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Old 08-08-2006, 05:05 AM   #1
basak
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Registered: Jul 2006
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a file without comments


Hi..İs there an easy way of cleaning out the comments in a bash shell script file?(cleaning out the lines beginning with "#" may be even helpful enough)Thank u..
 
Old 08-08-2006, 05:08 AM   #2
b0uncer
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Registered: Aug 2003
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I believe you can achieve this with cat and grep:
Code:
cat scriptfile | grep -v "#" > scriptfile_new
That reads the script file (scriptfile), then prints out every line except those with comment marks # and outputs it to scriptfile_new

EDIT: oh but I just took it off the top of my head so don't delete the original script before checking out you didn't miss anything.. i.e. if there are lines that contain the #-mark but that are NOT completely comments, those lines get drawn off totally anyway. Shortly said, every line containing # gets removed. So a line like

somecode(); # this does somecode

would get removed too, though the comment is only after the code piece (but still on the same line).

You might want something a bit more clever to wipe out comments that are on the same line than some non-comment code..maybe sed or some other alike tool could read the text and when it encounters a # it would not read text until the next linefeed..well, I'm not sure. Better keep comments at their own lines

Last edited by b0uncer; 08-08-2006 at 05:13 AM.
 
Old 08-08-2006, 06:20 AM   #3
timmeke
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Registered: Nov 2005
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1. cat some_file | grep ...
can be replaced by:
grep ... some_file

2. As indicated, the given grep won't do. It will also omit lines that start with code and end with a comment.
Rather, use a regular expression like this:
Code:
grep -E -v '^#' scriptfile
That will only throw out lines that start off with a '#'.

3. If you want to clean out comments from lines like
some_code; #some_comment
then you'll need to look at something else than grep. 'cut' and 'awk' seem to be good choices. I wouldn't recommend 'sed' for this one.
ie
Code:
cut -d'#' -f1 scriptfile;
should read the file and keep only the text that's before the first '#'.
 
  


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