Toby Stuart <[EMAIL PROTECTED]> writes: > Check the length of second, minute and hour. > If the length == 1 then add a leading zero eg. $sec = "0" . $sec > Maybe there is some other "magical" way of doing this, if there is i'm not > aware of it :)
Good idea... thanks. "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); $mnth = ($lta[4] + 1); print " $mnth/$lta[3]/" . ($lta[5] += 1900) . " $lta[2]:$lta[1]:$lta[0]\n"; print localtime() . "\n"; printf " %02d/%02d/%04d %02d:%02d:%02d\n", $lta[4] + 1, $lta[3], $lta[5] + 1900, @lta[2,1,0]; OUTPUT: $ ./test2 12/15/2002 20:16:38 Mon Dec 16 20:16:38 2002 12/15/3902 20:16:38 -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
