All of the lines with comments in a file begin with #
. How can I delete all of the lines (and only those lines) which begin with #
? Other lines containing #
, but not at the beginning of the line should be ignored.
#blah \<nl>blah
counts as a single "logical line" because the backslash escapes the newline?
make
, which utilities use the 'backslash splices lines before ending a comment'? The shells (bash and ksh tested) don't. C and C++ do handle newline splicing before other processing of preprocessor directives, but they're directives rather than comments.
\<nl>
escaping would also work on comments. But wow I was wrong. I haven't been able to find another example yet... :) Thanks!
This can be done with a sed one-liner:
sed '/^#/d'
This says, "find all lines that start with # and delete them, leaving everything else."
I'm a little surprised nobody has suggested the most obvious solution:
grep -v '^#' filename
This solves the problem as stated.
But note that a common convention is for everything from a #
to the end of a line to be treated as a comment:
sed 's/#.*$//' filename
though that treats, for example, a #
character within a string literal as the beginning of a comment (which may or may not be relevant for your case) (and it leaves empty lines).
A line starting with arbitrary whitespace followed by #
might also be treated as a comment:
grep -v '^ *#' filename
if whitespace is only spaces, or
grep -v '^[ ]#' filename
where the two spaces are actually a space followed by a literal tab character (type "control-v tab").
For all these commands, omit the filename
argument to read from standard input (e.g., as part of a pipe).
grep -v "^#" filename
The opposite of Raymond's solution:
sed -n '/^#/!p'
"don't print anything, except for lines that DON'T start with #"
you can directly edit your file with
sed -i '/^#/ d'
If you want also delete comment lines that start with some whitespace use
sed -i '/^\s*#/ d'
Usually, you want to keep the first line of your script, if it is a sha-bang, so sed
should not delete lines starting with #!
. also it should delete lines, that just contain only a hash but no text. put it all together:
sed -i '/^\s*\(#[^!].*\|#$\)/d'
To be conform with all sed variants you need to add a backup extension to the -i
option:
sed -i.bak '/^\s*#/ d' $file
rm -Rf $file.bak
You can use the following for an awk solution -
awk '/^#/ {sub(/#.*/,"");getline;}1' inputfile
This answer builds upon the earlier answer by Keith.
egrep -v "^[[:blank:]]*#"
should filter out comment lines.
egrep -v "^[[:blank:]]*(#|$)"
should filter out both comments and empty lines, as is frequently useful.
For information about [:blank:]
and other character classes, refer to https://en.wikipedia.org/wiki/Regular_expression#Character_classes.
egrep
supports that syntax; older versions might not.
You also might want to remove empty lines as well
sed -E '/(^$|^#)/d' inputfile
Delete all empty lines and also all lines starting with a #
after any spaces:
sed -E '/^$|^\s*#/d' inputfile
For example, see the following 3 deleted lines (including just line numbers!):
1. # first comment
2.
3. # second comment
After testing the command above, you can use option -i
to edit the input file in place.
Just this!
Here is it with a loop for all files with some extension:
ll -ltr *.filename_extension > list.lst
for i in $(cat list.lst | awk '{ print $8 }') # validate if it is the 8 column on ls
do
echo $i
sed -i '/^#/d' $i
done
Success story sharing
sed /^#/d
sed '/^#/ d' < inputFile.txt > outputFile.txt
sed -i '/^#/d' filepath
.sed -i '' '/^#/d' filepath
on Mac (because the -i suffix is mandatory)awk '/^#/ && !first { first=1 ; next } { print $0}'