date & time difference calculation

[Lorry Litman]

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Hi Joe,

The email attachment got lost in the replies.
Here's the info. At the end is the problem I'm still trying to work out.

> Lorry Litman wrote:
>
> Hi,
>=20
> How to do a date/time difference calculation?
>=20
> VMS 7.2, PH 7.10G, RDB 7.0
> In the RDB database the DATE field is INTEGER and the TIME field is
SMALLINT.
> In the dictionary (century included), the DATE field is type D, size 8,
datatype ZU, and the TIME field is type N, size 4, datatype ZU.
>=20
> A new field is being created, call it DATE_DIFF, I suspect it will be
INTEGER in RDB and Numeric size 10, ZU in the dictionary.
> I need to store the DATE_DIFF result in format YYMMDDHHMM.
> The display of the DATE_DIFF field on the screen is to be in the format=
 of
YY:MM:DD:HH:MM.
>=20
> Any help/suggestions would be much appreciated.

> Chris replied
>
> decimaltime(vmsdatetime1, vmsdatetime2) will return (vmsdatetime1 -=20
> vmsdatetime2) as a signed, floating point, number of days.
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> You're on your own in deciding how many difference days in a difference=
=20
> month or year, it being variable. I'd've recommended the VMS delta time=
=20
> format of dddd-hh:mm (good for just under 30 years).
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> ; convert own-brand date & time into vms date/time format
> def vmsdt1 vmsdate =3D floor(mydate) + floor(mytime*0.01)/24 + &
>       mod(mytime,100)/1440
> ; calculate difference
> def fdays =3D decimaltime(vmsdt1,vmsdt2)
> def idays int*4 =3D floor(fdays)
> def ihours int*2 =3D floor((fdays - idays)*24)
> def imins int*2 =3D floor((fdays - idays)*1440 - 24 * ihours)
> def date_dif int*10 =3D idays*10000 + ihours*100 + imins


I tried Chris' suggestion which was very helpful (thank-you Chris).
I ran into a problem with the hour & minute calculation.
As an example,=20
 difference between 21-jan-1999-10:15 and 20-jan-1999-14:00
 resulted in 2735 (27 hrs, 35 min)
 I expected a result of 2015 (20 hrs, 15 min),=20
=20
 difference between 10:15 and 10:00 comes up as 14 instead of 15.=20

Because I'm not familiar with these functions (or the math) in Chris's
solution
I have changed the imins define.=20

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from: def imins int*2 = floor((fdays - idays)*1440 - 24 * ihours)
to:     def imins int*2 = floor((((fdays - idays)*24) - ihours)* 60 + .01)
MIME-Version: 1.0

I believe this now gives me the correct result in days,hours,minutes.
However, I need the result in years,months,days,hours,minutes.

As an example
 difference between 30-dec-2004-10:06 and 11-oct-2003-10:30
 results in 4452336 (445 days, 23 hours, 36 minutes)
 I need the result as 0102192336 (1 year, 2 months, 19 days, 23 hours, 36
minutes)

Any suggestions?
Maybe there are different powerhouse functions I can/should use.

Thanx
    Lorry
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