private int GetAge(int _year, int _month, int _day
{
    DateTime yourBirthDate= new DateTime(_year, _month, _day);

    DateTime todaysDateTime = DateTime.Today;
    int noOfYears = todaysDateTime.Year - yourBirthDate.Year;

    if (DateTime.Now.Month < yourBirthDate.Month ||
        (DateTime.Now.Month == yourBirthDate.Month && DateTime.Now.Day < yourBirthDate.Day))
    {
        noOfYears--;
    }

    return  noOfYears;
}

我对DateTime一无所知，但我能做的就是：

using System;
                    
public class Program
{
    public static int getAge(int month, int day, int year) {
        DateTime today = DateTime.Today;
        int currentDay = today.Day;
        int currentYear = today.Year;
        int currentMonth = today.Month;
        int age = 0;
        if (currentMonth < month) {
            age -= 1;
        } else if (currentMonth == month) {
            if (currentDay < day) {
                age -= 1;
            }
        }
        currentYear -= year;
        age += currentYear;
        return age;
    }
    public static void Main()
    {
        int ageInYears = getAge(8, 10, 2007);
        Console.WriteLine(ageInYears);
    }
}

有点困惑，但仔细看代码，这一切都是有意义的。

我只想这样做：

DateTime birthDay = new DateTime(1990, 05, 23);
DateTime age = DateTime.Now - birthDay;

这样你就可以计算出一个人的确切年龄，如果你愿意的话，可以精确到毫秒。

private int GetYearDiff(DateTime start, DateTime end)
{
    int diff = end.Year - start.Year;
    if (end.DayOfYear < start.DayOfYear) { diff -= 1; }
    return diff;
}
[Fact]
public void GetYearDiff_WhenCalls_ShouldReturnCorrectYearDiff()
{
    //arrange
    var now = DateTime.Now;
    //act
    //assert
    Assert.Equal(24, GetYearDiff(new DateTime(1992, 7, 9), now)); // passed
    Assert.Equal(24, GetYearDiff(new DateTime(1992, now.Month, now.Day), now)); // passed
    Assert.Equal(23, GetYearDiff(new DateTime(1992, 12, 9), now)); // passed
}

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);

下面是一个测试片段：

DateTime bDay = new DateTime(2000, 2, 29);
DateTime now = new DateTime(2009, 2, 28);
MessageBox.Show(string.Format("Test {0} {1} {2}",
                CalculateAgeWrong1(bDay, now),      // outputs 9
                CalculateAgeWrong2(bDay, now),      // outputs 9
                CalculateAgeCorrect(bDay, now),     // outputs 8
                CalculateAgeCorrect2(bDay, now)));  // outputs 8

这里有一些方法：

public int CalculateAgeWrong1(DateTime birthDate, DateTime now)
{
    return new DateTime(now.Subtract(birthDate).Ticks).Year - 1;
}

public int CalculateAgeWrong2(DateTime birthDate, DateTime now)
{
    int age = now.Year - birthDate.Year;

    if (now < birthDate.AddYears(age))
        age--;

    return age;
}

public int CalculateAgeCorrect(DateTime birthDate, DateTime now)
{
    int age = now.Year - birthDate.Year;

    if (now.Month < birthDate.Month || (now.Month == birthDate.Month && now.Day < birthDate.Day))
        age--;

    return age;
}

public int CalculateAgeCorrect2(DateTime birthDate, DateTime now)
{
    int age = now.Year - birthDate.Year;

    // For leap years we need this
    if (birthDate > now.AddYears(-age)) 
        age--;
    // Don't use:
    // if (birthDate.AddYears(age) > now) 
    //     age--;

    return age;
}