How write script to print uncomment line

Question

I have a very long config file with hardly comment and uncomment line Now I want to print out lines with uncomment

# LOCAL PATHNAME INFORMATION
#
# The queue_directory specifies the location of the Postfix queue.
# This is also the root directory of Postfix daemons that run chrooted.
# See the files in examples/chroot-setup for setting up Postfix chroot
# environments on different UNIX systems.
#
queue_directory = /var/spool/postfix

# The command_directory parameter specifies the location of all
# postXXX commands.
#
command_directory = /usr/sbin

I want the output is lines with no comment

queue_directory = /var/spool/postfix

command_directory = /usr/sbin

Answer 1

egrep -a -v '^[[:space:]]*#' config_file | egrep -a '[[:print:]]' | less

will account for spaces before the #. Pipe it to less and you can view it easily

replace config_file with $1 and put the line in /usr/local/bin/cless and chmod it +x and you have nice little script whenever you need it.

cless config_file

Answer 2

Assuming that commented lines begin with "#" in the first position:

grep -v '^#' config_file

will print all the lines that do not begin with "#" in the file "config_file".