How can I get parent branch name using simpleGit ?

[nvuillam]

I'm trying to implement a way to get the name of the branch that has been used to create the current branch

Solutions exists, but I don't succeed to implement them with simpleGit

Would someone have some tip to reproduce the following solution with simpleGit ? https://stackoverflow.com/a/17843908/7113625

[steveukx]

Based on the answer to the stackoverflow you linked to, you would use:

const outputFromGit = (await simpleGit({ trimmed: true }).raw('show-branch', '-a')).split('\n');
const rev = await simpleGit({ trimmed: true }).raw('rev-parse', '--abbrev-ref', 'HEAD');

To give you an array of each line of output from git, which is the first step in the piped chain. From there, you would:

outputFromGit
  .map(line => line.replace(/\].*/, '')) // remove branch commit message
  .filter(line => line.includes('*'))     // only lines with a star in them
  .filter(line => !line.includes(rev))  // only lines not including the specified revision
  .filter((line, index, all) => index < all.length - 1) // not the last line
  .map(line => line.replace(/^.*\[/, '')) // remove all but the branch name

Or thereabouts.

[nvuillam]

@steveukx thanks for your reply :)

It did not work exactly as expected (cf screenshot) , but I built some ugly code that seems to work :)

export async function getGitParentBranch() {
  try {
    const outputFromGit = (
      await simpleGit({ trimmed: true }).raw("show-branch", "-a")
    ).split("\n");
    const rev = await simpleGit({ trimmed: true }).raw(
      "rev-parse",
      "--abbrev-ref",
      "HEAD"
    );
    const allLinesNormalized = outputFromGit.map((line) =>
      line.trim().replace(/\].*/, "")
    );
    const indexOfCurrentBranch = allLinesNormalized.indexOf(`* [${rev}`);
    if (indexOfCurrentBranch > -1) {
      const parentBranch = allLinesNormalized[indexOfCurrentBranch + 1].replace(
        /^.*\[/,
        ""
      );
      return parentBranch;
    }
  } catch (e) {
    return null;
  }
  return null;
}