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;
}
其他回答
以下方法(从.NET类DateDiff的时间段库中提取)考虑区域性信息的日历:
// ----------------------------------------------------------------------
private static int YearDiff( DateTime date1, DateTime date2 )
{
return YearDiff( date1, date2, DateTimeFormatInfo.CurrentInfo.Calendar );
} // YearDiff
// ----------------------------------------------------------------------
private static int YearDiff( DateTime date1, DateTime date2, Calendar calendar )
{
if ( date1.Equals( date2 ) )
{
return 0;
}
int year1 = calendar.GetYear( date1 );
int month1 = calendar.GetMonth( date1 );
int year2 = calendar.GetYear( date2 );
int month2 = calendar.GetMonth( date2 );
// find the the day to compare
int compareDay = date2.Day;
int compareDaysPerMonth = calendar.GetDaysInMonth( year1, month1 );
if ( compareDay > compareDaysPerMonth )
{
compareDay = compareDaysPerMonth;
}
// build the compare date
DateTime compareDate = new DateTime( year1, month2, compareDay,
date2.Hour, date2.Minute, date2.Second, date2.Millisecond );
if ( date2 > date1 )
{
if ( compareDate < date1 )
{
compareDate = compareDate.AddYears( 1 );
}
}
else
{
if ( compareDate > date1 )
{
compareDate = compareDate.AddYears( -1 );
}
}
return year2 - calendar.GetYear( compareDate );
} // YearDiff
用法:
// ----------------------------------------------------------------------
public void CalculateAgeSamples()
{
PrintAge( new DateTime( 2000, 02, 29 ), new DateTime( 2009, 02, 28 ) );
// > Birthdate=29.02.2000, Age at 28.02.2009 is 8 years
PrintAge( new DateTime( 2000, 02, 29 ), new DateTime( 2012, 02, 28 ) );
// > Birthdate=29.02.2000, Age at 28.02.2012 is 11 years
} // CalculateAgeSamples
// ----------------------------------------------------------------------
public void PrintAge( DateTime birthDate, DateTime moment )
{
Console.WriteLine( "Birthdate={0:d}, Age at {1:d} is {2} years", birthDate, moment, YearDiff( birthDate, moment ) );
} // PrintAge
我已经创建了一个SQL Server用户定义函数来计算某人的年龄,给定他们的出生日期。当您需要它作为查询的一部分时,这很有用:
using System;
using System.Data;
using System.Data.Sql;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;
public partial class UserDefinedFunctions
{
[SqlFunction(DataAccess = DataAccessKind.Read)]
public static SqlInt32 CalculateAge(string strBirthDate)
{
DateTime dtBirthDate = new DateTime();
dtBirthDate = Convert.ToDateTime(strBirthDate);
DateTime dtToday = DateTime.Now;
// get the difference in years
int years = dtToday.Year - dtBirthDate.Year;
// subtract another year if we're before the
// birth day in the current year
if (dtToday.Month < dtBirthDate.Month || (dtToday.Month == dtBirthDate.Month && dtToday.Day < dtBirthDate.Day))
years=years-1;
int intCustomerAge = years;
return intCustomerAge;
}
};
要使用最近的年龄计算年龄:
var ts = DateTime.Now - new DateTime(1988, 3, 19);
var age = Math.Round(ts.Days / 365.0);
简单易懂的解决方案。
// Save today's date.
var today = DateTime.Today;
// Calculate the age.
var age = today.Year - birthdate.Year;
// Go back to the year in which the person was born in case of a leap year
if (birthdate.Date > today.AddYears(-age)) age--;
然而,这假设你在寻找西方的时代观念,而不是使用东亚的推算法。
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));
}
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