我想计算两个日期之间的差异，并希望根据差异将其转换为2年，5个月或仅3个月或2天，因为所有月份都是30天......
例如;
来自及包括：2009年3月12日 至，但不包括：2011年11月26日
输出必须是：2年，8个月，14天，不包括结束日期。
另一个例子;
开始时间：2010年1月26日 结束：2010年2月15日
输出：从开始日期到结束日期的20天，但不包括结束日期
我可以使用Datediff计算月，日或小时的差异，但问题是如何将其转换为年，月和日期。实际上这很复杂，因为我们不知道两个月之间有多少天（30,31或28天）
我使用这个经典ASP代码来转换差异，但有许多缺点。
Function Convert_Date_to_Text(tarih1,tarih2,useDates)
if (tarih1<>"" AND tarih2<>"") then
if Tarih_Araligi_Belirle(tarih1,tarih2,"day")>0 then
Date1_Year = Year(tarih1)
Date1_Month = Month(tarih1)
Date1_Day = Day(tarih1)
Date2_Year = Year(tarih2)
Date2_Month = Month(tarih2)
Date2_Day = Day(tarih2)
If (Date1_Month = 12) and (Date1_Day = 31) and
(Date2_Month = 1) and (Date2_Day = 1) Then
NoOfyears = Date2_Year - Date1_Year - 1
NoOfmonths = 0
NoOfdays = 1
Else
NoOfyears = Date2_Year - Date1_Year
NoOfmonths = Date2_Month - Date1_Month
NoOfdays = Date2_Day - Date1_Day
End If
If NoOfyears = 1 Then
FormatString = "1 year "
Else If NoOfyears <= 0 then
FormatString = ""
Else
FormatString = CStr(NoOfyears) & " years "
End If:End If
If NoOfmonths = 1 Then
FormatString = FormatString & "1 month"
Else If NoOfmonths <= 0 then
FormatString = FormatString
Else
FormatString = FormatString & CStr(NoOfmonths) & " months "
End If:End If
if useDates=1 then
If NoOfdays = 1 Then
FormatString = FormatString & "1 day"
Else If NoOfdays <= 0 Then
FormatString = FormatString
Else
FormatString = FormatString & CStr(NoOfdays) & " days"
End If:End If
end if
end if
end if
Convert_Date_to_Text = FormatString
End Function
此网站完美地计算差异。 TimeAndDate.Com
注意：我使用Classic ASP有几个原因（公司限制）。对不起，但我需要ASP功能。看起来ASP中不存在TimeSpan :(
亲切的问候
答案 0 :(得分：2)
如果您可以将输入字符串转换为DateTime个变量，可以尝试这样的事情：
DateTime starTime = //something;
DateTime endTime = //something;
TimeSpan oneDay = new TimeSpan(1, 0, 0, 0); //creates a timespan of 1 day
TimeSpan deltaTime = (endTime - startTime) - oneDay;
我认为asp有DateTime和TimeSpan变量类型。
答案 1 :(得分：1)
这个怎么样？ （没有TimeSpan但不确定是否经典的asp兼容）
DateTime dateTime1 = new DateTime(2003,2,2);
DateTime dateTime2 = new DateTime(2001,1,1);
int daysDiff = dateTime1.Day - dateTime2.Day;
int monthsDiff = dateTime1.Month - dateTime2.Month;
int yearsDiff = dateTime1.Year - dateTime2.Year;
if (daysDiff < 0)
{
daysDiff += DateTime.DaysInMonth(dateTime1.Year, dateTime1.Month);
monthsDiff--;
}
if (monthsDiff < 0)
{
monthsDiff += 12;
yearsDiff--;
}
Console.WriteLine(daysDiff);
Console.WriteLine(monthsDiff);
Console.WriteLine(yearsDiff);
答案 2 :(得分：1)
这是我过去使用过的一个功能。如果你测试它，我想你会发现它准确。 Here's where I got it from.
Function YearsMonthsDays(Date1 As Date, Date2 As Date, Optional ShowAll As _
Boolean = False, Optional Grammar As Boolean = True)
' This function returns a string "X years, Y months, Z days" showing the time
' between two dates. This function may be used in any VBA or VB project
' Date1 and Date2 must either be dates, or strings that can be implicitly
' converted to dates. If these arguments have time portions, the time portions
' are ignored. If Date1 > Date2 (after ignoring time portions), the function
' returns an empty string
' ShowAll indicates whether all portions of the string "X years, Y months, Z days"
' are included in the output. If ShowAll = True, all portions of the string are
' always included. If ShowAll = False, then if the year portion is zero the year
' part of the string is omitted, and if the year portion and month portion are both
' zero, than both year and month portions are omitted. The day portion is always
' included, and if at least one year has passed then the month portion is always
' included
' Grammar indicates whether to test years/months/days for singular or plural
' By definition, a "full month" means that the day number in Date2 is >= the day
' number in Date1, or Date1 and Date2 occur on the last days of their respective
' months. A "full year" means that 12 "full months" have passed.
' In Excel, this function is an alternative to the little-known DATEDIF. DATEDIF
' usually works well, but can create strange results when a date is at month end.
' Thus, this formula:
' =DATEDIF(A1,B1,"y") & " years, " & DATEDIF(A1,B1,"ym") & " months, " &
' DATEDIF(A1,B1,"md") & " days"
' will return "0 years, 1 months, -2 days" for 31-Jan-2006 and 1-Mar-2006.
' This function will return "0 years, 1 month, 1 day"
Dim TestYear As Long, TestMonth As Long, TestDay As Long
Dim TargetDate As Date, Last1 As Date, Last2 As Date
' Strip time portions
Date1 = Int(Date1)
Date2 = Int(Date2)
' Test for invalid dates
If Date1 > Date2 Then
YearsMonthsDays = ""
Exit Function
End If
' Test for whether the calendar year is the same
If Year(Date2) > Year(Date1) Then
' Different calendar year.
' Test to see if calendar month is the same. If it is, we have to look at the
' day to see if a full year has passed
If Month(Date2) = Month(Date1) Then
If Day(Date2) >= Day(Date1) Then
TestYear = DateDiff("yyyy", Date1, Date2)
Else
TestYear = DateDiff("yyyy", Date1, Date2) - 1
End If
' In this case, a full year has definitely passed
ElseIf Month(Date2) > Month(Date1) Then
TestYear = DateDiff("yyyy", Date1, Date2)
' A full year has not passed
Else
TestYear = DateDiff("yyyy", Date1, Date2) - 1
End If
' Calendar year is the same, so a full year has not passed
Else
TestYear = 0
End If
' Test to see how many full months have passed, in excess of the number of full
' years
TestMonth = (DateDiff("m", DateSerial(Year(Date1), Month(Date1), 1), _
DateSerial(Year(Date2), Month(Date2), 1)) + IIf(Day(Date2) >= _
Day(Date1), 0, -1)) Mod 12
' See how many days have passed, in excess of the number of full months. If the day
' number for Date2 is >= that for Date1, it's simple
If Day(Date2) >= Day(Date1) Then
TestDay = Day(Date2) - Day(Date1)
' If not, we have to test for end of the month
Else
Last1 = DateSerial(Year(Date2), Month(Date2), 0)
Last2 = DateSerial(Year(Date2), Month(Date2) + 1, 0)
TargetDate = DateSerial(Year(Date2), Month(Date2) - 1, Day(Date1))
If Last2 = Date2 Then
If TestMonth = 11 Then
TestMonth = 0
TestYear = TestYear + 1
Else
TestMonth = TestMonth + 1
End If
Else
TestDay = DateDiff("d", IIf(TargetDate > Last1, Last1, TargetDate), Date2)
End If
End If
If ShowAll Or TestYear >= 1 Then
YearsMonthsDays = TestYear & IIf(TestYear = 1 And Grammar, " year, ", _
" years, ") & TestMonth & IIf(TestMonth = 1 And Grammar, " month, ", _
" months, ") & TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
Else
If TestMonth >= 1 Then
YearsMonthsDays = TestMonth & IIf(TestMonth = 1 And Grammar, " month, ", _
" months, ") & TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
Else
YearsMonthsDays = TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
End If
End If
End Function
答案 3 :(得分：0)
您可以减去DateTime个对象以获取TimeSpan对象：
DateTime startDate = GetStartDate();
DateTime endDate = GetEndDate();
TimeSpan duration = endDate - startDate;
答案 4 :(得分：0)
此article包含 DateDiff 类：
// ----------------------------------------------------------------------
public void DateDiffSample()
{
DateTime date1 = new DateTime( 2009, 11, 8, 7, 13, 59 );
Console.WriteLine( "Date1: {0}", date1 );
// > Date1: 08.11.2009 07:13:59
DateTime date2 = new DateTime( 2011, 3, 20, 19, 55, 28 );
Console.WriteLine( "Date2: {0}", date2 );
// > Date2: 20.03.2011 19:55:28
DateDiff dateDiff = new DateDiff( date1, date2 );
// differences
Console.WriteLine( "DateDiff.Years: {0}", dateDiff.Years );
// > DateDiff.Years: 1
Console.WriteLine( "DateDiff.Quarters: {0}", dateDiff.Quarters );
// > DateDiff.Quarters: 5
Console.WriteLine( "DateDiff.Months: {0}", dateDiff.Months );
// > DateDiff.Months: 16
Console.WriteLine( "DateDiff.Weeks: {0}", dateDiff.Weeks );
// > DateDiff.Weeks: 70
Console.WriteLine( "DateDiff.Days: {0}", dateDiff.Days );
// > DateDiff.Days: 497
Console.WriteLine( "DateDiff.Weekdays: {0}", dateDiff.Weekdays );
// > DateDiff.Weekdays: 71
Console.WriteLine( "DateDiff.Hours: {0}", dateDiff.Hours );
// > DateDiff.Hours: 11940
Console.WriteLine( "DateDiff.Minutes: {0}", dateDiff.Minutes );
// > DateDiff.Minutes: 716441
Console.WriteLine( "DateDiff.Seconds: {0}", dateDiff.Seconds );
// > DateDiff.Seconds: 42986489
// elapsed
Console.WriteLine( "DateDiff.ElapsedYears: {0}", dateDiff.ElapsedYears );
// > DateDiff.ElapsedYears: 1
Console.WriteLine( "DateDiff.ElapsedMonths: {0}", dateDiff.ElapsedMonths );
// > DateDiff.ElapsedMonths: 4
Console.WriteLine( "DateDiff.ElapsedDays: {0}", dateDiff.ElapsedDays );
// > DateDiff.ElapsedDays: 12
Console.WriteLine( "DateDiff.ElapsedHours: {0}", dateDiff.ElapsedHours );
// > DateDiff.ElapsedHours: 12
Console.WriteLine( "DateDiff.ElapsedMinutes: {0}", dateDiff.ElapsedMinutes );
// > DateDiff.ElapsedMinutes: 41
Console.WriteLine( "DateDiff.ElapsedSeconds: {0}", dateDiff.ElapsedSeconds );
// > DateDiff.ElapsedSeconds: 29
// description
Console.WriteLine( "DateDiff.GetDescription(1): {0}", dateDiff.GetDescription( 1 ) );
// > DateDiff.GetDescription(1): 1 Year
Console.WriteLine( "DateDiff.GetDescription(2): {0}", dateDiff.GetDescription( 2 ) );
// > DateDiff.GetDescription(2): 1 Year 4 Months
Console.WriteLine( "DateDiff.GetDescription(3): {0}", dateDiff.GetDescription( 3 ) );
// > DateDiff.GetDescription(3): 1 Year 4 Months 12 Days
Console.WriteLine( "DateDiff.GetDescription(4): {0}", dateDiff.GetDescription( 4 ) );
// > DateDiff.GetDescription(4): 1 Year 4 Months 12 Days 12 Hours
Console.WriteLine( "DateDiff.GetDescription(5): {0}", dateDiff.GetDescription( 5 ) );
// > DateDiff.GetDescription(5): 1 Year 4 Months 12 Days 12 Hours 41 Mins
Console.WriteLine( "DateDiff.GetDescription(6): {0}", dateDiff.GetDescription( 6 ) );
// > DateDiff.GetDescription(6): 1 Year 4 Months 12 Days 12 Hours 41 Mins 29 Secs
} // DateDiffSample
答案 5 :(得分：0)
Dim intYears
Dim intMonths
Dim intDays
Dim strDate1
Dim strDate2
Dim strAnswer
strDate1 = "01/26/2010"
strDate2 = "02/15/2010"
intYears = DateDiff("yyyy",strDate1,strDate2)
intMonths = DateDiff("m",strDate1,strDate2)
intDays = DateDiff("d",strDate1,strDate2)
strAnswer = ""
if intYears > 0 then
strAnswer = strAnswer & CStr(intYears) & "years "
end if
if intMonths > 0 then
strAnswer = strAnswer & CStr(intMonths) & "months"
end if
if intDays > 0 then
strAnswer = strAnswer & CStr(intDays) & "days"
end if
Response.Write("The difference between these two dates is " & strAnswer)