'Regex branch name pattern conditional

I am trying to get the branch name using regex and then use it in a case.
Does anyone know how it could work?

Possible names are:

  • release/v1.1 -> release
  • master -> master
  • develop -> develop
BRANCH="release/v1.1";
#BRANCH="master";
#BRANCH="develop";


#branch_name=`expr "${BRANCH}" : '^\(release\)\/v[0-9]\.[0-9]$'`
branch_name=`expr "${BRANCH}" : '^\(master)|(release\)\/v[0-9]\.[0-9]$'`
echo $branch_name
bash


Solution 1:[1]

You can also use sed and put every alternative in it's own group

The part ((release)/v[0-9]+(\.[0-9]+)? matches release followed by / and 1+ digits, optionally followed by . and 1+ digits.

In the replacement use group 2 and group 4.

for BRANCH in release/v1.1 master develop; do
  sed -E 's~^((release)/v[0-9]+(\.[0-9]+)?|(develop|master))$~\2\4~' <<< "$BRANCH"
done

Output

release
master
develop

See the regex groups.

You could also use a bit shorter version of the pattern with awk, setting the field separator to / and when there is a match print field 1.

awk -F'/' '/^(release\/v[0-9]+(\.[0-9]+)?|develop|master)$/ {print $1}' <<< "$BRANCH"

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1

Related Questions