アクセスのdd：hh：mmの日付の差

Question

フィールドの日付と今日の日付の間のdatedfferenceをdd：hh：mm fromatで計算します。私は以下を試しました

Format(DateDiff("s",FieldA,Date())/86400,"dd:hh:nn") 

しかし、上記のコードは私に正しい違いを与えません。たとえば、FieldAの最初の値は04.11.2016 11:52:56です。違いは約11日間でなければなりません。しかし、私は09:12:07の結果を得ます。私は間違っていますか？

Answer 1

以下のような機能を使用できます。その後：

TimePassed: FormatYearDayHourMinuteSecondDiff(#2016/11/04 11:52:56#, Date()) 

'TimePassed -> 0 11 12:07:04 

それは外部関数Yearsを使用しますが、簡単にそれが年を表示しないように変更することができます。

Public Function FormatYearDayHourMinuteSecondDiff(_ 
    ByVal datTimeStart As Date, _ 
    ByVal datTimeEnd As Date, _ 
    Optional ByVal strSeparatorDate As String = " ", _ 
    Optional ByVal strSeparatorTime As String = ":") _ 
    As String 

' Returns count of years, days, hours, minutes and seconds of difference 
' between datTimeStart and datTimeEnd converted to 
' years, days, hours and minutes and seconds as a formatted string 
' with an optional choice of date and/or time separator. 
' 
' Should return correct output for a negative time span but 
' this is not fully tested. 
' 
' Example: 
' datTimeStart: #2006-05-24 10:03:02# 
' datTimeEnd : #2009-04-17 20:01:18# 
' returns  : 2 328 09:58:16 
' 
' 2007-11-06. Cactus Data ApS, CPH. 

    Const cintSecondsHour As Integer = 60& * 60& 

    Dim intYears  As Integer 
    Dim intDays  As Integer 
    Dim intSeconds As Integer 
    Dim intHours  As Integer 
    Dim datTime  As Date 
    Dim strDatePart As String 
    Dim strTimePart As String 
    Dim strYDHMS  As String 

    intYears = Years(datTimeStart, datTimeEnd) 
    datTimeStart = DateAdd("yyyy", intYears, datTimeStart) 
    intDays = DateDiff("h", datTimeStart, datTimeEnd) \ 24 
    datTimeStart = DateAdd("d", intDays, datTimeStart) 
    intHours = DateDiff("h", datTimeStart, datTimeEnd) 
    datTimeStart = DateAdd("h", intHours, datTimeStart) 
    intSeconds = DateDiff("s", datTimeStart, datTimeEnd) 

    ' Format year and day part. 
    strDatePart = CStr(intYears) & strSeparatorDate & CStr(intDays) 
    datTime = TimeSerial(intHours, 0, intSeconds Mod cintSecondsHour) 
    ' Format hour, minute and second part. 
    strTimePart = Format(datTime, "hh\" & strSeparatorTime & "nn\" & strSeparatorTime & "ss") 
    strYDHMS = strDatePart & " " & IIf(datTime < 0, "-", "") & strTimePart 

    FormatYearDayHourMinuteSecondDiff = strYDHMS 

End Function