Regex branch name pattern conditional

user_6191795 0
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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
2 Answers

What about cut?

| cut -f1 -d"/"

This uses a slash as a separator and only shows the first entry:

Prompt> echo "master" | cut -f1 -d"/"
master

Prompt> echo "develop" | cut -f1 -d"/"
develop

Prompt> echo "release/v1.1 you can put whatever here :-)" | cut -f1 -d"/"
release

Edit: In order to use this to assign to variable branch_name, this is what you need to do:

branch_name=$(echo $BRANCH | cut -f1 -d"/")
user_42791550

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"
user_54249880
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