Bill Akins wrote: > ... > Try this: > #!/usr/bin/perl -w > > my @now = localtime; > my $sec = ( $now[0] ); > my $min = ( $now[1] ); > my $hr = ( $now[2] ); > my $day = ( $now[3] ); > my $mth = ( $now[4] + 1 ); > my $yr = ( $now[5] + 1900 ); > > $day = ($day -1); > print "Yesterday was $mth/$day/$yr\nor if on the other side > of the pond, $day/$mth/$yr\n";
What if today is the first of the month? Better to use something like: ($d, $m, $y) = (localtime(time - 86_400))[3..5]; $m++; $y += 1900; IOW, have the system calculate local time for current epoch - 86,400 seconds (# of seconds in a day). -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
