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;
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