33

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?

| improve this question | |
  • Not seeing anything enlightening from apropos mode or apropos permissions on OS X and Debian. Do you know what platform/distribution you were using? Could it have been a site-local command? You could script such a tool using dirname and stat. – mrb Jul 9 '13 at 17:22
  • Using Arch Linux, and a pretty vanilla installation so not too many extra tools installed. Note that it was owners AND/OR permissions, I'm not sure at this point. Either would be useful. – please delete me Jul 10 '13 at 6:43
35

The command could have been:

namei -m /home/user/dir/child/file
| improve this answer | |
  • 6
    namei -mo is great, that it also gives you the owner. – earthmeLon Jun 16 '15 at 5:17
  • 2
    BEWARE: namei won't show linux ACL's or SELinux MAC's. I figured out that an ACL was blocking nginx by manually checking with sudo su nginx -s/bin/bash – Ray Foss Sep 29 '16 at 0:36
34

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

| improve this answer | |
  • 3
    I think it was asked to display the ancestors of /home/user/dir/child/file not the children. – Raphael Ahrens Jul 10 '13 at 5:12
  • @RaphaelAhrens - yes I realize that but there isn't a command that I'm aware of that does exactly that, there is however the tree command that does what I showed and it's suspiciously similar to what the OP was thinking of. – slm♦ Jul 10 '13 at 6:18
  • Is the -u really necessary? It seems it's default here (Tree ver. 1.6.0 under GNU bash 4.2.45). Is there an option to silence the "username/userid"? – Nikos Alexandris Dec 13 '13 at 23:42
  • 1
    @NikosAlexandris - if I drop that switch I do not get the username. My ver: tree v1.6.0. I'm on Fedora 19, GNU bash, version 4.2.45(1)-release (x86_64-redhat-linux-gnu) . – slm♦ Dec 14 '13 at 0:17
  • Different "defaults" between the bash versions? Thanks for informing. – Nikos Alexandris Dec 16 '13 at 10:56
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'
| improve this answer | |
  • 1
    You can reverse with | tac if you have it, or | sed '1!G;h;$!d' if not. – mrb Jul 9 '13 at 17:44
  • 2
    This will break for relative paths. Do [ "$l_path" != / -a "$l_path" != . ] instead. – user26112 Jul 9 '13 at 17:52
  • @Evan Teitelman dirname should gives the full path – Raphael Ahrens Jul 9 '13 at 17:57
  • 1
    @RaphaelAhrens no, it'll return a dot for relative path in the end. – rush Jul 9 '13 at 17:58
  • 1
    Also, stat -l should probably be stat or ls -ld. – user26112 Jul 9 '13 at 17:59
0

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}
| improve this answer | |
-1

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.

| improve this answer | |
  • 2
    I don't think that's what is asked for. Looks like the asker is looking for a list of permissions of directories that lead to a file. – Mat Jul 9 '13 at 17:13

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