# dtp value

I've got a date time picker control array.
When I run my app, (0) is set by default as 12:00 AM and (1) as 1:00 AM
But if I step through my code,
(0) contains the value of 11/30/1999 while (1) contains 1:00 AM as expected.

Any clue as to why?
LVL 67
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Commented:
12AM is always going to be an issue -- it all comes down to how VB processes dates:

Imagine every day represented as a number (take, for example, today is 5 -- it's not, but let's pretend)
Now each hour in that day is a fraction -- 1/24 (or 0.0416r) in this case.

Today = 5
Today, 1AM = 5.0416r
Today, 2AM = 5.083r
Today, 3AM = 5.125
etc...

Take Today, 2AM:  VB handles the 'whole number' part of these value (the 5) and says that _that_ is the date.  It then takes the 'decimal' part of the value (0.083r) and says that _that_ is the time.

In your case, VB is trying to give you a date/time variable from the value of 12AM.  The problem is, that as 12AM is the 0th hour of the day, the 'decimal' part that it's using to give you the time is 0.0.

0.0 is the same as 0, so when you are converting to date/time form, it is taking that 0 and saying that this value is a 'whole number', so this date/time is a 'whole date'.

When you ask VB to print 5.0000, it prints 5.  Likewise, when you ask it to print 12/05/2004 00:00:00, it prints 12/05/2004.

You may ask, why are you getting this effect when you are only trying to deal with time.

The answer is its a caveate of a by-design feature.  The handling of time only assumes a date 'value' of 0 -- if you do

Print CDate(0)

you will get the earliest date that your computer can handle (mine returns 31/12/1899).  If you do

Print CDate(0.041666666)

you will get the earliest date your computer can handle + 1hr (mine returns 31/12/1899 01:00:00).

dtpTime(0) = TimeValue(dtpTime(0))

is the correct way of disregarding dates when dealing with date/time values.  You should do this every time you handle a dtp value.

HTH

J.
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Commented:
Is the .Format property set to dtpTime?
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Author Commented:
Nope - they're both custom.
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Author Commented:
Sorry - to clarify that custom format:

hh:mm tt
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Commented:
If you convert the value returned by the dtp to a double, with CDbl(yourdtpvalue), is it an integer number or is it divisible by 86400?

There are 86400 seconds in a day and this can cause problems:

0 seconds = the exact date
1.0 days = an exact date
1 second = 12:01:00AM

(I'm not explaining myself too well here, am I)

Try calculating the following:

CDate (yourdtpvalue - Int(yourdtpvalue))

Does this return the correct times?

J.
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Author Commented:
debug.? CDate (dtpTime(0)- Int(dtpTime(0)))
returns
12:00:00 AM

as expected.

Right now, my workaround is:

dtpTime(0) = TimeValue(dtpTime(0))

Although I'd like to know why I need this only on the first element...
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Author Commented:
Thanx for the explanation.  Now, for clarification, although redundant - should I also set the ending time to be Timevalue(dtpTime(1))??
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Author Commented:
Sorry for the delay...I was hoping I'd get a response on my last clarification question, but oh well...