How to use regex capturing group in bash correctly? - regex

Question

I've loaded some strings into variable "result". The strings look like this:
school/proj_1/file1.txt
school/proj_1/file2.txt
school/proj_1/file3.txt
I try to get only the name after the last slash, so file1.txt, file2.txt and file3.txt is the desirable result for me. I use this piece of code
for i in $result
do
grep "school/proj_1/(.*)" $i
done
but it doesn't work. I feel that the regex would work for Python with the caputuring group I created, but I can't really wrap my head around how to use capturing groups in bash or if it is even possible at all.
I'm sorry if it's a dumb question, I'm very new to scripting in bash.

Answer 1

You may use a simple approach with a string manipulation operation:
echo "${i##*/}"
${string##substring}
Deletes longest match of $substring from front of $string.
Or using a regex in Bash, you may get the capturing groups like
result=("school/proj_1/file1.txt" "school/proj_1/file2.txt" "school/proj_1/file3.txt")
rx='school/proj_1/(.*)'
for i in "${result[#]}"; do
if [[ "$i" =~ $rx ]]; then
echo "${BASH_REMATCH[1]}"
fi
done
See the online demo. Here, ${BASH_REMATCH[1]} is the contents inside capturing group #1.

Answer 2

Try this :
variable declaration :
$ result="school/proj_1/file1.txt
school/proj_1/file2.txt
school/proj_1/file3.txt"
Commands :
(all as one-liners)
$ grep -oP "school/proj_1/\K.*" "$i" <<< "$result"
or
$ awk -F'/' '{print $NF}' <<< "$result
or
$ sed 's|.*/||' <<< "$result"
or if number of sub dirs are fixed :
$ cut -d'/' -f3 <<< "$result"
Output :
file1.txt
file2.txt
file3.txt