输入：自0001年1月1日起的秒数
输出：此期间的全年数
我开发了一种我认为不是最佳解决方案的算法。我认为应该有一个不涉及循环的解决方案。有关算法的信息，请参阅代码块1. A）确定天数和B）根据闰年从迭代年中逐步减去366或365，同时增加年总数
它并不像分割DayCount那么简单365.2425并截断，因为我们在1月1日遇到了失败点，0002（31536000秒/（365.2425 * 24 * 60 * 60））= 0.99934。
关于非循环方法从0001年1月1日上午12:00起提取年数的想法？
我需要解决这个问题，因为我需要一个长时间嵌入的日期（存储秒数），这样我就可以用1秒的精度跟踪年数超过1200万。
代码块1 - 从秒数（包括闰年）获得年数的低效算法
Dim Days, Years As Integer
'get Days
Days = Ticks * (1 / 24) * (1 / 60) * (1 / 60) 'Ticks = Seconds from Year 1, January 1
'get years by counting up from the beginning
Years = 0
While True
'if leap year
If (Year Mod 4 = 0) AndAlso (Year Mod 100 <> 0) OrElse (Year Mod 400 = 0) Then
If Days >= 366 Then 'if we have enough days left to increment the year
Years += 1
Days -= 366
Else
Exit While
End If
'if not leap year
Else
If Days >= 365 Then 'if we have enough days left to increment the year
Years += 1
Days -= 365
Else
Exit While
End If
End If
End While
Return Years
编辑：我的解决方案是跳过在8位内嵌入日期的内存节省，并将每个值（秒到年）存储在单独的整数中。这会以内存为代价导致即时检索。
Edit2：第一次编辑错误（8位）
答案 0 :(得分：20)
如果您需要 second 的准确性，您可能需要商业级日期时间套餐;使用简单的算法准确地完成它太复杂了。例如：
由于这些和更多的复杂性，你最好不要自己编写代码，除非你可以放松你需要准确度的约束第二次超过1200万年。
“1582年10月4日 - 阿维拉的圣特雷莎去世。第二天，10月15日，她被埋葬了。”
答案 1 :(得分：5)
维基百科有一篇关于Julian Date的文章，其中包含一个可以适应您需求的算法。
答案 2 :(得分：1)
Const TICKS_PER_YEAR As Long = 315360000000000
Function YearsSinceBeginningOfTimeUntil(ByVal d As DateTime) As Integer
Return Math.Floor(d.Ticks / TICKS_PER_YEAR)
End Function
答案 3 :(得分：1)
您不需要循环，计算从0001年1月1日到unix epoch start（1970年1月1日00:00:00）的秒数，并保存到某处。然后从输入中减去它，然后使用任何可用的工具将unix时间戳（1970年1月1日的秒）转换为年，然后添加1970.我不太了解VB编程以发布详细指南。
答案 4 :(得分：0)
我认为这对你有用
function foo(seconds):
count = seconds
year = 0
while (count > 0):
if leap_year(year)
count = count - 366
else
count = count - 365
year ++
return year
答案 5 :(得分：0)
以下假设公历将在即将到来的五百八十四亿五十亿年中继续生效。但是要准备好失望;随着我们的太阳开始膨胀，改变地球的轨道和一年的持续时间，日历可能最终被废弃，当地球落入太阳七十五亿年时，很可能会采取其他措施从现在开始。
顺便说一句，我甚至没有尝试在采用格里高利历之前处理日期。我只返回1582年10月15日之前发生日期的天数，并且需要能够表达那种返回值是GetDateFromSerial函数具有asString的唯一原因。参数。
Sub GetDateFromSerial(ByVal dateSerial As ULong, ByRef year As Long, ByRef month As Integer, ByRef dayOfMonth As Integer, ByRef secondsIntoDay As Integer, ByRef asString As String)
Const SecondsInOneDay As ULong = 86400 ' 24 hours * 60 minutes per hour * 60 seconds per minute
'Dim startOfGregorianCalendar As DateTime = New DateTime(1582, 10, 15)
'Dim startOfGregorianCalendarInSeconds As ULong = (startOfGregorianCalendar - New DateTime(1, 1, 1)).TotalSeconds
Const StartOfGregorianCalendarInSeconds As ULong = 49916304000
secondsIntoDay = dateSerial Mod SecondsInOneDay
If dateSerial < StartOfGregorianCalendarInSeconds Then
year = -1
month = -1
dayOfMonth = -1
Dim days As Integer = (StartOfGregorianCalendarInSeconds - dateSerial) \ SecondsInOneDay
asString = days & IIf(days = 1, " day", " days") & " before the adoption of the Gregorian calendar on October 15, 1582"
Else
'Dim maximumDateValueInSeconds As ULong = (DateTime.MaxValue - New DateTime(1, 1, 1)).TotalSeconds
Const MaximumDateValueInSeconds As ULong = 315537897600
If dateSerial <= MaximumDateValueInSeconds Then
Dim parsedDate As DateTime = DateTime.MinValue.AddSeconds(dateSerial)
year = parsedDate.Year
month = parsedDate.Month
dayOfMonth = parsedDate.Day
Else
' Move the date back into the range that DateTime can parse, by stripping away blocks of
' 400 years. Aim to put the date within the range of years 2001 to 2400.
Dim dateSerialInDays As ULong = dateSerial \ SecondsInOneDay
Const DaysInFourHundredYears As Integer = 365 * 400 + 97 ' Three multiple-of-4 years in each 400 are not leap years.
Dim fourHundredYearBlocks As Integer = dateSerialInDays \ DaysInFourHundredYears
Dim blocksToFactorInLater As Integer = fourHundredYearBlocks - 5
Dim translatedDateSerialInDays As ULong = dateSerialInDays - blocksToFactorInLater * CLng(DaysInFourHundredYears)
' Parse the date as normal now.
Dim parsedDate As DateTime = DateTime.MinValue.AddDays(translatedDateSerialInDays)
year = parsedDate.Year
month = parsedDate.Month
dayOfMonth = parsedDate.Day
' Factor back in the years we took out earlier.
year += blocksToFactorInLater * 400L
End If
asString = New DateTime(2000, month, dayOfMonth).ToString("dd MMM") & ", " & year
End If
End Sub
Function GetSerialFromDate(ByVal year As Long, ByVal month As Integer, ByVal dayOfMonth As Integer, ByVal secondsIntoDay As Integer) As ULong
Const SecondsInOneDay As Integer = 86400 ' 24 hours * 60 minutes per hour * 60 seconds per minute
If (year < 1582) Or _
((year = 1582) And (month < 10)) Or _
((year = 1582) And (month = 10) And (dayOfMonth < 15)) Then
Throw New Exception("The specified date value has no meaning because it falls before the point at which the Gregorian calendar was adopted.")
End If
' Use DateTime for what we can -- which is years prior to 9999 -- and then factor the remaining years
' in. We do this by translating the date back by blocks of 400 years (which are always the same length,
' even factoring in leap years), and then factoring them back in after the fact.
Dim fourHundredYearBlocks As Integer = year \ 400
Dim blocksToFactorInLater As Integer = fourHundredYearBlocks - 5
If blocksToFactorInLater < 0 Then blocksToFactorInLater = 0
year = year - blocksToFactorInLater * 400L
Dim dateValue As DateTime = New DateTime(year, month, dayOfMonth)
Dim translatedDateSerialInDays As ULong = (dateValue - New DateTime(1, 1, 1)).TotalDays
Const DaysInFourHundredYears As ULong = 365 * 400 + 97 ' Three multiple-of-4 years in each 400 are not leap years.
Dim dateSerialInDays As ULong = translatedDateSerialInDays + blocksToFactorInLater * DaysInFourHundredYears
Dim dateSerial As ULong = dateSerialInDays * SecondsInOneDay + secondsIntoDay
Return dateSerial
End Function
答案 6 :(得分：0)
我知道这个问题现在已经过时了，但我经常看到这样的问题，而且这里没有任何简单的答案。
我的解决方案使用了写两个日期的旧技巧，就像它们是数字一样（例如'2013年12月12日'作为20131212），然后从另一个中减去一个并丢弃最后四位数。我在F＃中挖出了我的实现，你可以将它粘贴到LinqPad来检查答案。也考虑到闰年等：
let dateFrom = new DateTime(1,1,1)
let dateTo = dateFrom.AddSeconds(100000000.0)
let yearsSince dateFrom dateTo =
let intRepresentation (date: DateTime) =
date.ToString "yyyy.MMdd" |> Convert.ToDouble
let dateToNum = intRepresentation dateTo
let dateFromNum = intRepresentation dateFrom
int (dateToNum - dateFromNum)
yearsSince dateFrom dateTo |> Dump
let dob = DateTime(1985, 4, 16)
let calculateAge = yearsSince dob
calculateAge DateTime.Today |> Dump
请注意，这非常天真：除了已经由.NET的DateTime类处理的更改之外，它不会考虑时区或历史时区更改。实际的grunt工作由DateTime.AddSeconds方法完成。希望这会有所帮助。