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
我对Mark Soen的答案做了一个小小的修改:我重写了第三行,以便可以更容易地解析表达式。
public int AgeInYears(DateTime bday)
{
DateTime now = DateTime.Today;
int age = now.Year - bday.Year;
if (bday.AddYears(age) > now)
age--;
return age;
}
为了清晰起见,我还将其转换为函数。
其他回答
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
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));
}
我有一个定制的计算年龄的方法,加上一条奖金验证消息,以防有帮助:
public void GetAge(DateTime dob, DateTime now, out int years, out int months, out int days)
{
years = 0;
months = 0;
days = 0;
DateTime tmpdob = new DateTime(dob.Year, dob.Month, 1);
DateTime tmpnow = new DateTime(now.Year, now.Month, 1);
while (tmpdob.AddYears(years).AddMonths(months) < tmpnow)
{
months++;
if (months > 12)
{
years++;
months = months - 12;
}
}
if (now.Day >= dob.Day)
days = days + now.Day - dob.Day;
else
{
months--;
if (months < 0)
{
years--;
months = months + 12;
}
days += DateTime.DaysInMonth(now.AddMonths(-1).Year, now.AddMonths(-1).Month) + now.Day - dob.Day;
}
if (DateTime.IsLeapYear(dob.Year) && dob.Month == 2 && dob.Day == 29 && now >= new DateTime(now.Year, 3, 1))
days++;
}
private string ValidateDate(DateTime dob) //This method will validate the date
{
int Years = 0; int Months = 0; int Days = 0;
GetAge(dob, DateTime.Now, out Years, out Months, out Days);
if (Years < 18)
message = Years + " is too young. Please try again on your 18th birthday.";
else if (Years >= 65)
message = Years + " is too old. Date of Birth must not be 65 or older.";
else
return null; //Denotes validation passed
}
方法调用此处并传递日期时间值(如果服务器设置为美国语言环境,则为MM/dd/yyyy)。将其替换为消息框或要显示的任何容器:
DateTime dob = DateTime.Parse("03/10/1982");
string message = ValidateDate(dob);
lbldatemessage.Visible = !StringIsNullOrWhitespace(message);
lbldatemessage.Text = message ?? ""; //Ternary if message is null then default to empty string
记住,您可以按任何方式格式化邮件。
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。
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);
推荐文章
- 如何从枚举中选择一个随机值?
- 驻留在App_Code中的类不可访问
- 在链式LINQ扩展方法调用中等价于'let'关键字的代码
- dynamic (c# 4)和var之间的区别是什么?
- Visual Studio: ContextSwitchDeadlock
- Sql Server字符串到日期的转换
- 返回文件在ASP。Net Core Web API
- 自定义HttpClient请求头
- 如果我使用OWIN Startup.cs类并将所有配置移动到那里,我是否需要一个Global.asax.cs文件?
- VS2013外部构建错误"error MSB4019: The imported project <path> was not found"
- 从另一个列表id中排序一个列表
- 等待一个无效的异步方法
- 无法加载文件或程序集…参数不正确
- c#中枚举中的方法
- 如何从字符串中删除新的行字符?