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.
}
其他回答
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);
还有一个答案:
public static int AgeInYears(DateTime birthday, DateTime today)
{
return ((today.Year - birthday.Year) * 372 + (today.Month - birthday.Month) * 31 + (today.Day - birthday.Day)) / 372;
}
这已经过广泛的单元测试。它看起来确实有点“神奇”。数字372是如果每个月有31天,一年中会有多少天。
其工作原理的解释(此处省略)如下:
让我们设置Yn=DateTime.Now.Year,Yb=生日.Year,Mn=DateTime.Now.Month,Mb=生日.Month、Dn=DateTime.Now.Day,Db=生日.Day年龄=Yn-Yb+(31*(Mn-Mb)+(Dn-Db))/372我们知道,如果日期已经到达,我们需要的是Yn-Yb,如果日期尚未到达,则需要Yn-Yb-1。a) 如果Mn<Mb,我们有-341<=31*(Mn-Mb)<=-31和-30<=Dn-Db<=30-371<=31*(锰-Mb)+(Dn-Db)<=-1带整数除法(31*(Mn-Mb)+(Dn-Db))/372=-1b) 如果Mn=Mb和Dn<Db,则我们有31*(Mn-Mb)=0和-30<=Dn Db<=-1再次使用整数除法(31*(Mn-Mb)+(Dn-Db))/372=-1c) 如果Mn>Mb,我们有31<=31*(Mn-Mb)<=341和-30<=Dn-Db<=301<=31*(Mn-Mb)+(Dn-Db)<=371带整数除法(31*(Mn-Mb)+(Dn-Db))/372=0d) 如果Mn=Mb且Dn>Db,则我们有31*(Mn-Mb)=0且1<=Dn Db<=30再次使用整数除法(31*(Mn-Mb)+(Dn-Db))/372=0e) 如果Mn=Mb,Dn=Db,我们有31*(Mn-Mb)+Dn Db=0因此(31*(Mn-Mb)+(Dn-Db))/372=0
非常简单的答案
DateTime dob = new DateTime(1991, 3, 4);
DateTime now = DateTime.Now;
int dobDay = dob.Day, dobMonth = dob.Month;
int add = -1;
if (dobMonth < now.Month)
{
add = 0;
}
else if (dobMonth == now.Month)
{
if(dobDay <= now.Day)
{
add = 0;
}
else
{
add = -1;
}
}
else
{
add = -1;
}
int age = now.Year - dob.Year + add;
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.
}
保持简单(可能是愚蠢的:)。
DateTime birth = new DateTime(1975, 09, 27, 01, 00, 00, 00);
TimeSpan ts = DateTime.Now - birth;
Console.WriteLine("You are approximately " + ts.TotalSeconds.ToString() + " seconds old.");
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