {
age -= 1;
} else if (currentMonth == month) {
if (currentDay < day) {
age -= 1;
}
}
currentYear -= year;
age += currentYear;
return age;
}
public static void Main()
{
int ageInYears = getAge(8, 10, 2007);
Console.WriteLine(ageInYears);
}
}
有点困惑,但仔细看代码,这一切都是有意义的。