converting seconds since 1980 to current date

[carolyn sorensen]

Hi, I have a database with dates in "HP/Apollo seconds since 1980." How can I convert this (under Windows) to date/time, with leap years, days, seconds etc. considered?

Thanks MUCH in advance for help.

Carolyn

[Hein van den Heuvel · Accepted solution]

Perl can readily solve this on Windows and Unix alike. Just figure out 'since 1980' once, and then add that as base to any Apollo date.

Base:

$ perl -e 'use Time::Local; print timelocal(0,0,0,1,0,1980) '
315550800

Use:

$ perl -e 'print scalar localtime(315550800 + 3650)'
Tue Jan 1 01:00:50 1980

$ perl -e 'print scalar localtime(315550800 + 403650)'
Sat Jan 5 16:07:30 1980

Check out the perl documenation to see how to get the components of such new time:

http://perldoc.perl.org/functions/localtime.html


Hope this helps some,
Hein van den Heuvel (at gmail dot com)
HvdH Performance Consulting

[Steven E. Protter]

Shalom,

Great scripts for figuring things like this out.

http://hpux.ws/merijn/caljd-2.25.sh

http://hpux.ws/merijn/caljd-2.2.pl

I use these as production scripts. Written by A. Clay Stephenson

SEP

[carolyn sorensen]

Thanks guys! I needed the hint about Julian dates. I got the date for 1/1/1980 using the script cal_jdate(). When added to our 'date' value & converted back to gregorian, it works beautifully. (Poor documentation on top of all else -- the doc says seconds since 1980, but its clearly not.)

Thanks again!
Carolyn