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Thread: houres counting

042713, 13:48 #1Registered User
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 Apr 2013
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Unanswered: houres counting
please i do want to do somme calculation like : 123h34  or + 3456h12min and have result in a text box and a table. all that in order to control MTBF(mean time between failer) of a mecanical componenent
thanks

042713, 15:53 #2Moderator
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 Mar 2009
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Provided Answers: 14Search in Access help for the DateAdd() and DateDiff functions. They also work for timeonly values.
Have a nice day!

043013, 18:01 #3Registered User
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 Apr 2013
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thanks
i did but no answer to my question

050113, 04:21 #4Jaded Developer
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 Nov 2004
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Provided Answers: 59how is your data stored
is it stored as decimal hours, minutes or as a string (999h99)
the smartedt woudlbe as noMinutes, and then format that as required
however the fact that you are asking the question suggests to me is as a string to do that I'd probably write a function that returned the number of minutes or hours
eg
Code:Public Function ConvertStringTime(TimeText As String) As Double 'ConvertStringTime converts a string representation of hours and minute (TimeText) 'and returns then decimal number of hours that string represents 'TimeText inputformat is xhyy 'where x is the number of hours can be any size and be signed, if not signed assume +ve 'y is two digits reprsenting the number of minutes 'the function returns a representation of that time as decimal hours 'ferinstance 123h45 returns 123.75 ConvertStringTime = 0 'set up our default return value 'lets filter out duff data 'first off if we have a null string or a string comprising less than 3 characters its duff, so can it If IsNull(TmeText) Or Len(TimeText) < 3 Then Exit Function 'have they been dumb enough to supply a decimal value already If (Mid(TimeText, Len(TimeText)  2, 1)) = "." Then If IsNumeric(TimeText) = True Then ConvertStringTime = CDbl(TimeText) End If Exit Function End If 'OK so is it a valid format? IE has an h in 3rd right position If UCase(Mid(TimeText, Len(TimeText)  2, 1)) <> "H" Then Exit Function 'OK so are the last two characters (the minutes portion) numeric If IsNumeric(Right(TimeText, 2)) = False Then Exit Function 'AND is the number of minutes valid If CInt(Right(TimeText, 2)) > 59 Then Exit Function 'just in case we have 0 hours eg h56 force that to be 0h56 If Len(TimeText) = 3 Then TimeText = "0" & TimeText 'are the first n characters (the hours portion) numeric If IsNumeric(Left(TimeText, Len(TimeText)  3)) = False Then Exit Function 'phew, so we got to here we think timetext contains valid data 'so convert timetext into decimal hours ConvertStringTime = CInt(Left(TimeText, Len(TimeText)  3)) 'if the hour component is negative If ConvertStringTime < 0 Then 'subtract the minute component ConvertStringTime = ConvertStringTime  CDbl(Right(TimeText, 2)) / 60 Else 'otherwise add the minute component ConvertStringTime = ConvertStringTime + CDbl(Right(TimeText, 2)) / 60 End If End Function Public Function FormatToHoursMinutes(TimeValue As Double) As String 'converts decimal hours to xxhmm 'ferinstance 45.5 = 45h30 Dim NoHours As Integer Dim NoMinutes As Integer NoHours = Fix(TimeValue) 'use fix to round the hours component NoMinutes = (Abs(TimeValue)  Abs(NoHours)) * 60 'find the minutes component (but alwasy use a positive number FormatToHoursMinutes = NoHours & "H" & Format(NoMinutes, "00") End Function
I'd rather be riding on the Tiger 800 or the Norton