Bash 脚本仍在继续，即使我希望它早点退出

Question

一段时间对此感到困惑，这里是一个代表问题的精炼脚本：

#start    
# note that master does not exist, so this should fail, would like to exit on the next line
git branch -D master || (echo "no master branch" && exit 1);
git fetch origin &&
git checkout master &&

BRANCH=$(git rev-parse --abbrev-ref HEAD)
if [[ "$BRANCH" != "master" ]]; then
  echo 'Aborting script because you are not on the right git branch (master).';
  exit 1;
fi

echo "done"
#end

当我运行上面的脚本时，我得到以下输出：

error: branch 'master' not found.
no master branch
error: Your local changes to the following files would be overwritten by checkout:
        publish-to-NPM.sh
Please, commit your changes or stash them before you can switch branches.
Aborting
Aborting script because you are not on the right git branch (master).

请注意，“done”不会得到回显，因此脚本会在第二次 exit 1 调用时退出。但是为什么脚本在第一次 exit 1 调用时不退出呢？对此很困惑。

Answer 1

git branch -D master || (echo "no master branch" && exit 1);

在子流程环境中运行条件的 RHS。exit 退出该子进程。如果您想退出主脚本，请不要在子进程中运行它。也就是说，写：

git branch -D master || { echo "no master branch" && exit 1; }