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Properly calculating time differences

Posted on 2004-10-12
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Last Modified: 2008-03-06
Before I get too involved in 'the wrong way' here, let me here some advice from you guys.

I have a time recording application that simply tracks time for my users.
I have two dtp items (dtpTime(0) & (1) - start and stop respectively).

Now, in my click event, I want to calculate the difference between the two.
However, if dtpTime(0) is valued at 12:00 AM, I simply want my total to be whatever is in dtpTime(1)

So, if a user sets
(0) = 8:00 AM
(1) = 8:30 AM

I want (I think) the return to be .5

There will be really rare occasions (and I haven't convinced myself yet to place it back in) where a timer function will allow these values to be set according to a stopwatch type recording for both start and stop values.
However, most values will probably be full, half or quarter hour (1, 1.25, 1.5, 1.75, etc)

Am I going about this the right way here?

The local db will probably store the actual Times (1:00PM and 1:30PM) as well as the difference (.5 or 30 mins).
But the master database is only concerned with the total (.5 or 30 mins).

I need to keep the range local in case the user ever wants to change the timeframe - but once submitted to the master, I have no need, nor will allow, these changes...

Any advice before I get my hands dirty? : )
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Question by:sirbounty
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Assisted Solution

by:mladenovicz
mladenovicz earned 62 total points
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datediff("n","8:00", "8:30") will return you difference in minutes
datediff("n","8:00", "8:30")/60 will return you 0.5
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by:sirbounty
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Right, but am I going about the right method for storing this, or should I take a different approach?
Just looking for advice here really, not so much on the coding methods to be used - I'm familiar with most of them (I think) ;)_
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Accepted Solution

by:
jimbobmcgee earned 63 total points
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Before you start, execute the following statements (in the Immediate window, perhaps):

    MsgBox Now
    MsgBox Int(Now)
    MsgBox CDbl(Now)
    MsgBoc CLong(Now)
    MsgBox Now - Int(CDbl)
   
In VB, a date is a number.  Everything before the decimal point is the date, everything after is the time.

Now try:

    Dim n1 As Date
    Dim n2 As Date
    Dim i As Long

    n1 = Now

    For i = 1 To 500000000: Next i 'WAIT A LONG TIME

    n2 = Now

    MsgBox n1 & vbCrLf & n2 & vbCrLf & CDate(n2 - n1)

You should see the difference in time between n2 and n1...

HTH

J.
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Expert Comment

by:jimbobmcgee
Comment Utility
I guess what I'm saying above is; I'd imagine your database could store them in a date/time format.  If not, a double precision number field should do the trick...
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Author Comment

by:sirbounty
Comment Utility
Oh sure, I can store them in date/time format, and I like that cdate function - never used that one.

Guess that's it then.
Thanx for the info gentlemen.
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