Numeric file permissions?

[Kris Spander]

Hello,

Is there an easy way to display the permissions of a file in numeric format like "775"?

Thanks,
Kris

[A. Clay Stephenson · Accepted solution]

Hi Kris:

Probably about the easiest method is to use Perl's stat() function.

Try this:

fmode.pl myfile1 myfile2
to output
myfile1 775
myfile2 600

I already had this one written.

Regards, Clay

[Sridhar Bhaskarla]

Hi Kris,

I tried to write a script in shell just for sake of it and ofcourse for my fun. I made it bit eloborate so that I wouldn't get confused myself while writing it. This lists the permissions in four digit format. For ex., 4555 or 0555 so that you would know if the file has suid/sgid/sticky bits.

Call it get_perms.ksh and run it with

---start---

#!/usr/bin/ksh
if [ $# -ne 1 ]
then
echo "Usage: $0 file"
exit
fi

PERM=$(ll $1|sed 's/-/0/g'| awk '{print $1}')

A=0

B=$(echo $PERM|cut -c 4)

if [[ "$B" = "s" || "$B" = "S" ]]
then
(( A = $A + 4 ))
fi

C=$(echo $PERM|cut -c 7)

if [[ "$C" = "s" || "$C" = "S" ]]
then
(( A = $A + 2 ))
fi
D=$(echo $PERM|cut -c 10)

if [[ "$C" = "t" || "$C" = "T" ]]
then
(( A = $A + 1 ))
fi

PERM=$(echo $PERM|sed -e 's/r/4/g' -e 's/w/2/g' -e 's/x/1/g' -e 's/s/1/g' -e 's/
t/1/g' -e 's/S/0/g' -e 's/T/0/g')



printf "$A"

printit()
{
c=0
K=$1
for i in $K
do
VAL=$(echo $PERM|cut -c $i)
((c = $c + $VAL ))
done
printf "$c"
}

printit "2 3 4"
printit "5 6 7"
printit "8 9 10 "

printf " $1\n"

---End--

-Sri
PS: Becareful while cutting pasting the sed statement. It should be in one line.

[James R. Ferguson]

Hi Kris:

No argument that perl is shorter/faster here, but here's another shell solution I offered sometime ago when this question was asked:

The script returns the octal numeric mode of the file passed to it. The script handles the setuid, setgid and sticky bits in addition to the standard permissions:

#!/usr/bin/sh
#
typeset -i PERMFCT=0 #...permission mode factor
typeset -i MISCFCT=0 #...miscellaneous mode factor
typeset -i BITSNBR=2 #...bit number (1-relative)
typeset -i BITSVAL=0 #...simple bit value
typeset -i OCTMODE=0 #...final, octal numeric mode
#
[ "$#" != 1 ] && { echo "Usage: $0 file"; exit 1 ; }
#
SMODE=`ls -l $1|awk '{print $1}`
#
while (( BITSNBR <= 10 ))
do
case `expr \( $BITSNBR - 2 \) / 3` in #...establish weighted values...
0 )
PERMFCT=100
MISCFCT=4000
;;
1 )
PERMFCT=10
MISCFCT=2000
;;
2 )
PERMFCT=1
MISCFCT=1000
;;
esac
#
case `expr substr $SMODE $BITSNBR 1` in
r )
BITSVAL=4
MISCFCT=0
;;
w )
BITSVAL=2
MISCFCT=0
;;
x )
BITSVAL=1
MISCFCT=0
;;
s|t )
BITSVAL=1
;;
S|T )
BITSVAL=0
;;
* )
BITSVAL=0
MISCFCT=0
;;
esac
#
OCTMODE=`expr $PERMFCT \* $BITSVAL + $MISCFCT + $OCTMODE`
#
let BITSNBR=$BITSNBR+1
done
#
echo "mode = $OCTMODE"
exit 0

Regards!

...JRF...

[Kris Spander]

Hello,

I decided to use A. Clay's method. It was was much easier.

Thanks to everybody,
Kris