sed - Commenting a line matching a specific string AND that is not already commented out

问题

I have the following test file

AAA
BBB
CCC

Using the following sed I can comment out the BBB line.

# sed -e '/BBB/s/^/#/g' -i file

I'd like to only comment out the line if it does not already has a # at the begining.

# sed -e '/^#/! /BBB/s/^/#/g' file

sed: -e expression #1, char 7: unknown command: `/'

Any ideas how I can achieve this?

回答1:

Assuming you don't have any lines with multiple #s this would work:

sed -e '/BBB/ s/^#*/#/' -i file

Note: you don't need /g since you are doing at most one substitution per line.

回答2:

I find this solution to work the best.

sed -i '/![^#]/ s/\(^.*BBB.*$\)/#\ \1/' file

It doesn't matter how many "#" symbols there are, it will never add another one. If the pattern you're searching for does not include a "#" it will add it to the beginning of the line, and it will also add a trailing space.

If you don't want a trailing space

sed -i '/![^#]/ s/\(^.*BBB.*$\)/#\1/' file

回答3:

Another solution with the & special character which references the whole matched portion of the pattern space. It's a bit simpler/cleaner than capturing and referencing a regexp group.

sed -i 's/^[^#]*BBB/#&/' file

回答4:

sed -i '/![^#]/ s/$^.*BBB.*$$/#\ \1/' file

This doesn't work for me with the keyword *.sudo, no comments at all...

Ony the syntax below works: sed -e '/sudo/ s/^#*/#/' file

回答5:

Assuming the BBB is at the beginning of a line, I ended up using an even simpler expression:

sed -e '/^BBB/s/^/#/' -i file

One more note for the future me. Do not overlook the -i. Because this won't work: sed -e "..." same_file > same_file.

回答6:

Actually, you don't need the exclamation sign (!) as the caret symbol already negates whatever is inside the square brackets and will ignore all hash symbol from your search. This example worked for me:

sed -i '/[^#]/ s/$^.*BBB.*$$/#\ \1/' file