Convert ls -l output format to chmod format

Question

Say I have the following output from ls -l:

drwxr-xr-x 2 root root 4096 Apr  7 17:21 foo

How can I automatically convert this to the format used by chmod?

For example:

$ echo drwxr-xr-x | chmod-format
755

I'm using OS X 10.8.3.

Answer 1

Some systems have commands to display the permissions of a file as a number, but unfortunately, nothing portable.

zsh has a stat (aka zstat) builtin in the stat module:

zmodload zsh/stat
stat -H s some-file

Then, the mode is in $s[mode] but is the mode, that is type + perms.

If you want the permissions expressed in octal, you need:

perms=$(([##8] s[mode] & 8#7777))

BSDs (including Apple OS/X) have a stat command as well.

stat -f %Lp some-file

(without the L, the full mode is returned, in octal)

GNU find (from as far back as 1990 and probably before) can print the permissions as octal:

find some-file -prune -printf '%m\n'

Later (2001, long after zsh stat (1997) but before BSD stat (2002)) a GNU stat command was introduced with again a different syntax:

stat -c %a some-file

Long before those, IRIX already had a stat command (already there in IRIX 5.3 in 1994) with another syntax:

stat -qp some-file

Again, when there's no standard command, the best bet for portability is to use perl:

perl -e 'printf "%o\n", (stat shift)[2]&07777' some-file

Answer 2

You can ask GNU stat to output the permissions in octal format by using the -c option. From man stat:

       -c  --format=FORMAT
              use the specified FORMAT instead of the default; output a
              newline after each use of FORMAT
⋮
       %a     access rights in octal
⋮
       %n     file name

So in your case:

bash-4.2$ ls -l foo
-rw-r--r-- 1 manatwork manatwork 0 Apr  7 19:43 foo

bash-4.2$ stat -c '%a' foo
644

Or you can even automate it by formatting stat's output as valid command:

bash-4.2$ stat -c "chmod %a '%n'" foo
chmod 644 'foo'

bash-4.2$ stat -c "chmod %a '%n'" foo > setpermission.sh

bash-4.2$ chmod a= foo

bash-4.2$ ls -l foo
---------- 1 manatwork manatwork 0 Apr  7 19:43 foo

bash-4.2$ sh setpermission.sh 

bash-4.2$ ls -l foo
-rw-r--r-- 1 manatwork manatwork 0 Apr  7 19:43 foo

The above solution will also work for multiple files if using a wildcard:

stat -c "chmod -- %a '%n'" -- *

Will work correctly with file names containing whitespace characters, but will fail on file names containing single quotes.

Answer 3

To convert from the symbolic to octal notation, I once came up with:

chmod_format() {
  sed 's/.\(.........\).*/\1/
    h;y/rwsxtSTlL-/IIIIIOOOOO/;x;s/..\(.\)..\(.\)..\(.\)/|\1\2\3/
    y/sStTlLx-/IIIIIIOO/;G
    s/\n\(.*\)/\1;OOO0OOI1OIO2OII3IOO4IOI5IIO6III7/;:k
    s/|\(...\)\(.*;.*\1\(.\)\)/\3|\2/;tk
    s/^0*\(..*\)|.*/\1/;q'
}

Expanded:

#! /bin/sed -f
s/.\(.........\).*/\1/; # extract permissions and discard the rest

h; # store a copy on the hold space

# Now for the 3 lowest octal digits (rwx), translates the flags to
# binary where O means 0 and I means 1.
# l, L are for mandatory locking (a regular file that has 02000 on
# and not 010 on some systems like Linux). Some ls implementations
# like GNU ls confusingly use S there like for directories even though 
# it has nothing to do with setgid in that case. Some ls implementations 
# use L, some others l (against POSIX which requires an uppercase
# flag for extra flags when the execution bit is not set).
y/rwsxtSTlL-/IIIIIOOOOO/

x; # swap hold and pattern space, to do a second processing on those flags.

# now only consider the "xXlLsStT" bits:
s/..\(.\)..\(.\)..\(.\)/|\1\2\3/

y/sStTlLx-/IIIIIIOO/; # make up the 4th octal digit as binary like before

G; # append the hold space so we now have all 4 octal digits as binary

# remove the extra newline and append a translation table
s/\n\(.*\)/\1;OOO0OOI1OIO2OII3IOO4IOI5IIO6III7/

:k
  # translate the OOO -> 0 ... III -> 7 in a loop
  s/|\(...\)\(.*;.*\1\(.\)\)/\3|\2/
tk

# trim leading 0s and our translation table.
s/^0*\(..*\)|.*/\1/;q

That returns the octal number from the output of ls -l on one file.

$ echo 'drwSr-sr-T' | chmod_format
7654

Answer 4

This command on Mac under sh

stat -f "%Lp %N" your_files

if you only want the numeric permission, use %Lp only.

for example:

stat -f "%Lp %N" ~/Desktop
700 Desktop

The 700 is the numeric permission which can be used in chmod, and Desktop is the filename.

Answer 5

Here's an answer to question Y (ignoring question X), inspired by the OP's attempt:

#!/bin/bash
LC_COLLATE=C
while read ls_out
do
        extra=0
        perms=0
        for i in {1..9}
        do
                # Shift $perms to the left one bit, so we can always just add the LSB.
                let $((perms*=2))
                this_char=${ls_out:i:1}
                # If it's different from its upper case equivalent,
                # it's a lower case letter, so the bit is set.
                # Unless it's "l" (lower case L), which is special.
                if [ "$this_char" != "${this_char^}" ]  &&  [ "$this_char" != "l" ]
                then
                        let $((perms++))
                fi
                # If it's not "r", "w", "x", or "-", it indicates that
                # one of the high-order (S/s=4000, S/s/L/l=2000, or T/t=1000) bits
                # is set.
                case "$this_char" in
                  ([^rwx-])
                        let $((extra += 2 ** (3-i/3) ))
                esac
        done
        printf "%o%.3o\n" "$extra" "$perms"
done

The above contains a few bashisms.  The following version seems to be POSIX-compliant:

#!/bin/sh
LC_COLLATE=C
while read ls_out
do
        extra=0
        perms=0
        for i in $(seq 1 9)
        do
                # Shift $perms to the left one bit, so we can always just add the LSB.
                : $((perms*=2))
                this_char=$(expr "$ls_out" : ".\{$i\}\(.\)")
                # Lower case letters other than "l" indicate that permission bits are set.
                # If it's not "r", "w", "x", or "-", it indicates that
                case "$this_char" in
                  (l)
                        ;;
                  ([a-z])
                        : $((perms+=1))
                esac
                # If it's not "r", "w", "x", or "-", it indicates that
                # one of the high-order (S/s=4000, S/s/L/l=2000, or T/t=1000) bits
                # is set.
                case "$this_char" in
                  ([!rwx-])
                        : $((extra += 1 << (3-i/3) ))
                esac
        done
        printf "%o%.3o\n" "$extra" "$perms"
done

Notes:

• The LC_COLLATE=C tells the shell to treat letter sequence range patterns as using the ASCII order, so [a-e] is equivalent to [abcde].  In some locales (e.g., en_US), [a-e] is equivalent to [aAbBcCdDeE] (i.e., [abcdeABCDE]) or perhaps [abcdeABCD] — see Why is the bash case statement not case-sensitive …?)

• In the second version (the POSIX-compliant one):

  • The first case statement could be rewritten:

            case "$this_char" in
              ([a-km-z])
                    : $((perms+=1))
            esac
    

    but I think the way I have it now makes it easier to see that l is the letter that's being handled differently.  Alternatively, it could be rewritten:

            case "$this_char" in
              ([rwxst])
                    : $((perms+=1))
            esac
    

    since r, w, x, s, and t are the only letters that should ever appear in a mode string (other than l).

  • The second case statement could be rewritten:

            case "$this_char" in
              ([rwx])
                    ;;
              ([A-Za-z])
                    : $((extra += 1 << (3-i/3) ))
             esac
    

    to enforce the rule that only letters are valid for specifying mode bits.  (By contrast, the more succinct version in the full script is lazy, and will accept -rw@rw#rw% as equivalent to rwSrwSrwT.)  Alternatively, it could be rewritten:

            case "$this_char" in
              ([SsTtLl])
                    : $((extra += 1 << (3-i/3) ))
            esac
    

    since S, s, T, t, L, and l are the only letters that should ever appear in a mode string (other than r, w, and x).

Usage:

$ echo drwxr-xr-x | chmod-format
0755
$ echo -rwsr-sr-x | chmod-format
6755
$ echo -rwSr-Sr-- | chmod-format
6644
$ echo -rw-r-lr-- | chmod-format
2644
$ echo ---------- | chmod-format
0000

And, yes, I know it's better not to use echo with text that might begin with -; I just wanted to copy the usage example from the question.  Note, obviously, that this ignores the 0th character (i.e., the leading d/b/c/-/l/p/s/D) and the 10th (+/./@).  It assumes that the maintainers of ls will never define r/R or w/W as valid characters in the third, sixth, or ninth position (and, if they do, they should be beaten with sticks).

---

Also, I just found the following code, by cas, under How to restore default group/user ownership of all files under /var:

        let perms=0

        [[ "${string}" = ?r???????? ]]  &&  perms=$(( perms +  400 ))
        [[ "${string}" = ??w??????? ]]  &&  perms=$(( perms +  200 ))
        [[ "${string}" = ???x?????? ]]  &&  perms=$(( perms +  100 ))
        [[ "${string}" = ???s?????? ]]  &&  perms=$(( perms + 4100 ))
        [[ "${string}" = ???S?????? ]]  &&  perms=$(( perms + 4000 ))
        [[ "${string}" = ????r????? ]]  &&  perms=$(( perms +   40 ))
        [[ "${string}" = ?????w???? ]]  &&  perms=$(( perms +   20 ))
        [[ "${string}" = ??????x??? ]]  &&  perms=$(( perms +   10 ))
        [[ "${string}" = ??????s??? ]]  &&  perms=$(( perms + 2010 ))
        [[ "${string}" = ??????S??? ]]  &&  perms=$(( perms + 2000 ))
        [[ "${string}" = ???????r?? ]]  &&  perms=$(( perms +    4 ))
        [[ "${string}" = ????????w? ]]  &&  perms=$(( perms +    2 ))
        [[ "${string}" = ?????????x ]]  &&  perms=$(( perms +    1 ))
        [[ "${string}" = ?????????t ]]  &&  perms=$(( perms + 1001 ))
        [[ "${string}" = ?????????T ]]  &&  perms=$(( perms + 1000 ))

I have tested this code (but not thoroughly), and it seems to work, except for the fact that it doesn't recognize l or L in the sixth position.  Note, though, that while this answer is superior in terms of simplicity and clarity, mine is actually shorter (counting only the code inside the loop; the code that handles a single -rwxrwxrwx string, not counting comments), and it could be made even shorter by replacing if condition; then … with condition && ….

---

Of course, you should not parse the output of ls.

Answer 6

An alternative, if you want to save the permissions away, to restore them later on, or on a different file is to use setfacl/getfacl, and it will also restore (POSIX-draft) ACLs as a bonus.

getfacl some-file > saved-perms
setfacl -M saved-perms some-other-file

(on Solaris, use -f instead of -M).

However, though they are available on some BSDs, they are not on Apple OS/X where the ACLs are manipulated with chmod only.

Answer 7

If your goal is to take permissions from one file and give them to another as well, GNU chmod already has a "reference" option for that.

Answer 8

On Mac OS X (10.6.8) you have to use stat -f format (because it is actually NetBSD / FreeBSD stat).

# using Bash

mods="$(stat -f "%p" ~)"    # octal notation
mods="${mods: -4}"
echo "$mods"

mods="$(stat -f "%Sp" ~)"  # symbolic notation
mods="${mods: -9}"
echo "$mods"

To just translate a symbolic permission string produced by ls -l into octal (using only shell builtins) see: showperm.bash.

# from: showperm.bash
# usage: showperm modestring
#
# example: showperm '-rwsr-x--x'