mkdir recursion with permissions - bash

Question

It's a simple question, I'm writing a bash script called from cron grouping files in tar file and classifing into a dir structure.
These dir needs a special owner and permissions, and I call mkdir command thru su:
#!/bin/bash
... # shortened code
$PERMS=750
$DIR=/home/luser/0/01/012/0123
$OWNER=luser
... # shortened code
su -c "mkdir -m $PERMS -p $DIR" $OWNER
Output for ll -R /home/luser/0
/home/luser/0:
total 4
drwxr-xr-x 3 luser luser 4096 Jan 7 18:13 01
/home/luser/0/01:
total 4
drwxr-xr-x 3 luser luser 4096 Jan 7 18:13 012
/home/luser/0/01/012:
total 4
drwxr-x--- 2 luser luser 4096 Jan 7 18:13 0123
/home/luser/0/01/012/0123:
total 0
Only the deepest dir has permissions (750) setting rightly.
I don't know how deep it's the last directory and set permissions for all home's file it's too hard (too much files).
PS: I'm googled about that, but I find nothing.

Answer 1

You can restrict the permissions on the parent directories via umask. Here is an example:
PERMS=750
UMASK=$(echo "$PERMS" | tr "01234567" "76543210")
DIR=/home/luser/0/01/012/0123
OWNER=luser
su -c "umask $UMASK; mkdir -m $PERMS -p $DIR" $OWNER
In action:
> PERMS=750
> UMASK=$(echo "$PERMS" | tr "01234567" "76543210")
> (umask $UMASK; mkdir -m $PERMS -p 1/2/3/4)
> ll -R .
.:
drwxr-x--- 3 luser luser 4096 Jan 7 1:38 1/
./1:
drwxr-x--- 3 luser luser 4096 Jan 7 1:38 2/
./1/2:
drwxr-x--- 3 luser luser 4096 Jan 7 1:38 3/
./1/2/3:
drwxr-x--- 2 luser luser 4096 Jan 7 1:38 4/