Bash how to replace comment backslash with nothing

Question

I have a strign in bash with the following format:

// comment.

I want to obtain a new variable with comment alone (no backslash) and I don't want to depend on the // begin the first two characters in the string. How can I do this?

I have tried this:

  nline=${line/%/////}
  echo $nline

To use string substitution but it doesn't work.

Answer 1

Perhaps you want the # substitution?

$ a='// this is a comment'
$ printf "%s\n" "${a#// }"
this is a comment
$ a='not a comment'
$ printf "%s\n" "${a#// }"
not a comment

And as SergA pointed out, a little better patterns for our variable extraction can save us the need for the sed solution below:

$ a="first //a comment"
$ printf "%s" "${a##*//}"

If you just want to get the comment part of a line anywhere it is you could use sed like so:

$ a="first //a comment"
$ printf "%s\n" "$a" | sed -e 's,^.*// \?,,'
a comment

which of course you could store in another variable:

nline=$(printf "%s" "$a" | sed -e 's,^.*// \?,,')

(note also that I remved the \\n from the printf )

Answer 2

删除前两个字符：

echo ${nline:2}

Answer 3

% matches the end of the string. # matches the beginning of the string.

Since you said you wanted neither of those you don't want either % or # in there.

Also you need to escape / in a / -delimited pattern.

nline=${line/\/\/}
echo "$nline"

This will remove the first // from the string no matter where it is or what comes before it. So foo // comment will become foo comment , etc.

If you want to also remove any surrounding spaces from the // string then you need to do a bit more work and can't so easily use string substitution for it.