Can the Unix list command 'ls' output numerical chmod permissions?

Question

it almost can ..

 ls -l | awk '{k=0;for(i=0;i<=8;i++)k+=((substr($1,i+2,1)~/[rwx]/) \
             *2^(8-i));if(k)printf("%0o ",k);print}'

Closest I can think of (keeping it simple enough) is stat, assuming you know which files you're looking for. If you don't, * can find most of them:

/usr/bin$ stat -c '%a %n' *
755 [
755 a2p
755 a2ps
755 aclocal
...

It handles sticky, suid and company out of the box:

$ stat -c '%a %n' /tmp /usr/bin/sudo
1777 /tmp
4755 /usr/bin/sudo

you can just use GNU find.

find . -printf "%m:%f\n"

You can use the following command

stat -c "%a %n" *

Also you can use any filename or directoryname instead of * to get a specific result.

On Mac, you can use

stat -f '%A %N' *

@The MYYN

wow, nice awk! But what about suid, sgid and sticky bit?

You have to extend your filter with s and t, otherwise they will not count and you get the wrong result. To calculate the octal number for this special flags, the procedure is the same but the index is at 4 7 and 10. the possible flags for files with execute bit set are ---s--s--t amd for files with no execute bit set are ---S--S--T

ls -l | awk '{
    k = 0
    s = 0
    for( i = 0; i <= 8; i++ )
    {
        k += ( ( substr( $1, i+2, 1 ) ~ /[rwxst]/ ) * 2 ^( 8 - i ) )
    }
    j = 4 
    for( i = 4; i <= 10; i += 3 )
    {
        s += ( ( substr( $1, i, 1 ) ~ /[stST]/ ) * j )
        j/=2
    }
    if ( k )
    {
        printf( "%0o%0o ", s, k )
    }
    print
}'  

For test:

touch blah
chmod 7444 blah

will result in:

7444 -r-Sr-Sr-T 1 cheko cheko   0 2009-12-05 01:03 blah

and

touch blah
chmod 7555 blah

will give:

7555 -r-sr-sr-t 1 cheko cheko   0 2009-12-05 01:03 blah

Use this to display the Unix numerical permission values (octal values) and file name.

stat -c '%a %n' *

Use this to display the Unix numerical permission values (octal values) and the folder's sgid and sticky bit, user name of the owner, group name, total size in bytes and file name.

stat -c '%a %A %U %G %s %n' *

Add %y if you need time of last modification in human-readable format. For more options see stat.

Better version using an Alias

Using an alias is a more efficient way to accomplish what you need and it also includes color. The following displays your results organized by group directories first, display in color, print sizes in human readable format (e.g., 1K 234M 2G) edit your ~/.bashrc and add an alias for your account or globally by editing /etc/profile.d/custom.sh

Typing cls displays your new LS command results.

alias cls="ls -lha --color=always -F --group-directories-first |awk '{k=0;s=0;for(i=0;i<=8;i++){;k+=((substr(\$1,i+2,1)~/[rwxst]/)*2^(8-i));};j=4;for(i=4;i<=10;i+=3){;s+=((substr(\$1,i,1)~/[stST]/)*j);j/=2;};if(k){;printf(\"%0o%0o \",s,k);};print;}'"

Folder Tree

While you are editing your bashrc or custom.sh include the following alias to see a graphical representation where typing lstree will display your current folder tree structure

alias lstree="ls -R | grep ":$" | sed -e 's/:$//' -e 's/[^-][^\/]*\//--/g' -e 's/^/   /' -e 's/-/|/'"

It would display:

   |-scripts
   |--mod_cache_disk
   |--mod_cache_d
   |---logs
   |-run_win
   |-scripts.tar.gz

You don't use ls to get a file's permission information. You use the stat command. It will give you the numerical values you want. The "Unix Way" says that you should invent your own script using ls (or 'echo *') and stat and whatever else you like to give the information in the format you desire.