使用JGit获得所有分支

Question

如何使用JGit获取存储库中的所有分支？我们来看一个示例存储库.我们可以看到,它有5个分支.
在这里我找到了这个例子:

int c = 0;
List<Ref> call = new Git(repository).branchList().call();
for (Ref ref : call) {
    System.out.println("Branch: " + ref + " " + ref.getName() + " "
            + ref.getObjectId().getName());
    c++;
}
System.out.println("Number of branches: " + c);

但我得到的只是:

Branch: Ref[refs/heads/master=d766675da9e6bf72f09f320a92b48fa529ffefdc] refs/heads/master d766675da9e6bf72f09f320a92b48fa529ffefdc
Number of branches: 1
Branch: master

Answer 1

如果它是你缺少的远程分支,你必须设置ListMode的ListBranchCommand对ALL或REMOTE.默认的ListMode(ListMode)仅返回本地分支.

new Git(repository).branchList().setListMode(ListMode.ALL).call();

Answer 2

我将下面的方法用于 git 分支，而不使用 Jgit 克隆 repo

这在 pom.xml 中

    <dependency>
        <groupId>org.eclipse.jgit</groupId>
        <artifactId>org.eclipse.jgit</artifactId>
        <version>4.0.1.201506240215-r</version>
    </dependency>

方法

public static List<String> fetchGitBranches(String gitUrl)
            {
                Collection<Ref> refs;
                List<String> branches = new ArrayList<String>();
                try {
                    refs = Git.lsRemoteRepository()
                            .setHeads(true)
                            .setRemote(gitUrl)
                            .call();
                    for (Ref ref : refs) {
                        branches.add(ref.getName().substring(ref.getName().lastIndexOf("/")+1, ref.getName().length()));
                    }
                    Collections.sort(branches);
                } catch (InvalidRemoteException e) {
                    LOGGER.error(" InvalidRemoteException occured in fetchGitBranches",e);
                    e.printStackTrace();
                } catch (TransportException e) {
                    LOGGER.error(" TransportException occurred in fetchGitBranches",e);
                } catch (GitAPIException e) {
                    LOGGER.error(" GitAPIException occurred in fetchGitBranches",e);
                }
                return branches;
            }