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
哇,我不得不在这里回答。。。这么简单的问题有很多答案。
private int CalcularIdade(DateTime dtNascimento)
{
var nHoje = Convert.ToInt32(DateTime.Today.ToString("yyyyMMdd"));
var nAniversario = Convert.ToInt32(dtNascimento.ToString("yyyyMMdd"));
double diff = (nHoje - nAniversario) / 10000;
var ret = Convert.ToInt32(Math.Truncate(diff));
return ret;
}
其他回答
看看这个:
TimeSpan ts = DateTime.Now.Subtract(Birthdate);
age = (byte)(ts.TotalDays / 365.25);
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
}
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);
试试这个解决方案,它奏效了。
int age = (Int32.Parse(DateTime.Today.ToString("yyyyMMdd")) -
Int32.Parse(birthday.ToString("yyyyMMdd rawrrr"))) / 10000;
我使用这个:
public static class DateTimeExtensions
{
public static int Age(this DateTime birthDate)
{
return Age(birthDate, DateTime.Now);
}
public static int Age(this DateTime birthDate, DateTime offsetDate)
{
int result=0;
result = offsetDate.Year - birthDate.Year;
if (offsetDate.DayOfYear < birthDate.DayOfYear)
{
result--;
}
return result;
}
}
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