如何在Bash中仅打印父路径并排除所有子目录路径

背景 #

之前解决过反向问题，现在要实现相反逻辑：从find输出的文件路径里，只保留最顶层的父目录，把所有子目录路径都排除掉。

原反向逻辑命令 #

原来用来排除父路径、保留子路径的命令是：

cat file | sort | tac | awk '{sub("/[^/]+$","")} index(prev,$0"/") != 1{print} {prev=$0}'

其中file是这条find命令的输出结果：

find src/services/ -type f -name 'BUILD.bazel'

示例输入（file内容） #

src/services/sam-agent/BUILD.bazel
src/services/sam-agent/auth/BUILD.bazel
src/services/sam-agent/certs/BUILD.bazel
src/services/sam-agent/server/BUILD.bazel
src/services/jam-controller/BUILD.bazel
src/services/jam-controller/api/BUILD.bazel
src/services/jam-controller/client/BUILD.bazel
src/services/wam-controller/api/BUILD.bazel
src/services/wam-controller/api/client/BUILD.bazel
src/services/wam-controller/api/server/BUILD.bazel

期望输出 #

只保留最顶层的父路径，其他子目录全部排除：

src/services/sam-agent
src/services/jam-controller
src/services/wam-controller

也就是只打印下面标记的路径，剩下的子目录都不输出：

src/services/sam-agent #< 需打印
src/services/sam-agent/auth
src/services/sam-agent/certs
src/services/sam-agent/server
src/services/jam-controller #< 需打印
src/services/jam-controller/api
src/services/jam-controller/client
src/services/wam-controller/api #< 需打印
src/services/wam-controller/api/client
src/services/wam-controller/api/server

尝试的错误命令及问题 #

试了下面的命令，但还是会输出子目录（比如src/services/rams/store/postgres），没能只留下最顶层路径：

cat /tmp/t.log | sort | tac | awk '{if ($0 in seen) print; else {sub("/[^/]+$", ""); seen[$0]}}'

/tmp/t.log的内容示例：

src/services/rams
src/services/rams/integration
src/services/rams/keys
src/services/rams/migrations
src/services/rams/mocks
src/services/rams/server
src/services/rams/smoke
src/services/rams/store
src/services/rams/store/postgres
src/services/rams/store/postgres/mocks
src/services/rams/tools/keyrotate
src/services/rams/tools/secretmanager
src/services/rams/vault

错误输出结果：

src/services/rams/store/postgres
src/services/rams

解决方案 #

方法1：处理已有文件的高效实现 #

先对路径排序，再用awk筛选出顶层目录：

sort file | awk '{
    dir = $0
    sub("/[^/]+$", "", dir)  # 去掉末尾的文件名，得到目录路径
    if (!prev || index(dir, prev "/") != 1) {
        if (prev) print prev
        prev = dir
    }
} END { print prev }'

方法2：直接处理find输出（无需中间文件） #

如果不需要保存file中间文件，直接把find和处理逻辑连起来：

find src/services/ -type f -name 'BUILD.bazel' | sort | awk '{
    dir = $0
    sub("/[^/]+$", "", dir)
    if (!prev || index(dir, prev "/") != 1) {
        if (prev) print prev
        prev = dir
    }
} END { print prev }'

方法3：简化版sed+awk实现 #

先提取所有目录路径，排序后只保留非子目录的顶层路径：

find src/services/ -type f -name 'BUILD.bazel' | sed 's/\/[^/]*$//' | sort | awk '
    !prev || !index($0, prev "/") { print $0; prev = $0 }
'

方案说明 #

1. 先对路径按字典序排序，这样子目录会自动跟在对应的父目录后面
2. 对每一行路径，去掉最后的文件名，得到纯目录路径
3. 检查当前目录是否是上一个目录的子目录：如果不是，说明上一个目录是顶层路径，输出它；如果是，则跳过继续往下
4. 最后在命令结束时输出最后一个顶层目录

内容的提问来源于stack exchange，提问作者AhmFM