[lug] Output every xth line?

[Tkil]

>>>>> "Joey" == Joseph McDonald <joem at uu.net> writes:

Joey> This should get you started..

Joey> cat /etc/passwd | perl -e '$start = 2; $stop = 6; $c = 0; while(<>) { $c++; if (($c >= $start) && ($c <= $stop)) { print } }'

If you really want to abuse yourself, take a look at the special case
on the ".." operator:

   If either operand of scalar ".." is a constant expression, that
   operand is implicitly compared to the $. variable, the current line
   number.

Which means that you could do:

   perl -nwe 'print if 2 .. 6;' /etc/passwd

Sub-lessons:

1. Your "$c" is Perl's built-in "$."

2. The "-n" flag replaces your "while (<>) { ... }" loop.

   a. You can still do one-time init with BEGIN { ... } blocks if you
      need to.

There's nothing wrong with your code; consider this more a suggestion
to read over the docs for command-line flags (in "perlrun" man page)
and the built-in variables ("perlvar"), and the corner cases of some
operators ("perlop").

t.