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

这是一种奇怪的方法,但如果您将日期设置为yyyymmdd,并从当前日期中减去出生日期,然后删除您获得的年龄的最后4位数字:)

我不知道C#,但我相信这在任何语言中都适用。

20080814 - 19800703 = 280111 

删除最后4位=28。

C#代码:

int now = int.Parse(DateTime.Now.ToString("yyyyMMdd"));
int dob = int.Parse(dateOfBirth.ToString("yyyyMMdd"));
int age = (now - dob) / 10000;

或者,也可以不进行扩展方法形式的所有类型转换。忽略错误检查:

public static Int32 GetAge(this DateTime dateOfBirth)
{
    var today = DateTime.Today;

    var a = (today.Year * 100 + today.Month) * 100 + today.Day;
    var b = (dateOfBirth.Year * 100 + dateOfBirth.Month) * 100 + dateOfBirth.Day;

    return (a - b) / 10000;
}

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;
    }
};

2需要解决的主要问题有:

1.计算准确年龄-以年、月、日等为单位。

2.计算人们普遍认为的年龄——人们通常不关心自己到底多大,他们只关心自己当年的生日是什么时候。


1的解决方案显而易见:

DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today;     //we usually don't care about birth time
TimeSpan age = today - birth;        //.NET FCL should guarantee this as precise
double ageInDays = age.TotalDays;    //total number of days ... also precise
double daysInYear = 365.2425;        //statistical value for 400 years
double ageInYears = ageInDays / daysInYear;  //can be shifted ... not so precise

2的解决方案在确定总年龄时并不那么精确,但人们认为它是精确的。当人们“手动”计算年龄时,通常也会使用它:

DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today;
int age = today.Year - birth.Year;    //people perceive their age in years

if (today.Month < birth.Month ||
   ((today.Month == birth.Month) && (today.Day < birth.Day)))
{
  age--;  //birthday in current year not yet reached, we are 1 year younger ;)
          //+ no birthday for 29.2. guys ... sorry, just wrong date for birth
}

注释2.:

这是我的首选解决方案我们不能使用DateTime.DayOfYear或TimeSpans,因为它们会在闰年中改变天数为了可读性,我只增加了几行

还有一个提示。。。我将为它创建两个静态重载方法,一个用于通用,另一个用于使用友好:

public static int GetAge(DateTime bithDay, DateTime today) 
{ 
  //chosen solution method body
}

public static int GetAge(DateTime birthDay) 
{ 
  return GetAge(birthDay, DateTime.Now);
}

简单代码

 var birthYear=1993;
 var age = DateTime.Now.AddYears(-birthYear).Year;

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