Harry Putnam wrote: > > "John W. Krahn" <[EMAIL PROTECTED]> writes: > > > printf " %02d/%02d/%04d %02d:%02d:%02d\n", $lta[4] + 1, $lta[3], $lta[5] + 1900, >@lta[2,1,0]; > > Thanks.. the tips work good. And I overlooked the part about mnths > being 0-11. > > I get the idea from your posted printf line and learned a new trick > with the @ar[2,4,7] syntax you used but couldn't really understand why it > gave a screwball year in my script until I noticed the `=' missing in > $lta[5] + 1900. > > Still not sure I understand why it needs to be there. > > cat test2 > > #!/usr/local/bin/perl -w > $now = time; > > $date_spec = ($now - 86400); > # local time adjusted > @lta = localtime($date_spec);
The value in $lta[5] is now 102. > $mnth = ($lta[4] + 1); > print " $mnth/$lta[3]/" . ($lta[5] += 1900) . " $lta[2]:$lta[1]:$lta[0]\n"; The value in $lta[5] has been modified to 2002, the += operator modifies the original value. > print localtime() . "\n"; > printf " %02d/%02d/%04d %02d:%02d:%02d\n", $lta[4] + 1, $lta[3], $lta[5] + 1900, >@lta[2,1,0]; You are now adding 1900 to 2002 so it prints 3902. Either don't modify $lta[5] and just add 1900 to it each time or modify it once and use that value everywhere. > OUTPUT: > $ ./test2 > 12/15/2002 20:16:38 > Mon Dec 16 20:16:38 2002 > 12/15/3902 20:16:38 @lta = localtime($date_spec); $lta[4] += 1; # modify month $lta[5] += 1900; # modify year print " $lta[4]/$lta[3]/$lta[5] $lta[2]:$lta[1]:$lta[0]\n"; print localtime() . "\n"; printf " %02d/%02d/%04d %02d:%02d:%02d\n", @lta[4,3,5,2,1,0]; John -- use Perl; program fulfillment -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
