Bash test for capturing # (hash) character

Question

How do I capture/escape # in a test? I'm trying to use the clever answer from How to reference captures in bash regex replacement to color everything after a # green

foo='hello # world'
foo="$([[ $foo =~ (#.*) ]] && echo -e ${foo/$BASH_REMATCH/\\e[32m${BASH_REMATCH[1]}\\e[0m} || echo $foo)"
echo -e $foo

I have tried using (#.*), (\#.*), and (\\#.*). The first two apparently successfully test hello # world, but don't replace it correctly...?

[[ $foo =~ (#.+) ]] && echo -e "${foo/$BASH_REMATCH/! ${BASH_REMATCH[1]} !}"
# prints: hello # world

[[ $foo =~ ( #.+) ]] && echo -e "${foo/$BASH_REMATCH/! ${BASH_REMATCH[1]} !}"
# prints: hello!  # world !

As seen above, if I precede the # with a space, it works fine, but I want a generic solution, and lookbehinds don't appear to work.

I can get this to work using sed, but for my purposes, it's very slow (on the order of seconds). I'm sure there's a faster way that I could do this using sed, but I'd really like a vanilla bash solution.

...if your goal is speed, you should try to avoid any fork operations. — Charles Duffy, Oct 26 '20 at 18:23
As opposed to replacing `echo -e ` with `foo=`? It's a remnant of a failed experiment. — dx_over_dt, Oct 26 '20 at 18:23

score 2 · Accepted Answer · answered Oct 26 '20 at 18:27

2

foo='hello # world'
esc_start=$'\e[32m'
esc_end=$'\e[0m'
 
if [[ $foo =~ (#.*) ]]; then
  foo="${foo/"$BASH_REMATCH"/"${esc_start}${BASH_REMATCH}${esc_end}"}"
fi

answered Oct 26 '20 at 18:27

Charles Duffy

280,126
43
390
441

I copied and pasted that and got nothing – dx_over_dt Oct 26 '20 at 18:29
‍♀️ I didn't `echo $foo`. – dx_over_dt Oct 26 '20 at 18:31

score 0 · Answer 2 · answered Oct 26 '20 at 19:02

0

Maybe consider replacing bash-native text replacement with sed which is more suitable for complex strings.

foo="hello # world" char='#' foo=$([[ $foo =~ (($char).*) ]] && echo -e $( echo $foo | sed "s/$BASH_REMATCH/\\\\e[32m${BASH_REMATCH[1]}\\\\e[0m/") || echo $foo); echo -e $foo

answered Oct 26 '20 at 19:02

Marcin Piechota

71
4

1

There's a major performance impact that comes with using `sed` here: It's faster once it's running, but _much_ slower to start up; starting a copy of `sed` just to process a single line is basically always counterproductive. – Charles Duffy Oct 26 '20 at 19:18
Also, note that `echo -e $foo` is several flavors of buggy. `echo -e "$foo"` is unreliable -- its behavior can be modified by both compile-time and runtime configuration, which is not true of the POSIX-compliant equivalent `printf '%b\n' "$foo"` -- and `echo -e $foo` additionally has the bugs described in [I just assigned a variable, but `echo $variable` shows something else!](https://stackoverflow.com/questions/29378566/i-just-assigned-a-variable-but-echo-variable-shows-something-else) – Charles Duffy Oct 26 '20 at 19:20
...for more details on the `printf`-vs-`echo` thing, see the excellent answer to [Why is printf better than echo?](https://unix.stackexchange.com/a/65819/3113) over at [unix.se]. – Charles Duffy Oct 26 '20 at 19:21

Bash test for capturing # (hash) character

2 Answers2