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);

我在这个问题上使用了以下内容。我知道它不太优雅，但它很管用。

DateTime zeroTime = new DateTime(1, 1, 1);
var date1 = new DateTime(1983, 03, 04);
var date2 = DateTime.Now;
var dif = date2 - date1;
int years = (zeroTime + dif).Year - 1;
Log.DebugFormat("Years -->{0}", years);

SQL版本：

declare @dd smalldatetime = '1980-04-01'
declare @age int = YEAR(GETDATE())-YEAR(@dd)
if (@dd> DATEADD(YYYY, -@age, GETDATE())) set @age = @age -1

print @age  

只需使用：

(DateTime.Now - myDate).TotalHours / 8766.0

当前日期-myDate=TimeSpan，获取总小时数并除以每年的总小时数，得到确切的年龄/月/日。。。

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

非常简单的答案

        DateTime dob = new DateTime(1991, 3, 4); 
        DateTime now = DateTime.Now; 
        int dobDay = dob.Day, dobMonth = dob.Month; 
        int add = -1; 
        if (dobMonth < now.Month)
        {
            add = 0;
        }
        else if (dobMonth == now.Month)
        {
            if(dobDay <= now.Day)
            {
                add = 0;
            }
            else
            {
                add = -1;
            }
        }
        else
        {
            add = -1;
        } 
        int age = now.Year - dob.Year + add;