Date variable minus 60 days - how to calculate?

[Maureen Gunkel]

Please ignore the previous - hit the Enter key by mistake.
Anyhoo, I'm trying to manipulate the date a user inputs while running a shell script to subtract 60 days from it. I have a VERY klutzy way of doing it, but my way won't work with months <> 30 days (approx. half of them!). I don't know C or Perl, and am tring to do this in C-shell, perhaps with awk?

Thanks to anyone who can help!!

[Patrick Wallek]

There is a set of GNU Utilities that has a date utility that will do exactly what you want. I currently use it to do date calculations. It should handle months without a problem.

Try to following url to get the bundle of utilities: http://www.gnu.org/gnulist/production/shellutils.html

[Patrick Wallek]

You can also try this url at the HPUX Software Porting Center for precompiled versions of the utilities. These are easier to install than having to compile the stuff you get from the gnu site.

http://hpux.cae.wisc.edu/hppd/hpux/Gnu/sh_utils-2.0/

[Madhu Sudhan_1]

Maureen:

Here is the URL which talks about date comparisons using perl.

http://webmaster.indiana.edu/perl56/pod/perlfaq4.html#data:%20dates

Hope this helps.

......Madhu

[Maureen Gunkel]

Although all those URL's are very helpful, they don't address my need (I don't think). I'm not using the current date, I'm using a date input by the user. For example, I ask them what date, they enter '10/26/00', which is assigned to dte1, and I need to find out what $dte1 - 60 would be. I hope this is clearer.

[Al Langen · Accepted solution]

Copy the lines after the *'s into a file and call it whatever.awk. In your script you might have:
read curdate
olddate=`echo $curdate | awk -f whatever.awk`

Hope this helps.

***************************************

BEGIN{ FS="/"
split( "Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec", month," ");
split( "31 28 31 30 31 30 31 31 30 31 30 31", days, " ");
}
{
year = $3
leapyear = year % 4;
if( leapyear == 0 )
days[2] = 29
else
days[2] = 28;
doy = 0;
for ( i = 1; i< $1; i++ )
doy = doy + days[i]
doy = doy + $2;
if ( doy <= 60 )
{
doy = doy + 365;
year = year - 1;
leapyear = year %4;
if( leapyear == 0 )
{
days[2] = 29;
doy = doy + 1;
}
}
doy = doy - 60;
for ( i = 1; i <= 12; i++ )
{
doy = doy - days[i];
if ( doy <= 0 )
{
doy = doy + days[i];
break;
}
}
printf( "%d/%d/%dn", i, doy, year )
}

[Ralph Grothe]

If you are prepared to call an external script/prog from your script this can easily be done with a Perl scriplet (although you mentioned your dislike for Perl).
Of course, one could also use the ctime() standard C func and Co., maybe slightly less comfortably though.
I'm not particularily fond of the C-shell and thus didn't care much of what the C-shell has to offer in this respect.

What is the date format read as user input from your script (because it seems more of a formatting issue)?

[Maureen Gunkel]

Al -
Bee-yoo-ti-ful! I have a slight problem with the format of the date returned, but I can fix that. You've saved my bacon! Thanks again!

[Al Langen]

Maureen,

You might want to change the printf format in order to print leading 0's or not.