Setting default permissions for newly created files and sub-directories under a directory in Linux?

LinuxPermissionsAclUmask

Linux Problem Overview


I have a bunch of long-running scripts and applications that are storing output results in a directory shared amongst a few users. I would like a way to make sure that every file and directory created under this shared directory automatically had u=rwxg=rwxo=r permissions.

I know that I could use umask 006 at the head off my various scripts, but I don't like that approach as many users write their own scripts and may forget to set the umask themselves.

I really just want the filesystem to set newly created files and directories with a certain permission if it is in a certain folder. Is this at all possible?

Update: I think it can be done with POSIX ACLs, using the Default ACL functionality, but it's all a bit over my head at the moment. If anybody can explain how to use Default ACLs it would probably answer this question nicely.

Linux Solutions


Solution 1 - Linux

To get the right ownership, you can set the group setuid bit on the directory with

chmod g+rwxs dirname

This will ensure that files created in the directory are owned by the group. You should then make sure everyone runs with umask 002 or 007 or something of that nature---this is why Debian and many other linux systems are configured with per-user groups by default.

I don't know of a way to force the permissions you want if the user's umask is too strong.

Solution 2 - Linux

Here's how to do it using default ACLs, at least under Linux.

First, you might need to enable ACL support on your filesystem. If you are using ext4 then it is already enabled. Other filesystems (e.g., ext3) need to be mounted with the acl option. In that case, add the option to your /etc/fstab. For example, if the directory is located on your root filesystem:

/dev/mapper/qz-root   /    ext3    errors=remount-ro,acl   0  1

Then remount it:

mount -oremount /

Now, use the following command to set the default ACL:

setfacl -dm u::rwx,g::rwx,o::r /shared/directory

All new files in /shared/directory should now get the desired permissions. Of course, it also depends on the application creating the file. For example, most files won't be executable by anyone from the start (depending on the mode argument to the open(2) or creat(2) call), just like when using umask. Some utilities like cp, tar, and rsync will try to preserve the permissions of the source file(s) which will mask out your default ACL if the source file was not group-writable.

Hope this helps!

Solution 3 - Linux

It's ugly, but you can use the setfacl command to achieve exactly what you want.

On a Solaris machine, I have a file that contains the acls for users and groups. Unfortunately, you have to list all of the users (at least I couldn't find a way to make this work otherwise):

user::rwx
user:user_a:rwx
user:user_b:rwx
...
group::rwx
mask:rwx
other:r-x
default:user:user_a:rwx
default:user:user_b:rwx
....
default:group::rwx
default:user::rwx
default:mask:rwx
default:other:r-x

Name the file acl.lst and fill in your real user names instead of user_X.

You can now set those acls on your directory by issuing the following command:

setfacl -f acl.lst /your/dir/here

Solution 4 - Linux

in your shell script (or .bashrc) you may use somthing like:

umask 022

umask is a command that determines the settings of a mask that controls how file permissions are set for newly created files.

Solution 5 - Linux

I don't think this will do entirely what you want, but I just wanted to throw it out there since I hadn't seen it in the other answers.

I know you can create directories with permissions in a one-liner using the -m option:

mkdir -m755 mydir

and you can also use the install command:

sudo install -C -m 755 -o owner -g group /src_dir/src_file /dst_file

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionDavid DeanView Question on Stackoverflow
Solution 1 - LinuxNorman RamseyView Answer on Stackoverflow
Solution 2 - LinuxpelleView Answer on Stackoverflow
Solution 3 - LinuxinnaMView Answer on Stackoverflow
Solution 4 - Linuxuser3270784View Answer on Stackoverflow
Solution 5 - LinuxA. HendryView Answer on Stackoverflow