How do I recursively check permissions in reverse?

Question

There's a command, I think it comes with apache, or is somehow related to it, that checks permissions, all the way down. So if I have /home/foo/bar/baz it will tell me what the permissions are for baz, bar, foo, and home. Does anyone know what this command is or another way of doing this? The command basically starts at the argument, and works it's way up to / letting you know what the permissions are along the way so you can see if you have a permission problem.

Answer 1

The utility you may be thinking of is the namei command. According to the manual page:

Namei uses its arguments as pathnames to any type of Unix file (symlinks, files, directories, and so forth). Namei then follows each pathname until a terminal point is found (a file, directory, char device, etc). If it finds a symbolic link, we show the link, and start following it, indenting the output to show the context.

The output you desire can be received as follows:

$ namei -l /usr/src/linux-headers-2.6.35-22/include/
f: /usr/src/linux-headers-2.6.35-22/include/
drwxr-xr-x root root /
drwxr-xr-x root root usr
drwxrwsr-x root src  src
drwxr-xr-x root root linux-headers-2.6.35-22
drwxr-xr-x root root include

The namei command is part of the linux-util-ng software package. See the manual page for more details.

Answer 2

I'm not aware of any commands, but it is quite easy to write a script:

#!/bin/bash    
ARG=$1
while [[ "$ARG" != "." && "$ARG" != "/" ]]
do
        ls -ld -- "$ARG"
        ARG=`dirname -- "$ARG"`      
done

Example:

$ perms.sh /tmp/1/2/3/hello.txt

-rw-rw-r--    1 user    group          0 Jan 14 16:59 /tmp/1/2/3/hello.txt
drwxrwxr-x    2 user    group       4096 Jan 14 16:59 /tmp/1/2/3
drwxrwxr-x    3 user    group       4096 Jan 14 16:43 /tmp/1/2
drwxrwxr-x    3 user    group       4096 Jan 14 16:43 /tmp/1
drwxrwxrwt   12 root     root       4096 Jan 14 17:02 /tmp

Answer 3

I think the command you where looking for is:

namei -l `pwd`

which, when executed in /tmp/foo/bar, gives you a listing like:

drwxr-xr-x root   root   /
drwxrwxrwt root   root   tmp
drwxr-xr-x user   group  foo
drwxr-xr-x user   group  bar

Answer 4

How about a recursive bash function for a fun solution:

[bash#] function uptree { ( \ls -ld -- "$PWD"; [ "$PWD" = '/' ] && return; cd ..; uptree ) | column -t ; } 

[bash#] pwd
/home/frielp/bin/dev

[bash#] uptree
drwxrwxr-x.  2   frielp  frielp  4096  Dec  14  14:50  /home/frielp/bin/dev
drwxr-xr-x.  15  frielp  frielp  4096  Aug  23  10:48  /home/frielp/bin
drwxr-xr-x.  60  frielp  frielp  4096  Jan  14  16:48  /home/frielp
drwxr-xr-x.  4   root    root    4096  Dec  1   09:14  /home
dr-xr-xr-x.  23  root    root    4096  Jan  14  08:18  /

[bash#] pwd
/home/frielp/bin/dev
[bash#] 

Answer 5

This could easily be made a one-liner. This is not recursive and should be a relatively fast way of doing this in bash. Calling pwd in each loop isn't particularly fast, so avoid if you can.

#!/bin/bash
if [ -n "$1" ]; then
    echo "Usage: $0 [FILE]"
    exit 1
fi
cd -P -- "$1"
IFS="/"
set -f # turn off globbing for the expansion of $PWD below
for dir in $PWD; do
    stat -c "$PWD %A" .
    cd ..
done

---

Alternative, a one-liner for the current directory.

(IFS="/"; set -f; for dir in $PWD; do stat -c "$PWD %A" .; cd ..; done)

Answer 6

alternately, consider using find with tac

find /path -printf '%M %u %g %p\n' | tac