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# Properly calculating time differences

Posted on 2004-10-12
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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)

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? : )
0
Question by:sirbounty
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LVL 8

Assisted Solution

mladenovicz earned 62 total points
datediff("n","8:00", "8:30") will return you difference in minutes
datediff("n","8:00", "8:30")/60 will return you 0.5
0

LVL 67

Author Comment

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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LVL 16

Accepted Solution

jimbobmcgee earned 63 total points
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.
0

LVL 16

Expert Comment

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...
0

LVL 67

Author Comment

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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