How to get the name of the current git branch into a variable in a shell script?
bash, git, linux, shell, ubuntu
Solution
The * is expanded, what you can do is use sed instead of grep and get the name of the branch immediately:
branch=$(git branch | sed -n -e 's/^\* \(.*\)/\1/p')
And a version using git symbolic-ref, as suggested by Noufal Ibrahim
branch=$(git symbolic-ref HEAD | sed -e 's,.*/\(.*\),\1,')
To elaborate on the expansion, (as marco already did,) the expansion happens in the echo, when you do `echo $test` with `$test` containing `* master` then the `*` is expanded according to the normal expansion rules. To suppress this one would have to quote the variable, as shown by marco: `echo "$test"`. Alternatively, if you get rid of the asterisk before you echo it, all will be fine, e.g. `echo ${test:2}` will just echo `master`. Alternatively you could assign it anew as you already proposed:
branch=${test:2}
echo $branch
This will echo `master`, like you wanted.
Problem
I am new to shell scripting and can't figure this out. If you are unfamiliar, the command git branch returns something like ``` * develop master ``` , where the asterisk marks the currently checked out branch. When I run the following in the terminal: ``` git branch | grep "*" ``` I get: ``` * develop ``` as expected. However, when I run ``` test=$(git branch | grep "*") ``` or ``` test=`git branch | grep "*"` ``` And then ``` echo $test ``` , the result is just a list of files in the directory. How do we make the value of test="* develop"? Then the next step (once we get "* develop" into a variable called test), is to get the substring. Would that just be the following? ``` currentBranch=${test:2} ``` I was playing around with that substring function and I got "bad substitution" errors a lot and don't know why.