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。

我对Mark Soen的答案做了一个小小的修改：我重写了第三行，以便可以更容易地解析表达式。

public int AgeInYears(DateTime bday)
{
    DateTime now = DateTime.Today;
    int age = now.Year - bday.Year;            
    if (bday.AddYears(age) > now) 
        age--;
    return age;
}

为了清晰起见，我还将其转换为函数。

我想添加希伯来文日历计算（或其他系统。全球化日历可以以相同的方式使用），使用此线程中的重写函数：

Public Shared Function CalculateAge(BirthDate As DateTime) As Integer
    Dim HebCal As New System.Globalization.HebrewCalendar ()
    Dim now = DateTime.Now()
    Dim iAge = HebCal.GetYear(now) - HebCal.GetYear(BirthDate)
    Dim iNowMonth = HebCal.GetMonth(now), iBirthMonth = HebCal.GetMonth(BirthDate)
    If iNowMonth < iBirthMonth Or (iNowMonth = iBirthMonth AndAlso HebCal.GetDayOfMonth(now) < HebCal.GetDayOfMonth(BirthDate)) Then iAge -= 1
    Return iAge
End Function

我认为TimeSpan包含了我们所需要的一切，而不必求助于365.25（或任何其他近似值）。扩展Aug的示例：

DateTime myBD = new DateTime(1980, 10, 10);
TimeSpan difference = DateTime.Now.Subtract(myBD);

textBox1.Text = difference.Years + " years " + difference.Months + " Months " + difference.Days + " days";

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  

这里有一个解决方案。

DateTime dateOfBirth = new DateTime(2000, 4, 18);
DateTime currentDate = DateTime.Now;

int ageInYears = 0;
int ageInMonths = 0;
int ageInDays = 0;

ageInDays = currentDate.Day - dateOfBirth.Day;
ageInMonths = currentDate.Month - dateOfBirth.Month;
ageInYears = currentDate.Year - dateOfBirth.Year;

if (ageInDays < 0)
{
    ageInDays += DateTime.DaysInMonth(currentDate.Year, currentDate.Month);
    ageInMonths = ageInMonths--;

    if (ageInMonths < 0)
    {
        ageInMonths += 12;
        ageInYears--;
    }
}

if (ageInMonths < 0)
{
    ageInMonths += 12;
    ageInYears--;
}

Console.WriteLine("{0}, {1}, {2}", ageInYears, ageInMonths, ageInDays);