How to remove the first character in a line which contains a specific string from Linux command line?

I need to frequently comment/uncomment a line containing a string with "#". For example:

Don't touch this line.
Touch this LINE instead.

<==>

Don't touch this line.
#Touch this LINE instead.

I was able to comment with sed, but had problem to uncomment. I tried command

sed 's/^#\(.*\)LINE/\1/' hosts

It did put a "#" at the beginning, but also removed LINE and the following characters, which made it:

Don't touch this line.
#Touch this 

What's the proper way to do it?

1 Answer

The proper way is to make s work only on lines that match a certain pattern, the substitution can be quite simple then.

This will remove leading # from lines that contain LINE:

sed '/LINE/ s/^#//'

This will remove all leading # characters:

sed '/LINE/ s/^#*//'

This will add one # character:

sed '/LINE/ s/^/#/'

Obviously you cannot remove # that isn't there, but you can add # while a leading # is already there. An improvement to the above command that doesn't modify already commented lines:

sed '/^\([^#].*\|\)LINE/ s/^/#/'

It detects ^LINE or ^[^#].*LINE where ^ matches the beginning of the line and [^#] matches a non-#. It's possible to improve the command further, so # with leading blanks at the beginning of a line will not be commented out again; I won't do this here.

Anyway, use /regex/ s…. If your regex happens to contain slashes, escape them with \; or choose another character to embrace the regex, but then you need to escape the opening one (e.g. \@regex@ s…).