bash script to delete all comments after '#'??

dr44mon · 01-30-2009, 11:46 AM

i need to make bash script than can delete all comment (all char after '#' in each line) in a single file.... I think it using grep? how to do make that? thanks a lot.....

David the H. · 01-30-2009, 12:20 PM

Since there's only a single matching reqirement this is a very easy task. You can do it with a simple sed replacement:

sed "s/#.*//" textfile.txt

dr44mon · 01-30-2009, 01:02 PM

Quote:

Originally Posted by David the H.

Since there's only a single matching reqirement this is a very easy task. You can do it with a simple sed replacement:

sed "s/#.*//" textfile.txt

may i ask again.... how we can do that if we have to use grep in the scripts? and how to replace the file that has been modified above (the comments has been removed) in the original file? thx a lot...

arizonagroovejet · 01-30-2009, 01:59 PM

Quote:

Originally Posted by dr44mon

may i ask again.... how we can do that if we have to use grep in the scripts? and how to replace the file that has been modified above (the comments has been removed) in the original file? thx a lot...

Sounds like homework. Run 'man grep' and look at the -v option. The other hint I'll give you is: regular expressions, specifically matching the start and end of a line.

pixellany · 01-30-2009, 03:12 PM

"have to use grep"----I assume because the instructor said so?

In addition to the advice above, you'll also be interested in the redirection operators (< and >), and the -i option in SED.

Depending on what texts are used in the class, you may want to get a copy of the Bash Guide for Beginners---- free at http://tldp.org

reduardo7 · 01-03-2020, 01:04 PM

Example input

Code:

cat example.sh

Code:

#!/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

Code:

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

Code:

cat example.sh

Code:

#!/bin/bash

echo "# test";

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

Source: https://unix.stackexchange.com/a/560198/22180

rtmistler · 01-03-2020, 08:34 PM

@reduardo7,

Thanks for contributing.

Please note that the age of the thread is in excess of 11 years.

Also seems based on the manner of questioning, it was an assignment of some type.

1. As shown, there were specific restrictions for how to accomplish it.
2. Rather than offer a full solution, we try to guide an OP by asking them to show their efforts and help them to learn how they can implement their own solution. They always could've performed a web search, they don't appear to have done so. This OP doesn't appear to have shown any effort, even the little you've put forth with your web search.

Also please note that the LQ site does ask you if you really intend to post new content to threads with no posts newer than six months ago. Not trying to he critical, but it's a good guide to consider and gauge whether or not any new information will be of any help.

MadeInGermany · 01-04-2020, 10:21 AM

Time to let the cat out of the bag: GNU grep can print partial lines by means of the -o option (print what matches).

Code:

grep -o '^[^#]*'

Good for the exercise.
In practice you'll use sed.