给定代表某人生日的DateTime,我如何计算他们的年龄(以年为单位)?


当前回答

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  

其他回答

保持简单(可能是愚蠢的:)。

DateTime birth = new DateTime(1975, 09, 27, 01, 00, 00, 00);
TimeSpan ts = DateTime.Now - birth;
Console.WriteLine("You are approximately " + ts.TotalSeconds.ToString() + " seconds old.");

以下是使用DateTimeOffset和手动数学的答案:

var diff = DateTimeOffset.Now - dateOfBirth;
var sinceEpoch = DateTimeOffset.UnixEpoch + diff;

return sinceEpoch.Year - 1970;

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";

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.
}

这很简单,似乎符合我的需要。我为闰年的目的做了一个假设,即无论一个人选择什么时候庆祝生日,从技术上讲,他们都不会比自己大一岁,直到他们的上一个生日过去365天(即2月28日不会使他们大一岁)。

DateTime now = DateTime.Today;
DateTime birthday = new DateTime(1991, 02, 03);//3rd feb

int age = now.Year - birthday.Year;

if (now.Month < birthday.Month || (now.Month == birthday.Month && now.Day < birthday.Day))//not had bday this year yet
  age--;

return age;