View permission / owner of full directory tree

Question

I remember doing someting like "XXX /home/user/dir/child/file" and it returned the owner and/or permission of:

/home
/home/user
/home/user/dir
/home/user/child
/home/user/child/file

But I don't remember what this command was. Anybody any idea?

Answer 1

I think you might be thinking of the tree command. For example:

$ tree -pufid apps/glassfish3/ | less
apps/glassfish3
[drwxr-xr-x saml    ]  apps/glassfish3/bin
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/bin
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/config
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/doc-files
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/javax
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/javax/annotation
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/javax/annotation/security
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/javax/annotation/sql
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/javax/decorator
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/javax/ejb
[drwxr-xr-x saml    ]  apps/glassfish3/glassfish/docs/api/javax/ejb/embeddable
...
...

The above switches do the following:

• -p - permissions
• -u - username/userid
• -f - full path
• -i - don't print indentation lines
• -d - print directories only

References

• tree man page

Answer 2

The command could have been:

namei -m /home/user/dir/child/file

Answer 3

After giving it some thougth I came up with this

#!/bin/sh
l_path=$1
while [ "$l_path" != / -a "$l_path" != . ]; do
     ls -ld $l_path
     l_path=$(dirname -- "$l_path")
done

The output looks like this

-rw------- 1 tant tant 181016423 Jun 25 23:49:17 2013 /home/tant/test_file
drwxr-xr-x 85 tant tant 5632 Jul  9 19:40:11 2013 /home/tant
lrwxr-xr-x 1 root wheel 8 Sep  4 23:53:27 2012 /home -> usr/home

I hope it is ok that it is in reverse order.

Based on the comments, here's a way to list from the root downwards:

#!/bin/sh
l_path=$1
while [ "$l_path" != / -a "$l_path" != . ]; do
     ls -ld $l_path
     l_path=$(dirname -- "$l_path")
done | sed '1!G;h;$!d'

Answer 4

In the directory of which you want to know the ancester's permissions and owners:

for i in $(seq 0 $(pwd | tr -cd / | wc -c)) ; do pwd ; ls -lad ; cd .. ; done

Note that after that, you'll be in / :) if you want to go back to where you were, wrap the command inside

HERE=$(pwd)
...
cd ${HERE}

Answer 5

List permissions of every directory in a directory path (up or down)

alias dirls='_dirls() { path="$(realpath $1)";_path="";for i in " " ${path//\// }; do _path="$_path$i/"; ls -ld $_path; done; unset _path; };_dirls'

alias dirls2='_dirls2() { path="$(realpath $1)"; while [[ "$path" != "/" ]]; do ls -ld "$path"; path="$(dirname "$path")"; done; ls -ldh /; };_dirls2'

---

eg:

# dirls /usr/local/bin
dr-xr-xr-x. 18 root root 236 Nov 11 19:01 /
drwxr-xr-x. 15 root root 181 Dec  7 11:50 /usr/
drwxr-xr-x. 15 root root 205 Nov 22 14:06 /usr/local/
drwxr-xr-x. 2 root root 4096 Dec  5 12:19 /usr/local/bin/

# dirls2 /usr/local/bin
drwxr-xr-x. 2 root root 4096 Dec  5 12:19 /usr/local/bin
drwxr-xr-x. 15 root root 205 Nov 22 14:06 /usr/local
drwxr-xr-x. 15 root root 181 Dec  7 11:50 /usr
dr-xr-xr-x. 18 root root 236 Nov 11 19:01 /

Answer 6

The command you are looking for listing permissions and owners/groups is ls -l.

-l option is used for long listing format.

ls -l /path/to/list

In addition, if you want to list and the hidden files then add the -a(all) option.

ls -al /path/to/list

Also, if you want to list permissions in your subdirectories use -R (recursive) option.

ls -Rl /path/to/list

The first column displays the permissions(read(r), write(w), execute(x)) and some special permissions(directories(d), -(regular file)) and the 3rd and 4th column shows you the file/directory owner and group respectively.