Bash one-liner to replace all [:punct:] characters with those same characters escaped

Question

my current kind of unsolved problem with bash is how to replace all the characters in the [:punct:] class, that is

[!"#$%&'()*+,\-./:;<=>?@[\\\]^_`{|}~]

, each with the same character escaped by a backslash. For example, I want ^h?e.l*l+o: to become \^h\?e\.l\*l\+o\:.

I can do it in a multi-liner fashion with loops. Is it possible to do it in a more succint and elegant way?

Thanks, Luca

Are you sure you need to escape *all* punctuation, or just ones that have special meaning in some context? What's the use case? (There may be various simple alternatives depending on exactly *why* they need to be escaped.) — chepner, Sep 18 '15 at 12:24
Does [this question](https://stackoverflow.com/questions/29613304/is-it-possible-to-escape-regex-metacharacters-reliably-with-sed) help any? — Etan Reisner, Sep 18 '15 at 13:25
@chepner You may be right. I tried to escape all punctuation just for simplicity, but I have a problem with the apostrophe ' (and maybe some other characters that I haven't tested yet) that doesn't match anymore with itself if escaped. I basically want to escape regex and glob metacharacters. The simplest solution is to remove them as in `[!#$%&()*+,\-./:;<=>?@[\\\]^_{|}~]`. Is there any other approach? — the_eraser, Sep 19 '15 at 09:21
Yes, there are many other approaches as in the post quoted by @EtanReisner two comments above. Thanks, I've learned a lot. — the_eraser, Sep 20 '15 at 14:28

choroba · Answer 1 · 2015-09-18T12:37:01.243

5

Can you use sed?

echo '^h?e.l*l+o:' | sed 's/\([[:punct:]]\)/\\\1/g'

Without sed, I can't think of anything else than looping over the characters:

s='^h?e.l*l+o:'
o=''
for ((i=0; i<${#s}; i++)) ; do
    c=${s:i:1}
    [[ $c =~ [[:punct:]] ]] && o+=\\
    o+=$c
done
echo "$o"

Another possibility is to use Perl's quotemeta:

echo '^h?e.l*l+o:' | perl -lne 'print quotemeta'

-l handles newlines. -n process the input line by line.

edited Sep 18 '15 at 12:37

answered Sep 18 '15 at 12:17

choroba

yeah, I believed it could be done with sed and that's why I asked, but I couldn't figure out how since I'm quite new to bash, sed, awk etc. Can you add a bit of explanation on how this works and what does `\1` mean? This was the fastest reply ever, thanks! – the_eraser Sep 18 '15 at 12:21
@the_eraser: Check the update - was that your solution? – choroba Sep 18 '15 at 12:24
@the_eraser: `\1` is the first matching group, i.e. the substring matched by the first pair of `$...$`. – choroba Sep 18 '15 at 12:25
ok, thanks. Yes, I solved with a for loops, slicing the string and `=~`, but using sed is a lot nicer. – the_eraser Sep 18 '15 at 12:37
Can just use `&` and no matching group. – 123 Sep 18 '15 at 13:11
yes `echo '^h?e.l*l+o:' | sed 's/[[:punct:]]/\\&/g'` works. What does & mean? – the_eraser Sep 18 '15 at 13:22
`&` refers to the entire string matched by the regex – the_eraser Sep 20 '15 at 14:31

1 Answers1