Question

datediff from a yyyymmdd #### fields

Asked by: epicazo

I am not sure how to get started with this
DATEDIFF(PPADDT PPADTM, PPDSDT PPDSTM) ... I would like results to show when the difference is greater than 24 hours as well.  (dd:hh:mm) or ex 40:mm)
 
PPADDT      PPADTM      PPDSDT      PPDSTM
20091001      2310      20091002      450
20091001      2338      20091002      500
20091002      20      20091002      545
20091002      40      20091002      520
20091002      258      20091002      447
20091002      609      20091002      820
20091002      637      20091002      825
20091002      730      20091002      930

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Asked On
2009-11-01 at 00:29:07ID24861527
Tags

microsoft access 2007

Topic

Microsoft Access Database

Participating Experts
3
Points
250
Comments
9

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Answers

 

by: peter57rPosted on 2009-11-01 at 00:46:36ID: 25712930

The datediff expression is to give the difference in minutes is ..

datediff("n", cdate(format(ppaddt,"00/00/00 ") & format(ppadtm, "00:00")), cdate(format(ppdsdt,"00/00/00 ") & format(ppdstm, "00:00")))

 

by: peter57rPosted on 2009-11-01 at 00:49:34ID: 25712939

Sorry - need 4 digit year..
datediff("n", cdate(format(ppaddt,"0000/00/00 ") & format(ppadtm, "00:00")), cdate(format(ppdsdt,"0000/00/00 ") & format(ppdstm, "00:00")))

 

by: GrahamMandenoPosted on 2009-11-01 at 02:08:34ID: 25713178

No, sorry, you can't use the Format function here.  Format turns a date/time into a string - it cannot do the reverse!

I'm not clear whether your data is text or numeric, and this would make a difference to the approach.  For example, to convert a string (s) in the form yyyymmdd to a date, you would use:

DateSerial( Left(s, 4), Mid(s, 5, 2), Right(s, 2) )

If the yyyymmdd was, say, a long integer (d) then you would use an arithmetic approach:

DateSerial( d \ 10000, (d \ 100) Mod 100, d Mod 100)

Looking at your time values, it looks like all your data is numeric - for example, 258 is presumably 2:58am and 20 is presumably 00:20, or 12:20am.

So, the first thing to do is write a function to take the date field and the time field and convert them into a date/time data type:

Public Function MakeDate( d as Long, t as Long) as Date
MakeDate = DateSerial( d \ 10000, (d \ 100) Mod 100, d Mod 100) _
   + TimeSerial( t \ 100, t Mod 100, 0 )
End Function

To find the difference between the two dates and times, use the smallest granularity in the source data for the unit - in this case, minutes:

DateDiff( "n", MakeDate(PPADDT, PPADTM), MakeDate(PPDSDT, PPDSTM) )

This will give you the difference, in minutes, between the two times.

To format this, once again you will need a custom function (the Format function is great for formatting points in time, but not durations).  The following function will format a number of minutes as hh:mm, even if it is more than 24 hours:

Public Function MinsToHHMM( mins as Long ) as String
MinsToHHMM = mins \ 60 & ":" & Format( mins Mod 60, "00" )
End Function

Putting all this together, this should do the trick:

MinsToHHMM( DateDiff( "n", MakeDate(PPADDT, PPADTM), MakeDate(PPDSDT, PPDSTM) ) )

--
Graham

 

by: harfangPosted on 2009-11-01 at 03:29:35ID: 25713354

Hello Graham,

Format does in fact work, and I use this quite regularly. Peter's (modified) expression is:

DateDiff('n', Format([ppaddt],'0\-00\-00\ ')+Format([ppadtm],'0\:00'), Format([ppdsdt],'0\-00\-00\ ')+Format([ppdstm],'0\:00'))

Since Format() encounters a number format, it will, if needed, convert strings to numbers, and format them as numbers. This gives for example:

DateDiff('n', '2009-10-02 0:20', '2009-10-02 15:45')

DateDiff expects date/time parameters, and will convert them if needed. Since the strings are well-formed ISO dates, the conversion works.

I changed the format string slightly because '/' means "system date separator" (which can cause problems) and the space is my thousand separator (it needs to be escaped). Bottom line: escape all literals in format strings.

If PPADT is text, and if you want to avoid the implicit conversion to Long, you can format the string directly like this:

    Format(PPADT, "@@@@-@@-@@")

If PPADT is a number it also works, but this time with an implicit conversion to string before the formatting...

Text formatting occurs left-to-right, so four '@' are needed. Number formatting occurs right-to-left, so Peter's first solution works just as well (on US computers).

Cheers!
(°v°)

 

by: harfangPosted on 2009-11-01 at 03:36:23ID: 25713369

I just tested, for fun:

    CDate(Format(PPADDT*1e4+PPADTM,'0\-00\-00\ 00\:00'))

Works nicely, too.
(^v°)

 

by: epicazoPosted on 2009-11-01 at 08:01:30ID: 31648509

Thank you all.  i tried all of the code provided and this seemed to get me the exact output.  I was erroring out on some of the other code.

thanks experts!

 

by: harfangPosted on 2009-11-01 at 11:20:21ID: 25714980

(^v°)

 

by: GrahamMandenoPosted on 2009-11-01 at 11:45:14ID: 25715093

Hi Peter

Please accept my humble apologies!  I posted my answer around midnight after a long day and I completely missed the fact that you were using zeroes in your format string, and not yyyy/mm/dd etc.

Thanks, Markus, for pointing out my folly :-)

Best regards to you both
--
Graham

 

by: harfangPosted on 2009-11-01 at 13:27:15ID: 25715577

No problem. What's more, your solution was clearly the most intelligible in the end. No implicit conversions, no "magic", no worries!
(^v°)

20120131-EE-VQP-002

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