Simplifying release procedure

This commit is contained in:
Sławomir Śledź 2019-10-07 10:27:02 +02:00
parent 4c16fc52f7
commit 142e1bade4
2 changed files with 2 additions and 49 deletions

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@ -522,8 +522,8 @@ cd test
Feel free to ask questions in chat, open issues, or contribute by creating pull requests. Feel free to ask questions in chat, open issues, or contribute by creating pull requests.
In order to create a pull request In order to create a pull request
* checkout develop branch * checkout master branch
* introduce your changes * introduce your changes & bump version
* submit pull request * submit pull request
# Resources # Resources

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@ -1,47 +0,0 @@
#!/bin/bash
version=$(cat .version)
release_version=${version%%-SNAPSHOT}
new_version=$(echo $release_version+0.1 | bc)
[[ $? -ne 0 ]] && echo 'Error exiting.' && exit 1
snapshot_version=${new_version}-SNAPSHOT
cat <<EOF
current version : $version
release version : $release_version
new snapshot version: $snapshot_version
EOF
git fetch
echo start the release by creating a new release branch
git checkout -b release/$release_version origin/develop
echo $release_version > ./.version
git add ./.version
git commit -m "[release] prepare release v$release_version"
git tag v$release_version
echo $snapshot_version > ./.version
git add ./.version
git commit -m "[release] prepare for next development iteration"
echo merge the version back into develop
git checkout develop
git merge --no-ff -m "[release] merge release/$release_version into develop" release/$release_version
git checkout master
echo merge the version back into master but use the tagged version instead of the release/$releaseVersion HEAD
git merge --no-ff -m "[release] merge previous version into master to avoid the increased version number" release/$release_version~1
echo get back on the develop branch
git checkout develop
echo finally push everything
git push origin develop master
git push --tags
echo removing the release branch
git branch -D release/$release_version