How can I remove all comments from a file?

Question

I have a file which includes comments:

foo
bar
stuff
#Do not show this...
morestuff
evenmorestuff#Or this

I want to print the file without including any of the comments:

foo
bar
stuff
morestuff
evenmorestuff

There are a lot of applications where this would be helpful. What is a good way to do it?

Answer 1

One way to remove all comments is to use grep with -o option:

grep -o '^[^#]*' file

where

• -o: prints only matched part of the line
• first ^: beginning of the line
• [^#]*: any character except # repeated zero or more times

Note that empty lines will be removed too, but lines with only spaces will stay.

Answer 2

I believe sed can do a much better job of this than grep. Something like this:

sed '/^[[:blank:]]*#/d;s/#.*//' your_file

Explanation

• sed will by default look at your file line by line and print each line after possibly applying the transformations in the quotes. (sed '' your_file will just print all the lines unchanged).
• Here we're giving sed two commands to perform on each line (they're separated by a semicolon).
• The first command says: /^[[:blank:]]*#/d. In English, that means if the line matches a hash at its beginning (preceded by any number of leading blanks), delete that line (it will not be printed).
• The second command is: s/#.*//. In English that is, substitute a hash mark followed by as many things as you can find (till the end of the line, that is) with nothing (nothing is the empty space between the final two //).
• In summary, this will run through your file deleting lines that consist entirely of comments and any lines left after that will have the comments stricken out of them.

Answer 3

As others have pointed out, sed and other text-based tools won't work well if any parts of a script look like comments but actually aren't. For example, you could find a # inside a string, or the rather common $# and ${#param}.

I wrote a shell formatter called shfmt, which has a feature to minify code. That includes removing comments, among other things:

$ cat foo.sh
echo $# # inline comment
# lone comment
echo '# this is not a comment'
[mvdan@carbon:12] [0] [/home/mvdan]
$ shfmt -mn foo.sh
echo $#
echo '# this is not a comment'

The parser and printer are Go packages, so if you'd like a custom solution, it should be fairly easy to write a 20-line Go program to remove comments in the exact way that you want.

Answer 4

Example input

cat example.sh

#!/bin/bash
# example script

echo "# test";# echo "# test"

# check the first parameter
if [ "$1" = "#" ]; then 
  # test couple of different cases
  echo "#"; # output # character 
  echo '\#'; # output # character '#' for test purpose
  echo \#\#\#; # comment # comment # comment '# comment'
  echo \#
  echo \#;
  echo \#; # comment
fi
# end of the script

Remove comments

sed -e '1{/^#!/ {p}}; /^[\t\ ]*#/d;/\.*#.*/ {/[\x22\x27].*#.*[\x22\x27]/ !{:regular_loop s/\(.*\)*[^\]#.*/\1/;t regular_loop}; /[\x22\x27].*#.*[\x22\x27]/ {:special_loop s/\([\x22\x27].*#.*[^\x22\x27]\)#.*/\1/;t special_loop}; /\\#/ {:second_special_loop s/\(.*\\#.*[^\]\)#.*/\1/;t second_special_loop}}' example.sh

Result

cat example.sh

#!/bin/bash

echo "# test";

if [ "$1" = "#" ]; then 
  echo "#"; 
  echo '\#'; 
  echo \#\#\#; 
  echo \#
  echo \#;
  echo \#;
fi

Read how it works at source: https://blog.sleeplessbeastie.eu/2012/11/07/how-to-remove-comments-from-a-shell-script/

Answer 5

You can achieve the required output using sed command. The below command had done the trick for me.

sed 's/#.*$//g' FileName

Where

• #.*$ - Regexp will filter all the string that starts with # up to the end of the line

Here we need to remove those lines so we replaced with empty so skipping 'replacement' part.

• g - mentioning repeated search of the pattern until end of file is reached.

General syntax of sed: s/regexp/replacement/flags FileName

Answer 6

You can use invert match like this:

grep -v "#" filename

-v
Select lines not matching any of the specified patterns. (As specified by POSIX.)

Answer 7

Use a command like:

grep -E -v "^#|^$" <file-name>

• -v inverts the match, i.e. only outputs lines which do not match
• ^# matches lines starting with #
• ^$ matches empty lines

Answer 8

cat YOUR_FILE | cut -d'#' -f1

It uses # as column separator and keeps just the first column (that is everything before #).

Answer 9

I like Joseph R.'s answer but needed it to strip // comments too, so I modified it slightly & tested on RHEL:

# no comments alias
alias nocom="sed -E '/^[[:blank:]]*(\/\/|#)/d;s/#.*//' | strings"
example
cat SomeFile | nocom | less

I bet there's a better way to remove blank lines than using strings but it was the quick & dirty solution I used.

Answer 10

This worked for me

sed -i.old -E  "/^(#.*)$/d" file 

Answer 11

Following the 2nd answer of Joseph R., I add /^$/d to remove blank line.

sed '/^[[:blank:]]*#/d;s/#.*//;/^$/d'

Answer 12

The best solution would be to use the command:

sed -i.$(date +%F) '/^#/d;/^$/d' ntp.conf

The -i is the in-place edit but the prefix directly following tells sed to create a backup. In this case with a date extension (ntp.conf.date) We run two commands each with an address space, the first deletes the commented lines and the second, separated from the first by a semi-colon, deletes the blank lines.

I found this solution on : theurbanpenguin.com

Answer 13

None of the other answers seem to do this justice, they either leave in empty lines, or leave in lines where the comment isn't at the first character. I ended up using this:

cat << EOF >> ~/.bashrc
alias nocom='sed -e "/^\s*#/d" -e "/^\s*$/d"'
EOF

This sets up an alias, so that you don't have to memorize it (which is impossible to begin with). Open a new session, and you'll have the new nocom command. Then you can just

nocom /etc/foobar.conf

Cheers.

Answer 14

awk '!/^#/{gsub(/#.*/,"",$0);print}' filename

output

foo
bar
stuff
morestuff
evenmorestuff

Answer 15

I'm posting what works for me and seems to make the most sense, after reading through the others, with explanation. A couple of posts came close, but I could not comment yet (because I'm a newb):

grep -E -v "(^#.*|^$)" filename

• -E = interpret the following pattern as a regular expression, similar to using egrep
• -v = print the inversion of the pattern (lines that do not match the expression will be printed)
• "(^#.*|^$)" = this has a pipe which designates an OR statement. This expression says to print any line that starts with a # (and anything else after it) OR any line with zero characters between the beginning and end of the line.

The -v will print on the screen the inversion of that, which will be any line with characters that does not start with a #.