How to capture the longest match of a repeating pattern using BASH_REMATCH - bash

Question

I am trying to capture the longest match of a repeating pattern
do_run() {
local regex='.*((abc)+).*'
local str='_abcabcabc123_'
echo "regex=${regex}"$'\n'
echo "str=${str}"$'\n'
if [[ "${str}" =~ ${regex} ]]
then
for i in ${!BASH_REMATCH[#]}
do
echo "$i=${BASH_REMATCH[i]}"
done
else
echo "no match"
fi
}
I get the following output :
regex=.*((abc)+).*
str=_abcabcabc_
0=_abcabcabc123_
1=abc
2=abc
I am trying to get something like :
regex=.*((abc)+).*
str=_abcabcabc123_
0=_abcabcabc123_
x=abcabcabc
(Update : x is just here to indicate that the index of the matching group does not matter but I need to know what number to use to retrieve the matching group ...)
Update:
After reading comment, the following regex will work : ((abc)+)
However, I also need to capture what precedes and what follows ((abc)+).
I had not mentionned it earlier because I thought the same solution would be applied.
So the new code would be :
do_run() {
local regex='(.*)((abc)+)(.*)'
local str='_abcabcabc123_'
echo "regex=${regex}"$'\n'
echo "str=${str}"$'\n'
if [[ "${str}" =~ ${regex} ]]
then
for i in ${!BASH_REMATCH[#]}
do
echo "$i=${BASH_REMATCH[i]}"
done
else
echo "no match"
fi
}
I get then the following output :
regex=(.*)((abc)+)(.*)
str=_abcabcabc123_
0=_abcabcabc123_
1=_abcabc
2=abc
3=abc
4=123_
I want to be able to retrieve abcabcabc from a matching group but also what precedes it and what follows it

Answer 1

As a workaround you can do like this:
[STEP 101] $ cat foo.sh
v=_abcabcabc123_
if [[ $v =~ (abc)+ ]]; then
middle=${BASH_REMATCH[0]}
[[ $v =~ (.*)"$middle" ]]
before=${BASH_REMATCH[1]}
[[ $v =~ "$middle"(.*) ]]
after=${BASH_REMATCH[1]}
echo "before: $before"
echo "middle: $middle"
echo "after : $after"
fi
[STEP 102] $ bash foo.sh
before: _
middle: abcabcabc
after : 123_
[STEP 103] $

Answer 2

I also need to capture what precedes and what follows ((abc)+).
For that, typically you'll need a negative lookahead with perl regex, something along (?<!abc)((abs)+)(.*).
I am bad at perl regex, with perl-enabled grep I was able to this:
$ grep -oxP '(.*)(?<!abc)((abc)+)\K(.*)' <<<'_abcabcabc123_'
123_
$ grep -oP '((abc)+)' <<<'_abcabcabc123_'
abcabcabc
$ rev <<<'_abcabcabc123_' | grep -oP '(.*)(?<!cba)((cba)+)\K(.*)' | rev
_
Bash has no lookarounds and no perl regex. Consider using python or perl.
But you may use sed by splitting the part on the regex and then reading lines, which may be simpler:
$ readarray -t lines < <(<<<'_abcabcabc123_' sed -E 's/((abc)+)/\n&\n/'); declare -p lines
declare -a lines=([0]="_" [1]="abcabcabc" [2]="123_")
Another idea: you may use bash expansion to replace the abc parts by something unique, then split it on that separator:
$ IFS=' ' read -r before post < <(printf "%s\n" "${str//abc/ }") ; declare -p before post
declare -- before="_"
declare -- post="123_"
# or
$ IFS='#' read -r before post < <(<<<"${str//abc/#}" tr -s '#') ; declare -p before post
declare -- before="_"
declare -- post="123_"

Answer 3

For your given input this regex would work:
re='^([^a]|a[^b]*|ab[^c]*)((abc)+)(.*)'
str='_abcabcabc123_'
[[ $str =~ $re ]] && declare -p BASH_REMATCH
Output:
declare -ar BASH_REMATCH=([0]="_abcabcabc123_" [1]="_" [2]="abcabcabc" [3]="abc" [4]="123_")
So you can use:
"${BASH_REMATCH[1]}" # string before
"${BASH_REMATCH[2]}" # string containing all "abc"s
"${BASH_REMATCH[4]}" # string after
RegEx Demo