how to get all escaped literal symbols in a string using bash substitution

Question

I want to be able to return only the escaped literals in a string currently I'm doing this:

foo="\'\"\(foobar\)'another'[program]\[\$var\]()"
echo "${foo//[^\\\']/}"

but it outputs this:

\'\\''\\

desired output should be like this:

\'\"\(\)\[\$\]

I'm still just on a stage where I'm trying to get the literal single quotes, but somehow it's not working or is it really doable in bash expansion?

EDIT

string is from bash $READLINE_LINE so there will be no extra escaping like the tripple backslash for double quotes.

Answer 1

To set a variable to the literal value

\'\"\(foobar\)'another'[program]\[\$var\]()

using a double-quoted string, you would need to escape each literal backslash and each double quote or dollar sign that would otherwise trigger an expansion.

string="\\'\\\"\\(foobar\\)'another'[program]\\[\\\$var\\]()"

If using a single-quoted string, you only have to care about inserting the single quotes specially:

string='\'"'"'\"\(foobar\)'"'"'another'"'"'[program]\[\$var\]()'

Here, I've opted for breaking out of the single quoted string to add a double quoted single quote for each single quote, i.e. '"'"'. You could also have used an escaped single quote outside of the single quoted string, i.e. '\''.

You may also choose to use a quoted here-document if the quoteng becomes too cumbersome:

string=$( cat <<'END'
\'\"\(foobar\)'another'[program]\[\$var\]()
END
)

Note that this would trim off the trailing newline if a newline character was the last character in the string.

Your code then tries to delete all backslashes and single quotes, which does not seem right. Instead, use some tool to extract all instances of \ and the following character:

grep -o '\\.' <<<"$string"

This produces

\'
\"
\(
\)
\[
\$
\]

Or,

grep -o '\\.' <<<"$string" | paste -s -d '\0' -

to reproduce exactly the output in the question.

You could also do this in a loop in bash directly:

while [[ $string =~ \\. ]]; do
    printf '%s\n' "${BASH_REMATCH[0]}"
    string=${string#*\\?}
done

or,

while [[ $string =~ '\'. ]]; do
    printf '%s\n' "${BASH_REMATCH[0]}"
    string=${string#*'\'?}
done

This modifies the value of string by trimming off the bit up to the next match of a backslash and some other character, for as long as such a character sequence exists in the string. The bit matching the given regular expression is printed in each iteration.

Answer 2

With zsh instead of bash, you could do:

set -o extendedglob
print -r -- ${foo//(#b)((\\?)|?)/$match[2]}

Or with ksh93:

print -r -- "${foo//@(@(\\?)|@(?))/\2}"

(you'd think print -r -- "${foo//@(@(\\?)|?)/\2}" should also work but it doesn't, which looks like a bug)

With fish:

string join '' (string match -ar '\\\\.' $foo)

Answer 3

I don't think you can use the pattern replacement operation to find such strings, as you need to look at the previous character to know if to remove the current one. You'd need something like Perl regexes' negative look-behind to do that. It's even harder if you consider strings like \\x, where the first backslash should escape the second, but the second should not escape the x, at least far as backslash-escapes usually work.

It'd be easier to just find the matching parts in a loop, not that Bash really makes that very simple either. (There's the regex match operator [[ text =~ re ]] and the matches can be found in ${BASH_REMATCH[@]}, but I don't think there's a way to loop for multiple hits other than manually.)

But you can do it with e.g. grep. E.g. this would output the matching sequences, one per line:

foo="\'\\\"\(foobar\)'another'[program]\[\$var\]()"
echo "$foo" | grep -oe '\\.'

And you then pipe the output to tr -d '\n' to remove the newlines. Or use while IFS= read -r line; do... if you need to handle them in the shell, but if you're doing that, you probably should use some other tool; the shell doesn't bend to text processing too well.