看看这个：

TimeSpan ts = DateTime.Now.Subtract(Birthdate);
age = (byte)(ts.TotalDays / 365.25);

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：）

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;
    }
};

我经常用手指数。我需要看一下日历，以确定事情何时发生变化。这就是我在代码中要做的：

int AgeNow(DateTime birthday)
{
    return AgeAt(DateTime.Now, birthday);
}

int AgeAt(DateTime now, DateTime birthday)
{
    return AgeAt(now, birthday, CultureInfo.CurrentCulture.Calendar);
}

int AgeAt(DateTime now, DateTime birthday, Calendar calendar)
{
    // My age has increased on the morning of my
    // birthday even though I was born in the evening.
    now = now.Date;
    birthday = birthday.Date;

    var age = 0;
    if (now <= birthday) return age; // I am zero now if I am to be born tomorrow.

    while (calendar.AddYears(birthday, age + 1) <= now)
    {
        age++;
    }
    return age;
}

在LINQPad中运行此过程可获得以下结果：

PASSED: someone born on 28 February 1964 is age 4 on 28 February 1968
PASSED: someone born on 29 February 1964 is age 3 on 28 February 1968
PASSED: someone born on 31 December 2016 is age 0 on 01 January 2017

LINQPad中的代码在这里。

这是用一行文字回答这个问题的最简单方法。

DateTime Dob = DateTime.Parse("1985-04-24");
 
int Age = DateTime.MinValue.AddDays(DateTime.Now.Subtract(Dob).TotalHours/24 - 1).Year - 1;

这也适用于闰年。

以下是计算某人年龄的最简单方法。计算某人的年龄非常简单，下面是计算方法！为了使代码正常工作，您需要一个名为BirthDate的DateTime对象，该对象包含生日。

 C#
        // get the difference in years
        int years = DateTime.Now.Year - BirthDate.Year; 
        // subtract another year if we're before the
        // birth day in the current year
        if (DateTime.Now.Month < BirthDate.Month || 
            (DateTime.Now.Month == BirthDate.Month && 
            DateTime.Now.Day < BirthDate.Day)) 
            years--;
  VB.NET
        ' get the difference in years
        Dim years As Integer = DateTime.Now.Year - BirthDate.Year
        ' subtract another year if we're before the
        ' birth day in the current year
        If DateTime.Now.Month < BirthDate.Month Or (DateTime.Now.Month = BirthDate.Month And DateTime.Now.Day < BirthDate.Day) Then 
            years = years - 1
        End If