Capture group from regex in bash

Question

I have the following string /path/to/my-jar-1.0.jar for which I am trying to write a bash regex to pull out my-jar.

Now I believe the following regex would work: ([^\/]*?)-\d but I don't know how to get bash to run it.

The following: echo '/path/to/my-jar-1.0.jar' | grep -Po '([^\/]*?)-\d' captures my-jar-1

Answer 1

In BASH you can do:

s='/path/to/my-jar-1.0.jar'

[[ $s =~ .*/([^/[:digit:]]+)-[[:digit:]] ]] && echo "${BASH_REMATCH[1]}"

my-jar

Here "${BASH_REMATCH[1]}" will print captured group #1 which is expression inside first (...).