在多个分支中寻找第一次提交的shell函数:

# Get the first commit hash of a given branch.
# Uses `git branch --contains` to backward (starts from HEAD) check each commits
# and output that branch's name.
first_commit_of_branch() {
    if [ $# -ne 1 ] || [ -z "${1}" ] ; then
        (>&2 echo "Error: Missing or empty branch name.")
        (>&2 echo "Usage: $0 branch_to_test")
        return 2
    fi
    local branch_to_test="${1}"; shift
    local commit_index_to_test
    local maximum_number_of_commit_to_test
    local branch_count_having_tested_commit

    git rev-parse --verify --quiet "${branch_to_test}" 2>&1 > /dev/null || {
        (>&2 echo "Error: Branch \"${branch_to_test}\" does not exists.")
        return 2
    }

    commit_index_to_test=0
    maximum_number_of_commit_to_test=$(git rev-list --count "${branch_to_test}")

    while [ ${commit_index_to_test} -le ${maximum_number_of_commit_to_test} ] ; do
        # Testing commit $branch_to_test~$commit_index_to_test…

        # If it fails, it means we tested all commits from the most recent of
        # branch $branch_to_test to the very first of the git DAG. So it must be it.
        git rev-parse --verify --quiet ${branch_to_test}~${commit_index_to_test} 2>&1 > /dev/null || {
            git rev-list --max-parents=0 "${branch_to_test}"
            return 0
        }

        branch_count_having_tested_commit="$( \
            git --no-pager branch --no-abbrev --verbose \
                --contains ${branch_to_test}~${commit_index_to_test} \
            | cut -c 3- \
            | cut -d ' ' -f 2 \
            | wc -l \
        )"

        # Tested commit found in more than one branch
        if [ ${branch_count_having_tested_commit} -gt 1 ] ; then
            if [ ${commit_index_to_test} -eq 0 ]; then
                (>&2 echo "Error: The most recent commit of branch \"${branch_to_test}\" (${branch_to_test}~${commit_index_to_test}) is already in more than one branch. This is likely a new branch without any commit (yet). Cannot continue.")
                return 1
            else
                # Commit $branch_to_test~$commit_index_to_test is in more than
                # one branch, stopping there…
                git rev-parse ${branch_to_test}~$((commit_index_to_test-1))
                return 0
            fi
        fi
        # else: Commit $branch_to_test~$commit_index_to_test is still only in
        #       branch ${branch_to_test} continuing…"
        commit_index_to_test=$((commit_index_to_test+1))
    done
}

警告:当在一个有子分支的分支上执行并且没有新的提交时，它会失败。

A---B---C---D      <- "main" branch
 \   \
  \   E---F        <- "work1" branch
   \       \
    \       G---H  <- "work1-b" branch
     \
      I---J        <- "work2" branch

first_commit_of_branch main # C
first_commit_of_branch work1 # (Fails)
first_commit_of_branch work1-b # G
first_commit_of_branch work2 # I

下面是Mark Reed解决方案的PowerShell实现:

git show-branch -a | where-object { $_.Contains('*') -eq $true} | Where-object {$_.Contains($branchName) -ne $true } | select -first 1 | % {$_ -replace('.*\[(.*)\].*','$1')} | % { $_ -replace('[\^~].*','') }

我并不是说这是解决问题的好方法，但这似乎对我来说确实有效:

git branch --contains $(cat .git/ORIG_HEAD)

问题是，隐藏一个文件是窥探Git的内部工作，所以这并不一定是向前兼容(或向后兼容)。

git log -2 --pretty=format:'%d' --abbrev-commit | tail -n 1 | sed 's/\s(//g; s/,/\n/g';

(来源/母体名,母体名)

git log -2 --pretty=format:'%d' --abbrev-commit | tail -n 1 | sed 's/\s(//g; s/,/\n/g';

起源- parent-name

git log -2 --pretty=format:'%d' --abbrev-commit | tail -n 1 | sed 's/(.*,//g; s/)//';

母体名

Mark Reed的解决方案基本上是正确的。但是，请注意，提交行不仅应该包含星号，而且应该以星号开头!否则，包含星号的提交消息也包含在匹配的行中。所以它应该是:

git show-branch——| awk - f '[]^~[]' '/^\*/ && !/'"$ current_branch”/{打印2美元;退出}'

或者是更长的版本:

git show-branch -a           |
  awk '^\*'                  | # we want only lines that contain an asterisk
  awk -v "$current_branch"   | # but also don't contain the current branch
  head -n1                   | # and only the first such line
  sed 's/.*\[\(.*\)\].*/\1/' | # really, just the part of the line between []
  sed 's/[\^~].*//'            # and with any relative refs (^, ~n) removed`

Joe Chrysler的命令行魔法可以简化。下面是Joe的逻辑——为了简洁起见，我在两个版本中都引入了一个名为cur_branch的参数来代替命令替换' git rev-parse——abbrev-ref HEAD ';可以像这样初始化:

cur_branch=$(git rev-parse --abbrev-ref HEAD)

然后，这是Joe的管道:

git show-branch -a           |
  grep '\*'                  | # we want only lines that contain an asterisk
  grep -v "$cur_branch"      | # but also don't contain the current branch
  head -n1                   | # and only the first such line
  sed 's/.*\[\(.*\)\].*/\1/' | # really, just the part of the line between []
  sed 's/[\^~].*//'            # and with any relative refs (^, ~n) removed

我们可以在一个相对简单的awk命令中完成与所有这五个单独的命令过滤器相同的事情:

git show-branch -a |
  awk -F'[]^~[]' '/\*/ && !/'"$cur_branch"'/ {print $2;exit}'

具体情况是这样的:

-F'[]^~[]'

将行分割为以]、^、~和[为字符的字段。

/\*/

找出包含星号的行

&& !/'"$cur_branch"'/

.．.但不是当前的分支名称

{ print $2;

当您找到这样的一行时，打印它的第二个字段(即字段分隔符第一次和第二次出现之间的部分)。对于简单的分支名称，这将是括号之间的内容;对于具有相对跳转的引用，它将只是没有修饰符的名称。因此，我们的字段分隔符集处理了两个sed命令的意图。

  exit }

然后立即退出。这意味着它只处理第一个匹配的行，所以我们不需要通过head -n 1来输出。