Regex branch name pattern conditional-CodePudding

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

CodePudding user response：

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"/")

CodePudding user response：

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"