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
这里有一个单行线:
int age = new DateTime(DateTime.Now.Subtract(birthday).Ticks).Year-1;
其他回答
这是我们在这里使用的版本。它有效,而且相当简单。这与Jeff的想法相同,但我认为它更清晰一点,因为它分离了减法的逻辑,所以更容易理解。
public static int GetAge(this DateTime dateOfBirth, DateTime dateAsAt)
{
return dateAsAt.Year - dateOfBirth.Year - (dateOfBirth.DayOfYear < dateAsAt.DayOfYear ? 0 : 1);
}
如果你认为这类事情不清楚,你可以扩展三元运算符使其更清晰。
显然,这是作为DateTime上的一个扩展方法完成的,但很明显,您可以抓取一行代码来完成工作并将其放在任何位置。这里我们有另一个传入DateTime的Extension方法重载。
另一个功能,不是我做的,而是在网上找到的,并做了一些改进:
public static int GetAge(DateTime birthDate)
{
DateTime n = DateTime.Now; // To avoid a race condition around midnight
int age = n.Year - birthDate.Year;
if (n.Month < birthDate.Month || (n.Month == birthDate.Month && n.Day < birthDate.Day))
age--;
return age;
}
我只想到了两件事:来自不使用公历的国家的人呢?DateTime。我认为现在是服务器特定的文化。我对实际使用亚洲日历一无所知,我不知道是否有一种简单的方法来转换日历之间的日期,但以防万一,你想知道4660年的中国人:-)
我使用ScArcher2的解决方案来精确计算人的年龄,但我需要进一步计算他们的月和日以及年。
public static Dictionary<string,int> CurrentAgeInYearsMonthsDays(DateTime? ndtBirthDate, DateTime? ndtReferralDate)
{
//----------------------------------------------------------------------
// Can't determine age if we don't have a dates.
//----------------------------------------------------------------------
if (ndtBirthDate == null) return null;
if (ndtReferralDate == null) return null;
DateTime dtBirthDate = Convert.ToDateTime(ndtBirthDate);
DateTime dtReferralDate = Convert.ToDateTime(ndtReferralDate);
//----------------------------------------------------------------------
// Create our Variables
//----------------------------------------------------------------------
Dictionary<string, int> dYMD = new Dictionary<string,int>();
int iNowDate, iBirthDate, iYears, iMonths, iDays;
string sDif = "";
//----------------------------------------------------------------------
// Store off current date/time and DOB into local variables
//----------------------------------------------------------------------
iNowDate = int.Parse(dtReferralDate.ToString("yyyyMMdd"));
iBirthDate = int.Parse(dtBirthDate.ToString("yyyyMMdd"));
//----------------------------------------------------------------------
// Calculate Years
//----------------------------------------------------------------------
sDif = (iNowDate - iBirthDate).ToString();
iYears = int.Parse(sDif.Substring(0, sDif.Length - 4));
//----------------------------------------------------------------------
// Store Years in Return Value
//----------------------------------------------------------------------
dYMD.Add("Years", iYears);
//----------------------------------------------------------------------
// Calculate Months
//----------------------------------------------------------------------
if (dtBirthDate.Month > dtReferralDate.Month)
iMonths = 12 - dtBirthDate.Month + dtReferralDate.Month - 1;
else
iMonths = dtBirthDate.Month - dtReferralDate.Month;
//----------------------------------------------------------------------
// Store Months in Return Value
//----------------------------------------------------------------------
dYMD.Add("Months", iMonths);
//----------------------------------------------------------------------
// Calculate Remaining Days
//----------------------------------------------------------------------
if (dtBirthDate.Day > dtReferralDate.Day)
//Logic: Figure out the days in month previous to the current month, or the admitted month.
// Subtract the birthday from the total days which will give us how many days the person has lived since their birthdate day the previous month.
// then take the referral date and simply add the number of days the person has lived this month.
//If referral date is january, we need to go back to the following year's December to get the days in that month.
if (dtReferralDate.Month == 1)
iDays = DateTime.DaysInMonth(dtReferralDate.Year - 1, 12) - dtBirthDate.Day + dtReferralDate.Day;
else
iDays = DateTime.DaysInMonth(dtReferralDate.Year, dtReferralDate.Month - 1) - dtBirthDate.Day + dtReferralDate.Day;
else
iDays = dtReferralDate.Day - dtBirthDate.Day;
//----------------------------------------------------------------------
// Store Days in Return Value
//----------------------------------------------------------------------
dYMD.Add("Days", iDays);
return dYMD;
}
无分支解决方案:
public int GetAge(DateOnly birthDate, DateOnly today)
{
return today.Year - birthDate.Year + (((today.Month << 5) + today.Day - ((birthDate.Month << 5) + birthDate.Day)) >> 31);
}
2需要解决的主要问题有:
1.计算准确年龄-以年、月、日等为单位。
2.计算人们普遍认为的年龄——人们通常不关心自己到底多大,他们只关心自己当年的生日是什么时候。
1的解决方案显而易见:
DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today; //we usually don't care about birth time
TimeSpan age = today - birth; //.NET FCL should guarantee this as precise
double ageInDays = age.TotalDays; //total number of days ... also precise
double daysInYear = 365.2425; //statistical value for 400 years
double ageInYears = ageInDays / daysInYear; //can be shifted ... not so precise
2的解决方案在确定总年龄时并不那么精确,但人们认为它是精确的。当人们“手动”计算年龄时,通常也会使用它:
DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today;
int age = today.Year - birth.Year; //people perceive their age in years
if (today.Month < birth.Month ||
((today.Month == birth.Month) && (today.Day < birth.Day)))
{
age--; //birthday in current year not yet reached, we are 1 year younger ;)
//+ no birthday for 29.2. guys ... sorry, just wrong date for birth
}
注释2.:
这是我的首选解决方案我们不能使用DateTime.DayOfYear或TimeSpans,因为它们会在闰年中改变天数为了可读性,我只增加了几行
还有一个提示。。。我将为它创建两个静态重载方法,一个用于通用,另一个用于使用友好:
public static int GetAge(DateTime bithDay, DateTime today)
{
//chosen solution method body
}
public static int GetAge(DateTime birthDay)
{
return GetAge(birthDay, DateTime.Now);
}
推荐文章
- 实体框架核心:在上一个操作完成之前,在此上下文中开始的第二个操作
- 如何为构造函数定制Visual Studio的私有字段生成快捷方式?
- 为什么Visual Studio 2015/2017/2019测试运行器没有发现我的xUnit v2测试
- 如何使用JSON确保字符串是有效的JSON。网
- AppSettings从.config文件中获取值
- 通过HttpClient向REST API发布一个空体
- 如何检查IEnumerable是否为空或空?
- 自动化invokerrequired代码模式
- 解析日期字符串并更改格式
- 没有ListBox。SelectionMode="None",是否有其他方法禁用列表框中的选择?
- 在c#代码中设置WPF文本框的背景颜色
- 在c#中,什么是单子?
- c#和Java中的泛型有什么不同?和模板在c++ ?
- c#线程安全快速(est)计数器
- 如何将此foreach代码转换为Parallel.ForEach?