Linux – Octal representation of file/folder permissions using awk

awkbashfile-permissionslinux

We have various of machines here with all sorts of hardware and Operating Systems, most of them do regular tasks with bash that an agent executes.
However it came to my attention that some Solaris machines we own do not have stat util, and adding it would be a problem that might take time (ironic)

Meanwhile I was trying to imitate one sisyphean task which stat did in a magnificent way:
return the file permissions in Octal

I found some old example which prints the regular permissions in Octal – rwx, but not for the "special" permissions – sticky group, sticky user, etc

My basic method was this : first calculate the special bits Octal value and then add the regular 'rwx', but it doesn't seem to work well

ls -lah $file  | awk '{k=0;for(i=0;i<=8;i++) {if (substr($1,i+2,1)~/[s]/) k += ((substr($1,i+2,1)~/[s]/)*2^9);else k+=((substr($1,i+2,1)~/[rwx]/)*2^(8-i));}if (k)printf("%0o ",k);}'


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

can anyone hint me what would be a good solution ?

Best Answer

This should handle any set of permissions including sticky, setuid and setgid ("s" stands for "set" rather than "sticky" which is "t"). Leading zeros are printed.

ls -lahd "$file" | awk '{k = 0; for (g=2; g>=0; g--) for (p=2; p>=0; p--) {c = substr($1, 10 - (g * 3 + p), 1); if (c ~ /[sS]/) k += g * 02000; else if (c ~ /[tT]/) k += 01000; if (c ~ /[rwxts]/) k += 8^g * 2^p} if (k) printf("%05o ", k)}'

Here it is on multiple lines for readability:

awk '{k = 0; 
      for (g=2; g>=0; g--) 
          for (p=2; p>=0; p--) {
              c = substr($1, 10 - (g * 3 + p), 1); 
              if (c ~ /[sS]/) 
                  k += g * 02000; 
              else 
                  if (c ~ /[tT]/) 
                      k += 01000; 
              if (c ~ /[rwxts]/) 
                  k += 8^g * 2^p
      } 
      if (k) {
          printf("%05o ", k); 
     }'

Demo:

$ touch foo
$ chmod 7654 foo
$ ls -l foo
-rwSr-sr-T 1 user user 0 2012-01-29 13:21 foo
$ ls -l foo | awk '...'
07654

Related Solutions

Bash – using awk in a bash script with if

Awk is the wrong solution here. How about:

if [[ "$task" =~ ^Clean ]]; then
  echo hello
fi

There's also a sneaky substitution method:

if [ "${task/#Clean/}" != "$task" ]; then
  echo hello
fi

But I find that to be far less readable.

Chris reckons case is tidier... let's try that out:

case $task in
  Clean*)
    echo hello
    ;;
esac

My aesthetic sense says "hell no with bells on", and fiddling with the formatting (Clean*) echo hello;; or similar) won't help much, IMAO. If you want a conditional, use an if, I say.

Parsing a diff file using grep/awk

I am thinking that with AWK, I could tell it start at (the above regex) and stop at the first line starting with a digit or the first line ending with a --

Please give us an example if it is not what you want:

sed -n '/[0-9][0-9]d[0-9][0-9]/,/^[0-9]\|--$/p'

EDIT

Although you've accepted my answer but I still want to edit my post to share with you a regex that can help you solve your problem thoroughly. sed allows you excluding the matching lines with b - branch command:

sed -n '/[0-9][0-9]d[0-9][0-9]/,/^[0-9]\|--$/ { /^[0-9]/b; p }'

but with this regex, sed also remove the REGEX1. So, Lookahead appears in my mind:

sed -n '/[0-9][0-9]d[0-9][0-9]/,/^[0-9]\|--$/ { /^[0-9](?:(?![0-9]d[0-9][0-9]).*)$/b; p }'

but it not works because the sed, awk, grep uses the POSIX RE flavor which doesn't support negative lookahead. You should try with Python, Perl, Ruby, ...

Best Answer

Related Solutions

Bash – using awk in a bash script with if

Parsing a diff file using grep/awk

Related Topic