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      LinuxDeveloperNews- Date Math In Linux Shell Script 
                       June 26, 2007
             http://www.linuxdevelopernews.com

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Date Math In Linux Shell Script 
Dave Taylor | Expert Author

Within a Linux shell script to be used as a cron job, how do I
calculate the current date, the current date - n days, and the
current date + n days ?

This script is to be used to partition an oracle database, and
automatically drop old partitions (n+1) and create a new partition
(n+1).

While I'm sure this will amaze some people, I am going to actually
take the daring step of suggesting that date math is rather difficult
to accomplish in many Unix and Linux shell script environments, and
it's much easier to actually write a brief C program to accomplish
your task. Yes, I've written one of the most popular shell script
programming books ever written (Wicked Cool Shell Scripts), but even
I don't think that you can solve every problem with a script! :-)

There's a version of the date command (commonly known as GNU date)
that allows some rudimentary mathematical calculations, but that's
only useful if you have the more sophisticated version of this
particular utility.

Instead, the key to working with date math on a Unix system is to
remember that Unix (and Linux) work on what's called epoch time,
the number of seconds since a specific date back in 1970. As I write
this entry, for example, the "epoch time" is 1131948327. 

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Once you have that, and a utility that turns epoch time into a nice,
readable date, it's easy enough to add or delete days: just add or
subtract 60*60*24 seconds for each day in question. Here it is as
a C utility:

/** DATEMATH - Demonstration program that shows how to
    do date mathematics by utilizing Unix procedures.
    This is missing some error checking, etc., but will
    show how to do "+n" and "-n" date math well enough.
    (C) Copyright 2005 by Dave Taylor. Free to redistribute
    if this copyright is left intact. Thanks.
***/

#include <stdio.h>
#include <time.h>

#define ONEDAY       60*60*24

time_t time(time_t *tloc);
char *ctime(const time_t *clock);

main(int argc, char **argv)
{
      time_t theTime;
      int offset = 1;

      if (argc > 1) offset = atoi(argv[1]);

      theTime = time((time_t *) NULL);

      theTime += (long) (offset * ONEDAY);

      printf("Offset by %d, the date is: %s\n",
          offset, ctime(&theTime));
}

As you can see, the utility grabs the current epoch time by using
the time() function, adds any days necessary by multiplying the
requested offset by the seconds in a day ONEDAY, then uses the
ctime() function to output a standard date string.

Here's the utility at work:

$ ./a.out +41
Offset by 41, the date is: Sun Dec 25 06:16:44 2005

Yes, that means there are only 41 shopping days until Christmas!

$ ./a.out -41
Offset by -41, the date is: Tue Oct 4 06:16:50 2005

$ ./a.out -5000
Offset by -5000, the date is: Sat Mar 7 06:16:55 1992

$ ./a.out -999
Offset by -999, the date is: Wed Feb 19 06:17:13 2003

$ ./a.out 365
Offset by 365, the date is: Tue Nov 14 06:17:28 2006

I hope that helps you out!

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About the Author:
Dave Taylor has been involved with the Internet since 1980 and is
internationally known as an expert on both business and technology
issues. Holder of an MSEd and MBA, author of twenty books and founder
of four startups, he also runs a strategic marketing company and
consults with firms seeking the best approach to working with weblogs
and social networks. Dave is an award-winning speaker and frequent
guest on radio and podcast programs.

AskDaveTaylor.com
http://www.intuitive.com/blog/

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