Shell Programming and Scripting

Hi guys,

First of all, I would like to say this is my first post in the unix.com forums. I am a beginner in PERL and have only started writing my first scripts.

With that out of the way, I have a question regarding the calculation of time dates in PERL.

I have two scalar variables with the following dates, notice the format of YYYY:MM:dd:hh:mm

$date1 = 2012:05:10:08:55
$date2 = 2012:05:15:09:55

I would like my script to take $date2 and subtract it by $date1 in hours and minutes as follows:

$total = 121 hours

Any idea how I could do this?

Thanks in advance

Okay. You want to learn; I'll show you.
time/date in unix and in perl (which was written on UNIX originally) use a so-called broken down time/date. Everything in dates is based on the number of seconds (epoch time) since Jan 1 1970. VERY important. So most of what you do is turn dates into epoch seconds, subtract or add, then convert the seconds back into something humans like to see.

this returns time in seconds, epoch time:

Code:

use Time::Local;
$time = timelocal($sec,$min,$hour,$mday,$mon,$year);

The array is sort of self documenting == Broken down time is the array.
Read this to learn how to use timelocal and timegm:
Time::Local - perldoc.perl.org

pseudocode:

Code:

sec1 =  epoch seconds for date 1  (from timelocal)
sec2 = epoch seconds for date 2
subtract sec1 from sec2  to get diff
There are 86400 seconds in a day, so
days = diff /86400;
There are   3600 seconds per hour so,  (diff % 3600)/60 is minutes
60 seconds per minute    so seconds are diff % 60;

% is modulo operator - the remainder from division. You get to work out "hours".
hint: start with diff % 86400.

For your hours total just divide diff/3600 to get total=121, but you'll need to do this other stuff as well. Remember you want whole number answers, so perl's int is your friend.

have fun.