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
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();
其他回答
这是一个非常简单的方法:
int Age = DateTime.Today.Year - new DateTime(2000, 1, 1).Year;
因为闰年和所有事情,我知道的最好的方法是:
DateTime birthDate = new DateTime(2000,3,1);
int age = (int)Math.Floor((DateTime.Now - birthDate).TotalDays / 365.25D);
对此的简单答案是应用AddYears,如下所示,因为这是唯一一种将年份添加到闰年2月29日的本地方法,并获得普通年份2月28日的正确结果。
有些人认为3月1日是勒普林斯的生日,但.Net和任何官方规则都不支持这一点,也没有常见的逻辑解释为什么一些出生在2月的人应该在另一个月拥有75%的生日。
此外,Age方法可以作为DateTime的扩展添加。由此,您可以以最简单的方式获得年龄:
列表项目
int age=出生日期.age();
public static class DateTimeExtensions
{
/// <summary>
/// Calculates the age in years of the current System.DateTime object today.
/// </summary>
/// <param name="birthDate">The date of birth</param>
/// <returns>Age in years today. 0 is returned for a future date of birth.</returns>
public static int Age(this DateTime birthDate)
{
return Age(birthDate, DateTime.Today);
}
/// <summary>
/// Calculates the age in years of the current System.DateTime object on a later date.
/// </summary>
/// <param name="birthDate">The date of birth</param>
/// <param name="laterDate">The date on which to calculate the age.</param>
/// <returns>Age in years on a later day. 0 is returned as minimum.</returns>
public static int Age(this DateTime birthDate, DateTime laterDate)
{
int age;
age = laterDate.Year - birthDate.Year;
if (age > 0)
{
age -= Convert.ToInt32(laterDate.Date < birthDate.Date.AddYears(age));
}
else
{
age = 0;
}
return age;
}
}
现在,运行此测试:
class Program
{
static void Main(string[] args)
{
RunTest();
}
private static void RunTest()
{
DateTime birthDate = new DateTime(2000, 2, 28);
DateTime laterDate = new DateTime(2011, 2, 27);
string iso = "yyyy-MM-dd";
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
Console.WriteLine("Birth date: " + birthDate.AddDays(i).ToString(iso) + " Later date: " + laterDate.AddDays(j).ToString(iso) + " Age: " + birthDate.AddDays(i).Age(laterDate.AddDays(j)).ToString());
}
}
Console.ReadKey();
}
}
关键日期示例如下:
出生日期:2000-02-29出生日期:2011-02-28年龄:11
输出:
{
Birth date: 2000-02-28 Later date: 2011-02-27 Age: 10
Birth date: 2000-02-28 Later date: 2011-02-28 Age: 11
Birth date: 2000-02-28 Later date: 2011-03-01 Age: 11
Birth date: 2000-02-29 Later date: 2011-02-27 Age: 10
Birth date: 2000-02-29 Later date: 2011-02-28 Age: 11
Birth date: 2000-02-29 Later date: 2011-03-01 Age: 11
Birth date: 2000-03-01 Later date: 2011-02-27 Age: 10
Birth date: 2000-03-01 Later date: 2011-02-28 Age: 10
Birth date: 2000-03-01 Later date: 2011-03-01 Age: 11
}
2012年2月28日晚些时候:
{
Birth date: 2000-02-28 Later date: 2012-02-28 Age: 12
Birth date: 2000-02-28 Later date: 2012-02-29 Age: 12
Birth date: 2000-02-28 Later date: 2012-03-01 Age: 12
Birth date: 2000-02-29 Later date: 2012-02-28 Age: 11
Birth date: 2000-02-29 Later date: 2012-02-29 Age: 12
Birth date: 2000-02-29 Later date: 2012-03-01 Age: 12
Birth date: 2000-03-01 Later date: 2012-02-28 Age: 11
Birth date: 2000-03-01 Later date: 2012-02-29 Age: 11
Birth date: 2000-03-01 Later date: 2012-03-01 Age: 12
}
另一个功能,不是我做的,而是在网上找到的,并做了一些改进:
public static int GetAge(DateTime birthDate)
{
DateTime n = DateTime.Now; // To avoid a race condition around midnight
int age = n.Year - birthDate.Year;
if (n.Month < birthDate.Month || (n.Month == birthDate.Month && n.Day < birthDate.Day))
age--;
return age;
}
我只想到了两件事:来自不使用公历的国家的人呢?DateTime。我认为现在是服务器特定的文化。我对实际使用亚洲日历一无所知,我不知道是否有一种简单的方法来转换日历之间的日期,但以防万一,你想知道4660年的中国人:-)
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";
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