Unix Date Help

[ndoggy020]

Hey guys.

I need to manipulate the system date to move forward a day. So, if the system date was 02/14/08, I need the 'new' date to be 02/15/08 so my script will look for the date of 02/15/08. The thing is the file I will be looking for always has the date of the next day. Any simple unix command solution?

Any ideas,

Thanks

[WinblowsME]

Here's a Perl solution if you can't find anything simpler. You'll need to download the Time::Piece and Time::Seconds modules from CPAN.

[CODE=perl]use warnings;
use strict;
use Time::Piece;
use Time::Seconds;

&init;

sub init
{
my $today = localtime();
my $days_ahead = 1;
my $date = &get_days_ah ead ( $today, $days_ahead );

print "$date\n";
}

sub get_days_ahead
{
my ( $date, $days ) = @_;

for ( my $i = 0; $i < $days; $i++ )
{
$date += ONE_DAY;
}

return ( &get_mmddyy ( $date ) );
}

sub get_mmddyy
{
my ( $date ) = @_;
my ( $weekday, $month, $day, $time, $year ) = split ( / +/, uc ( $date ) );

if ( $month eq "JAN" ) { $month = "01"; }
elsif ( $month eq "FEB" ) { $month = "02"; }
elsif ( $month eq "MAR" ) { $month = "03"; }
elsif ( $month eq "APR" ) { $month = "04"; }
elsif ( $month eq "MAY" ) { $month = "05"; }
elsif ( $month eq "JUN" ) { $month = "06"; }
elsif ( $month eq "JUL" ) { $month = "07"; }
elsif ( $month eq "AUG" ) { $month = "08"; }
elsif ( $month eq "SEP" ) { $month = "09"; }
elsif ( $month eq "OCT" ) { $month = "10"; }
elsif ( $month eq "NOV" ) { $month = "11"; }
elsif ( $month eq "DEC" ) { $month = "12"; }

if ( length ( $day ) == 1 ) { $day = "0" . $day; }
$year = substr ( $year, 2 );

return ( "$month/$day/$year" );
}[/CODE]

[ghostdog74]

Hey guys.

I need to manipulate the system date to move forward a day. So, if the system date was 02/14/08, I need the 'new' date to be 02/15/08 so my script will look for the date of 02/15/08. The thing is the file I will be looking for always has the date of the next day. Any simple unix command solution?

Any ideas,

Thanks

if you have GNU date

# date +%Y/%m/%d -d "1 day"
2008/02/16