`ls -l` for all parent directories - linux

Question

I want to get a list of all directory permissions from current folder to /. For example, for the directory: /var/lib/program/subfolder, I want an output such as:
$ pwd
/var/lib/program/subfolder
$ magic_ls_-l_command somefile
drwxr-xr-x 10 root root 4096 May 15 20:20 var
drwxr-xr-x 10 root root 4096 May 15 20:20 lib
drwxrwxr-x 10 root user 4096 May 16 20:21 program
drwxrwxr-x 10 root user 4096 May 16 20:21 subfolder
-rwxrwxr-- 1 root user 4096 May 16 20:22 somefile
I don't care about the order (from /var to /subfolder or the other way around), the number of hard links or even the date. I just wrote them down to emulate the ls -l output. Also, I don't care how each filename in printed (/var and /lib, var and lib, or /var and /var/lib). I'm just interested in the ownership of each file/directory in the path from the choosen file or pwd to /.
In case I should install some program, I'm under Ubuntu 20.04.

Answer 1

This question has already been answered in superuser.com (I don't know if I can mark a question from one site as duplicate from another). The solution is as simple as writing (assuming I am in the same directory as the target filename):
$ namei -l $(pwd)/somefile ## or `namei -l $(realpath -s somefile)`
Because of -l, it lists basic permissions in long format for each parent directory.
I have to use pwd/realpath because namei doesn't resolve relative paths. If I'm not in the target directory, just write the full path.

Answer 2

I made this small script that does this. I use cd "$1"; pwd to get the current directory so that paths are not canonicalized (say, if you try magic-ls . and your current directory is /var/lib/postgres, but that is a symlink to /mnt/postgres, you will get /var, /var/lib and /var/lib/postgres, while using realpath you would get /mnt and /mnt/postgres)
magic-ls() {
local current=$(cd "$1"; pwd)
while [[ $current != '/' ]]; do
ls -ld "$current"
current=$(dirname "$current")
done
}
Here's an example output:
[leodag#desk ~]$ magic-ls
drwx------ 1 leodag leodag 2722 jun 21 13:49 /home/leodag
drwxr-xr-x 1 root root 18 mai 2 2019 /home
By the way it will also work with no argument since cd "" does not change your directory.
Edit: removed realpath from the while check, since that could lead to unexpected results if there was a link to / in the path, and was unneeded.

Answer 3

I wrote a bash script for you. It'll have some bugs, if you have space in names. If it bothers you, I'm happy for changes recommendations in the comments.
#!/bin/bash
if [ ! -z "$1" ] && [ -e "$1" ]
then
path=`realpath -s "$1"` # read argument as absolute path
else
path="$PWD" # No valid argument, so we take pwd
fi
paths=""
while [ "$path" != / ];do
paths+=" $path"
path=`dirname "$path"`
done
paths+=" $path" # Adding / to pathlist too
ls -ld $paths
With realpath -s you can catch the absolute path, but you wont follow the link. If no argument is given, we will use pwd as the file/directory to list.
We append each path to a list. This gives us the advantage of a better layout in the end, so that we get a nice table because we run ls only once.
Output:
bobafit:~$ magic_ls_-l_command /usr/bin/python3
drwxr-xr-x 21 root root 4096 Jun 20 10:07 /
drwxr-xr-x 14 root root 4096 Sep 5 2019 /usr
drwxr-xr-x 2 root root 110592 Jun 20 10:07 /usr/bin
lrwxrwxrwx 1 root root 9 Apr 7 12:43 /usr/bin/python3 -> python3.8

Answer 4

Just using parameter expansion:
#!/usr/bin/env bash
path="$1"
while test -n "$path"; do
ls -lLd "$path"
path="${path%/*}"
done
calling method :
bash test.sh /var/lib/program/subfolder/somefile
giving
-rw-r--r-- 1 root root 0 Jun 21 18:49 /var/lib/program/subfolder/somefile
drwxr-xr-x 1 root root 4096 Jun 21 18:49 /var/lib/program/subfolder
drwxr-xr-x 1 root root 4096 Jun 21 18:49 /var/lib/program
drwxr-xr-x 1 root root 4096 Jun 21 18:49 /var/lib
drwxr-xr-x 1 root root 4096 Jun 13 19:24 /var

Answer 5

#! /bin/bash
cur=""
IFS="/"
path=`pwd`
for dir in ${path:1}
do
cur=$cur/$dir
ls -lhd "$cur"
done
cur=$cur/$1
ls -lhd "$cur"
Terminal Session:
$ pwd
/tmp/dir_underscore/dir space/dir special #!)
$ ls
bash.sh test.txt
$ ./bash.sh test.txt
drwxrwxrwt 28 root root 36K Jun 21 22:45 /tmp
drwxr-xr-x 3 root root 4.0K Jun 21 22:27 /tmp/dir_underscore
drwxr-xr-x 3 root root 4.0K Jun 21 22:28 '/tmp/dir_underscore/dir space'
drwxr-xr-x 2 root root 4.0K Jun 21 22:54 '/tmp/dir_underscore/dir space/dir special #!)'
-rw-r--r-- 1 root root 0 Jun 21 22:29 '/tmp/dir_underscore/dir space/dir special #!)/test.txt'

Answer 6

This should possibly work:
pwd ; ls -lh ; while true ; do cd .. ; pwd ; ls -lh ; [[ "$PWD" == "/" ]] && break ; done
EDIT: I misunderstood the question at first. Try this:
(pwd ; ls -ldh ; while true ; do cd .. ; pwd ; ls -ldh ; [[ "$PWD" == "/" ]] &&
break ; done ; cd "$START")
EDIT2: fillipe's answer is probably the best, but here's my third and last attempt, which works on both files and directories:
magic_ls() {
fname="$1"
while true ; do
ls -lhd "$fname"
[[ "$fname" == "/" ]] && break ;
fname=$(dirname $(readlink -f "$fname"))
done
}

Answer 7

Just my 2 cents. My mac doesn't have the namei command (perhaps homebrew has a copy), but wanted to whip up a quick version that aligned the output in top-down order
#!/usr/bin/env bash
path="${1%/}"
DIRS=()
while test -n "$path"; do
DIRS=( "$path" "${DIRS[#]}" )
path="${path%/*}"
done
ls -ld "${DIRS[#]}"
Example output:
$ lspath $TMPDIR
lrwxr-xr-x# 1 root wheel 11 Oct 5 2018 /var -> private/var
drwxr-xr-x 7 root wheel 224 Jul 16 2020 /var/folders
drwxr-xr-x# 3 root wheel 96 Apr 5 2018 /var/folders/0c
drwxr-xr-x# 5 me staff 160 Apr 5 2018 /var/folders/0c/2_s_qxd11m3d1smzqdrs3qg40000gp
drwx------# 255 me staff 8160 Oct 7 09:18 /var/folders/0c/2_s_qxd11m3d1smzqdrs3qg40000gp/T