Is git ls-remote able to use GITHUB variables? Error: Process completed with exit code 2

Question

Actions code:

jobs:
  build:
    runs-on: ubuntu-latest
    steps:
      - name: Checkout codes
        uses: actions/checkout@v2
      - name: Test git for actions
        shell: bash
        run: |
          ## use GitHub variables, such as: GITHUB_REF,GITHUB_HEAD_REF..
          BRANCH=${GITHUB_REF##*/}
          branch=$BRANCH
          git ls-remote --heads --exit-code repo_url "$branch" >/dev/null
          if [ "$?" == "1" ]
          then 
            echo "Branch doesn't exist"
          else
            echo "Branch exist"
          fi

It will occur the following error:

  BRANCH=${GITHUB_REF##*/}
  branch=${BRANCH}
  echo $branch
  git ls-remote --heads --exit-code repo_url "$branch" >/dev/null
  if [ "$?" == "1" ]
  then 
    echo "Branch doesn't exist"
  else
    echo "Branch exist"
  fi

  shell: /usr/bin/bash --noprofile --norc -e -o pipefail {0}
main
Error: Process completed with exit code 2.

When I replace ${GITHUB_REF} with main, it works fine.

jobs:
  build:
    runs-on: ubuntu-latest
    steps:
      - name: Test git for actions
        shell: bash
        run: |
          ## use GitHub variables, such as: GITHUB_REF,GITHUB_HEAD_REF..
          BRANCH=${GITHUB_REF##*/}
          branch=main
          git ls-remote --heads --exit-code repo_url "$branch" >/dev/null
          if [ "$?" == "1" ]
          then 
            echo "Branch doesn't exist"
          else
            echo "Branch exist"
          fi

Output:

  BRANCH=${GITHUB_REF##*/}
  branch=main
  echo $branch
  git ls-remote --heads --exit-code repo_url "$branch" >/dev/null
  if [ "$?" == "1" ]
  then 
    echo "Branch doesn't exist"
  else
    echo "Branch exist"
  fi
main
Branch exist

Is the git ls-remote command not able to use variables?

I want to check whether a certain branch exists in the remote warehouse in GitHub Actions.

Answer 1

According to jobs.<job_id>.steps[*].shell, the default Bash invocation is:

bash --noprofile --norc -eo pipefail {0}

which makes it to fail fast as described under Exit codes and error action preference, for bash:

Fail-fast behavior using set -eo pipefail: This option is set when shell: bash is explicitly specified.

And, according to Bash manual, under The Set Builtin:

-e: Exit immediately if a pipeline (see Pipelines), which may consist of a single simple command (see Simple Commands), a list (see Lists of Commands), or a compound command (see Compound Commands) returns a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in an if statement, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command’s return status is being inverted with !. If a compound command other than a subshell returns a non-zero status because a command failed while -e was being ignored, the shell does not exit. A trap on ERR, if set, is executed before the shell exits.

and,

-o pipefail: If set, the return value of a pipeline is the value of the last (rightmost) command to exit with a non-zero status, or zero if all commands in the pipeline exit successfully. This option is disabled by default.

---

In your case, the possible solution could be to directly handle the exit status of the command in an if:

if command; then
  # succes
else
  # failure
end

i.e.

if ! git ls-remote --heads --exit-code repo_url "$branch" >/dev/null
then 
  echo "Branch doesn't exist"
else
  echo "Branch exist"
fi

or,

if git ls-remote --heads --exit-code repo_url "$branch" >/dev/null
then 
  echo "Branch exist"
else
  echo "Branch doesn't exist"
fi