HI (again):
> Can we print previous date using awk..
The HP-UX supplied 'awk' lacks some of the nice enhancements of GNU 'awk'. In either, you can call an external command ('date' for example) and in the GNU version you have 'systime()' and 'strftime' (the former to return Epoch seconds; the later to format timestamps).
If I understand you question, correctly, though, you need to use Perl.
I'm coping to copy-and-paste two responses I offered to another member in an earlier thread:
You might note that by using Perl and the 'strftime' function you can choose any date format you want. The format directives are equivalent to those you are familiar with in the 'date' command. Of course, see too the manpages for 'strftime(3C)'. For example:
Find the date 2-days ago from today:
# perl -MPOSIX -le 'print strftime "%d %b",localtime(time-(60*60*24*2))'
05 Jun
# perl -MPOSIX -le 'print strftime "%m/%d/%Y",localtime(time-(60*60*24*2))'
06/05/2011
You could also use this to find a date n-days from a specific starting date:
# cat ./fromwhen
#!/usr/bin/perl
use strict;
use warnings;
use POSIX qw(strftime);
use Time::Local;
my $date = shift or die "Date (YYYYMMDD) expected\n";
die "YYYMMDD expected\n" unless
my ( $yyyy, $mm, $dd ) = ( $date =~ /(\d{4})(\d\d)(\d\d)/ );
my $when = shift or die "Number of days in past expected\n";
my $secs = timelocal(0, 0, 0, $dd, $mm-1,$yyyy );
print strftime "%Y/%m/%d\n",localtime($secs-(60*60*24*$when));
1;
# ./fromwhen 20110607 7
2011/05/31
Regards!
...JRF...
