我找到的最简单的方法就是这样。它适用于美国和西欧地区。无法与其他地区通话，尤其是中国这样的地方。在最初计算年龄后，最多可额外进行4次比较。

public int AgeInYears(DateTime birthDate, DateTime referenceDate)
{
  Debug.Assert(referenceDate >= birthDate, 
               "birth date must be on or prior to the reference date");

  DateTime birth = birthDate.Date;
  DateTime reference = referenceDate.Date;
  int years = (reference.Year - birth.Year);

  //
  // an offset of -1 is applied if the birth date has 
  // not yet occurred in the current year.
  //
  if (reference.Month > birth.Month);
  else if (reference.Month < birth.Month) 
    --years;
  else // in birth month
  {
    if (reference.Day < birth.Day)
      --years;
  }

  return years ;
}

我仔细查看了答案，发现没有人提及闰日出生的监管/法律影响。例如，根据维基百科，如果你在2月29日出生在不同的司法管辖区，你的非闰年生日会有所不同：

在英国和香港：这是一年中的第几天，所以第二天，3月1日是你的生日。在新西兰：这是前一天，2月28日用于驾驶执照，3月1日用于其他目的。台湾：今天是2月28日。

据我所知，在美国，法规对此事保持沉默，这取决于普通法以及各个监管机构如何在其法规中定义事物。

为此，需要改进：

public enum LeapDayRule
{
  OrdinalDay     = 1 ,
  LastDayOfMonth = 2 ,
}

static int ComputeAgeInYears(DateTime birth, DateTime reference, LeapYearBirthdayRule ruleInEffect)
{
  bool isLeapYearBirthday = CultureInfo.CurrentCulture.Calendar.IsLeapDay(birth.Year, birth.Month, birth.Day);
  DateTime cutoff;

  if (isLeapYearBirthday && !DateTime.IsLeapYear(reference.Year))
  {
    switch (ruleInEffect)
    {
      case LeapDayRule.OrdinalDay:
        cutoff = new DateTime(reference.Year, 1, 1)
                             .AddDays(birth.DayOfYear - 1);
        break;

      case LeapDayRule.LastDayOfMonth:
        cutoff = new DateTime(reference.Year, birth.Month, 1)
                             .AddMonths(1)
                             .AddDays(-1);
        break;

      default:
        throw new InvalidOperationException();
    }
  }
  else
  {
    cutoff = new DateTime(reference.Year, birth.Month, birth.Day);
  }

  int age = (reference.Year - birth.Year) + (reference >= cutoff ? 0 : -1);
  return age < 0 ? 0 : age;
}

需要注意的是，该代码假设：

西方（欧洲）对年龄的推算，以及一种日历，如公历，在月底插入一个闰日。

对此的简单答案是应用AddYears，如下所示，因为这是唯一一种将年份添加到闰年2月29日的本地方法，并获得普通年份2月28日的正确结果。

有些人认为3月1日是勒普林斯的生日，但.Net和任何官方规则都不支持这一点，也没有常见的逻辑解释为什么一些出生在2月的人应该在另一个月拥有75%的生日。

此外，Age方法可以作为DateTime的扩展添加。由此，您可以以最简单的方式获得年龄：

列表项目

int age=出生日期.age（）；

public static class DateTimeExtensions
{
    /// <summary>
    /// Calculates the age in years of the current System.DateTime object today.
    /// </summary>
    /// <param name="birthDate">The date of birth</param>
    /// <returns>Age in years today. 0 is returned for a future date of birth.</returns>
    public static int Age(this DateTime birthDate)
    {
        return Age(birthDate, DateTime.Today);
    }

    /// <summary>
    /// Calculates the age in years of the current System.DateTime object on a later date.
    /// </summary>
    /// <param name="birthDate">The date of birth</param>
    /// <param name="laterDate">The date on which to calculate the age.</param>
    /// <returns>Age in years on a later day. 0 is returned as minimum.</returns>
    public static int Age(this DateTime birthDate, DateTime laterDate)
    {
        int age;
        age = laterDate.Year - birthDate.Year;

        if (age > 0)
        {
            age -= Convert.ToInt32(laterDate.Date < birthDate.Date.AddYears(age));
        }
        else
        {
            age = 0;
        }

        return age;
    }
}

现在，运行此测试：

class Program
{
    static void Main(string[] args)
    {
        RunTest();
    }

    private static void RunTest()
    {
        DateTime birthDate = new DateTime(2000, 2, 28);
        DateTime laterDate = new DateTime(2011, 2, 27);
        string iso = "yyyy-MM-dd";

        for (int i = 0; i < 3; i++)
        {
            for (int j = 0; j < 3; j++)
            {
                Console.WriteLine("Birth date: " + birthDate.AddDays(i).ToString(iso) + "  Later date: " + laterDate.AddDays(j).ToString(iso) + "  Age: " + birthDate.AddDays(i).Age(laterDate.AddDays(j)).ToString());
            }
        }

        Console.ReadKey();
    }
}

关键日期示例如下：

出生日期：2000-02-29出生日期：2011-02-28年龄：11

输出：

{
    Birth date: 2000-02-28  Later date: 2011-02-27  Age: 10
    Birth date: 2000-02-28  Later date: 2011-02-28  Age: 11
    Birth date: 2000-02-28  Later date: 2011-03-01  Age: 11
    Birth date: 2000-02-29  Later date: 2011-02-27  Age: 10
    Birth date: 2000-02-29  Later date: 2011-02-28  Age: 11
    Birth date: 2000-02-29  Later date: 2011-03-01  Age: 11
    Birth date: 2000-03-01  Later date: 2011-02-27  Age: 10
    Birth date: 2000-03-01  Later date: 2011-02-28  Age: 10
    Birth date: 2000-03-01  Later date: 2011-03-01  Age: 11
}

2012年2月28日晚些时候：

{
    Birth date: 2000-02-28  Later date: 2012-02-28  Age: 12
    Birth date: 2000-02-28  Later date: 2012-02-29  Age: 12
    Birth date: 2000-02-28  Later date: 2012-03-01  Age: 12
    Birth date: 2000-02-29  Later date: 2012-02-28  Age: 11
    Birth date: 2000-02-29  Later date: 2012-02-29  Age: 12
    Birth date: 2000-02-29  Later date: 2012-03-01  Age: 12
    Birth date: 2000-03-01  Later date: 2012-02-28  Age: 11
    Birth date: 2000-03-01  Later date: 2012-02-29  Age: 11
    Birth date: 2000-03-01  Later date: 2012-03-01  Age: 12
}

var birthDate = ... // DOB
var resultDate = DateTime.Now - birthDate;

使用resultDate，您可以应用TimeSpan财产来显示任何内容。

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);

试试这个解决方案，它奏效了。

int age = (Int32.Parse(DateTime.Today.ToString("yyyyMMdd")) - 
           Int32.Parse(birthday.ToString("yyyyMMdd rawrrr"))) / 10000;

这为这个问题提供了“更多细节”。也许这就是你要找的

DateTime birth = new DateTime(1974, 8, 29);
DateTime today = DateTime.Now;
TimeSpan span = today - birth;
DateTime age = DateTime.MinValue + span;

// Make adjustment due to MinValue equalling 1/1/1
int years = age.Year - 1;
int months = age.Month - 1;
int days = age.Day - 1;

// Print out not only how many years old they are but give months and days as well
Console.Write("{0} years, {1} months, {2} days", years, months, days);