我有一个定制的计算年龄的方法，加上一条奖金验证消息，以防有帮助：

public void GetAge(DateTime dob, DateTime now, out int years, out int months, out int days)
{
    years = 0;
    months = 0;
    days = 0;

    DateTime tmpdob = new DateTime(dob.Year, dob.Month, 1);
    DateTime tmpnow = new DateTime(now.Year, now.Month, 1);

    while (tmpdob.AddYears(years).AddMonths(months) < tmpnow)
    {
        months++;
        if (months > 12)
        {
            years++;
            months = months - 12;
        }
    }

    if (now.Day >= dob.Day)
        days = days + now.Day - dob.Day;
    else
    {
        months--;
        if (months < 0)
        {
            years--;
            months = months + 12;
        }
        days += DateTime.DaysInMonth(now.AddMonths(-1).Year, now.AddMonths(-1).Month) + now.Day - dob.Day;
    }

    if (DateTime.IsLeapYear(dob.Year) && dob.Month == 2 && dob.Day == 29 && now >= new DateTime(now.Year, 3, 1))
        days++;

}   

private string ValidateDate(DateTime dob) //This method will validate the date
{
    int Years = 0; int Months = 0; int Days = 0;

    GetAge(dob, DateTime.Now, out Years, out Months, out Days);

    if (Years < 18)
        message =  Years + " is too young. Please try again on your 18th birthday.";
    else if (Years >= 65)
        message = Years + " is too old. Date of Birth must not be 65 or older.";
    else
        return null; //Denotes validation passed
}

方法调用此处并传递日期时间值（如果服务器设置为美国语言环境，则为MM/dd/yyyy）。将其替换为消息框或要显示的任何容器：

DateTime dob = DateTime.Parse("03/10/1982");  

string message = ValidateDate(dob);

lbldatemessage.Visible = !StringIsNullOrWhitespace(message);
lbldatemessage.Text = message ?? ""; //Ternary if message is null then default to empty string

记住，您可以按任何方式格式化邮件。

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：）

可以这么简单：

int age = DateTime.Now.AddTicks(0 - dob.Ticks).Year - 1;

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;
}

我创建了一个Age结构，如下所示：

public struct Age : IEquatable<Age>, IComparable<Age>
{
    private readonly int _years;
    private readonly int _months;
    private readonly int _days;

    public int Years  { get { return _years; } }
    public int Months { get { return _months; } }
    public int Days { get { return _days; } }

    public Age( int years, int months, int days ) : this()
    {
        _years = years;
        _months = months;
        _days = days;
    }

    public static Age CalculateAge( DateTime dateOfBirth, DateTime date )
    {
        // Here is some logic that ressembles Mike's solution, although it
        // also takes into account months & days.
        // Ommitted for brevity.
        return new Age (years, months, days);
    }

    // Ommited Equality, Comparable, GetHashCode, functionality for brevity.
}

我认为这个问题可以用这样一种更简单的方法解决-

该类可以是-

using System;

namespace TSA
{
    class BirthDay
    {
        double ageDay;
        public BirthDay(int day, int month, int year)
        {
            DateTime birthDate = new DateTime(year, month, day);
            ageDay = (birthDate - DateTime.Now).TotalDays; //DateTime.UtcNow
        }

        internal int GetAgeYear()
        {
            return (int)Math.Truncate(ageDay / 365);
        }

        internal int GetAgeMonth()
        {
            return (int)Math.Truncate((ageDay % 365) / 30);
        }
    }
}

电话可以是这样的-

BirthDay b = new BirthDay(1,12,1990);
int year = b.GetAgeYear();
int month = b.GetAgeMonth();