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
这是一个非常适合我的功能。没有计算,非常简单。
public static string ToAge(this DateTime dob, DateTime? toDate = null)
{
if (!toDate.HasValue)
toDate = DateTime.Now;
var now = toDate.Value;
if (now.CompareTo(dob) < 0)
return "Future date";
int years = now.Year - dob.Year;
int months = now.Month - dob.Month;
int days = now.Day - dob.Day;
if (days < 0)
{
months--;
days = DateTime.DaysInMonth(dob.Year, dob.Month) - dob.Day + now.Day;
}
if (months < 0)
{
years--;
months = 12 + months;
}
return string.Format("{0} year(s), {1} month(s), {2} days(s)",
years,
months,
days);
}
这里是一个单元测试:
[Test]
public void ToAgeTests()
{
var date = new DateTime(2000, 1, 1);
Assert.AreEqual("0 year(s), 0 month(s), 1 days(s)", new DateTime(1999, 12, 31).ToAge(date));
Assert.AreEqual("0 year(s), 0 month(s), 0 days(s)", new DateTime(2000, 1, 1).ToAge(date));
Assert.AreEqual("1 year(s), 0 month(s), 0 days(s)", new DateTime(1999, 1, 1).ToAge(date));
Assert.AreEqual("0 year(s), 11 month(s), 0 days(s)", new DateTime(1999, 2, 1).ToAge(date));
Assert.AreEqual("0 year(s), 10 month(s), 25 days(s)", new DateTime(1999, 2, 4).ToAge(date));
Assert.AreEqual("0 year(s), 10 month(s), 1 days(s)", new DateTime(1999, 2, 28).ToAge(date));
date = new DateTime(2000, 2, 15);
Assert.AreEqual("0 year(s), 0 month(s), 28 days(s)", new DateTime(2000, 1, 18).ToAge(date));
}
其他回答
看看这个:
TimeSpan ts = DateTime.Now.Subtract(Birthdate);
age = (byte)(ts.TotalDays / 365.25);
下面是一个测试片段:
DateTime bDay = new DateTime(2000, 2, 29);
DateTime now = new DateTime(2009, 2, 28);
MessageBox.Show(string.Format("Test {0} {1} {2}",
CalculateAgeWrong1(bDay, now), // outputs 9
CalculateAgeWrong2(bDay, now), // outputs 9
CalculateAgeCorrect(bDay, now), // outputs 8
CalculateAgeCorrect2(bDay, now))); // outputs 8
这里有一些方法:
public int CalculateAgeWrong1(DateTime birthDate, DateTime now)
{
return new DateTime(now.Subtract(birthDate).Ticks).Year - 1;
}
public int CalculateAgeWrong2(DateTime birthDate, DateTime now)
{
int age = now.Year - birthDate.Year;
if (now < birthDate.AddYears(age))
age--;
return age;
}
public int CalculateAgeCorrect(DateTime birthDate, DateTime now)
{
int age = now.Year - birthDate.Year;
if (now.Month < birthDate.Month || (now.Month == birthDate.Month && now.Day < birthDate.Day))
age--;
return age;
}
public int CalculateAgeCorrect2(DateTime birthDate, DateTime now)
{
int age = now.Year - birthDate.Year;
// For leap years we need this
if (birthDate > now.AddYears(-age))
age--;
// Don't use:
// if (birthDate.AddYears(age) > now)
// age--;
return age;
}
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
}
这是一种奇怪的方法,但如果您将日期设置为yyyymmdd,并从当前日期中减去出生日期,然后删除您获得的年龄的最后4位数字:)
我不知道C#,但我相信这在任何语言中都适用。
20080814 - 19800703 = 280111
删除最后4位=28。
C#代码:
int now = int.Parse(DateTime.Now.ToString("yyyyMMdd"));
int dob = int.Parse(dateOfBirth.ToString("yyyyMMdd"));
int age = (now - dob) / 10000;
或者,也可以不进行扩展方法形式的所有类型转换。忽略错误检查:
public static Int32 GetAge(this DateTime dateOfBirth)
{
var today = DateTime.Today;
var a = (today.Year * 100 + today.Month) * 100 + today.Day;
var b = (dateOfBirth.Year * 100 + dateOfBirth.Month) * 100 + dateOfBirth.Day;
return (a - b) / 10000;
}
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;
}
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