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
MSDN帮助为什么没有告诉您这一点?看起来很明显:
System.DateTime birthTime = AskTheUser(myUser); // :-)
System.DateTime now = System.DateTime.Now;
System.TimeSpan age = now - birthTime; // As simple as that
double ageInDays = age.TotalDays; // Will you convert to whatever you want yourself?
其他回答
这个经典问题值得野田时间来解决。
static int GetAge(LocalDate dateOfBirth)
{
Instant now = SystemClock.Instance.Now;
// The target time zone is important.
// It should align with the *current physical location* of the person
// you are talking about. When the whereabouts of that person are unknown,
// then you use the time zone of the person who is *asking* for the age.
// The time zone of birth is irrelevant!
DateTimeZone zone = DateTimeZoneProviders.Tzdb["America/New_York"];
LocalDate today = now.InZone(zone).Date;
Period period = Period.Between(dateOfBirth, today, PeriodUnits.Years);
return (int) period.Years;
}
用法:
LocalDate dateOfBirth = new LocalDate(1976, 8, 27);
int age = GetAge(dateOfBirth);
您可能还对以下改进感兴趣:
将时钟作为IClock传递,而不是使用SystemClock.Instance,将提高可测试性。目标时区可能会更改,因此您也需要DateTimeZone参数。
另请参阅我关于这个主题的博客文章:处理生日和其他周年纪念日
通过较少的转换和UtcNow,这段代码可以照顾闰年2月29日出生的人:
public int GetAge(DateTime DateOfBirth)
{
var Now = DateTime.UtcNow;
return Now.Year - DateOfBirth.Year -
(
(
Now.Month > DateOfBirth.Month ||
(Now.Month == DateOfBirth.Month && Now.Day >= DateOfBirth.Day)
) ? 0 : 1
);
}
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。
这里有一个单行线:
int age = new DateTime(DateTime.Now.Subtract(birthday).Ticks).Year-1;
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