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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

I'm using OS X 10.8.3.

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Much easier with stat. Do you have it? (It's a GNU tool, so mostly available on Linux, not on Unix.) –  manatwork Apr 7 '13 at 15:55
@manatwork stat foo gives 16777219 377266 drwxr-xr-x 119 Tyilo staff 0 4046 "Apr 7 17:49:03 2013" "Apr 7 18:08:31 2013" "Apr 7 18:08:31 2013" "Nov 25 17:13:52 2012" 4096 0 0 /Users/Tyilo. I don't see 755 in it. –  Tyilo Apr 7 '13 at 16:08

8 Answers 8

up vote 10 down vote accepted

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.

mode=$(stat -f %p some-file)
perm=$(printf %o "$((mode & 07777))"

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
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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

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.

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My stat doesn't have a -c option. I'm using OS X 10.8.3. –  Tyilo Apr 7 '13 at 17:15
Thanks for the information, @Tyilo. And sorry, I can not help with OS X's tools. –  manatwork Apr 8 '13 at 6:25
Try reading manpage^W^W^W stat(1) on Mac OS X have -f flag for specifying output format, e.g. stat -f 'chmod %p "%N"' –  gelraen Apr 8 '13 at 13:56

I once came up with:

chmod_format() {
  sed 's/.\(.........\).*/\1/

(more as an exercise in sed style than a legible script...).

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

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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.

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The OP mentioned he was on Apple OS/X, so chmod will not be the GNU chmod there. –  Stéphane Chazelas Apr 7 '13 at 21:59
Ah yeah I'm seeing the comment on the other answer where they say their platform. Mine isn't the only one mentioning GNU though and you can get GNU Utilities on Mac OS X –  Bratchley Apr 7 '13 at 22:27

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

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


$ echo drwxr-xr-x | chmod-format
$ echo -rwsr-sr-x | chmod-format
$ echo -rwSr-Sr-- | chmod-format
$ echo ---------- | chmod-format

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 (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 3rd, 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 haven't tested it, but it looks like it should work.  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.

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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.

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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'
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